Chapter 14 Statistics

Given:

The marks obtained by 30 students and their corresponding frequencies are given in the table.

Marks obtained are denoted by $x_i$ and the number of students by $f_i$.

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To Find:

The mean of the marks obtained by the students.

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Solution:

We will use the direct method to find the mean of the given data.

The formula for the mean ($\bar{x}$) using the direct method is:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

First, let's prepare a table to calculate the sum of the products of marks and frequencies ($f_i \times x_i$).

Marks obtained ($x_i$)	Number of students ($f_i$)	$f_i \times x_i$
10	1	$10 \times 1 = 10$
20	1	$20 \times 1 = 20$
36	3	$36 \times 3 = 108$
40	4	$40 \times 4 = 160$
50	3	$50 \times 3 = 150$
56	2	$56 \times 2 = 112$
60	4	$60 \times 4 = 240$
70	4	$70 \times 4 = 280$
72	1	$72 \times 1 = 72$
80	1	$80 \times 1 = 80$
88	2	$88 \times 2 = 176$
92	3	$92 \times 3 = 276$
95	1	$95 \times 1 = 95$
Total	$\sum f_i$	$\sum (f_i \times x_i)$
	30	$10+20+108+160+150+112+240+280+72+80+176+276+95 = 1779$

Sum of frequencies, $\sum f_i = 30$

Sum of the products of frequencies and marks, $\sum (f_i \times x_i) = 1779$

Now, substitute these values into the mean formula:

$\bar{x} = \frac{1779}{30}$

$\bar{x} = 59.3$

The mean marks obtained by the students is $59.3$.

Given:

The percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India is given in the form of a grouped frequency distribution table.

The data is:

Percentage of female teachers	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65	65 - 75	75 - 85
Number of States/U.T. ($f_i$)	6	11	7	4	4	2	1

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To Find:

The mean percentage of female teachers using all three methods: Direct Method, Assumed Mean Method, and Step-deviation Method.

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Solution:

First, we calculate the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval.

Class mark ($x_i$) = $\frac{\text{Lower limit} + \text{Upper limit}}{2}$

Class marks: $(15+25)/2 = 20$, $(25+35)/2 = 30$, $(35+45)/2 = 40$, $(45+55)/2 = 50$, $(55+65)/2 = 60$, $(65+75)/2 = 70$, $(75+85)/2 = 80$.

The sum of frequencies is $\sum f_i = 6 + 11 + 7 + 4 + 4 + 2 + 1 = 35$.

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Method 1: Direct Method

The formula for the mean using the direct method is $\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$.

We prepare a table to calculate $f_i \times x_i$:

Percentage of female teachers	Number of States/U.T. ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
15 - 25	6	20	$6 \times 20 = 120$
25 - 35	11	30	$11 \times 30 = 330$
35 - 45	7	40	$7 \times 40 = 280$
45 - 55	4	50	$4 \times 50 = 200$
55 - 65	4	60	$4 \times 60 = 240$
65 - 75	2	70	$2 \times 70 = 140$
75 - 85	1	80	$1 \times 80 = 80$
Total	$\sum f_i = 35$		$\sum (f_i \times x_i) = 1390$

Mean ($\bar{x}$) = $\frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{1390}{35}$

$\bar{x} = \frac{278}{7} \approx 39.71$

By the direct method, the mean percentage of female teachers is approximately $39.71\%$.

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Method 2: Assumed Mean Method

The formula for the mean using the assumed mean method is $\bar{x} = a + \frac{\sum (f_i \times d_i)}{\sum f_i}$, where $d_i = x_i - a$ is the deviation from the assumed mean $a$.

Let's take the assumed mean $a = 50$ (a value from the class marks).

We prepare a table to calculate $d_i$ and $f_i \times d_i$:

Percentage of female teachers	Number of States/U.T. ($f_i$)	Class Mark ($x_i$)	Deviation $d_i = x_i - 50$	$f_i \times d_i$
15 - 25	6	20	$20 - 50 = -30$	$6 \times (-30) = -180$
25 - 35	11	30	$30 - 50 = -20$	$11 \times (-20) = -220$
35 - 45	7	40	$40 - 50 = -10$	$7 \times (-10) = -70$
45 - 55	4	50	$50 - 50 = 0$	$4 \times 0 = 0$
55 - 65	4	60	$60 - 50 = 10$	$4 \times 10 = 40$
65 - 75	2	70	$70 - 50 = 20$	$2 \times 20 = 40$
75 - 85	1	80	$80 - 50 = 30$	$1 \times 30 = 30$
Total	$\sum f_i = 35$			$\sum (f_i \times d_i) = -360$

Mean ($\bar{x}$) = $a + \frac{\sum (f_i \times d_i)}{\sum f_i} = 50 + \frac{-360}{35}$

$\bar{x} = 50 - \frac{360}{35} = 50 - \frac{72}{7}$

$\bar{x} = \frac{50 \times 7 - 72}{7} = \frac{350 - 72}{7} = \frac{278}{7}$

$\bar{x} \approx 39.71$

By the assumed mean method, the mean percentage of female teachers is approximately $39.71\%$.

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Method 3: Step-deviation Method

The formula for the mean using the step-deviation method is $\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$, where $u_i = \frac{x_i - a}{h}$, $a$ is the assumed mean, and $h$ is the class size.

Let's take the assumed mean $a = 50$ and the class size $h = 10$ (e.g., $25 - 15 = 10$).

We prepare a table to calculate $u_i$ and $f_i \times u_i$:

Percentage of female teachers	Number of States/U.T. ($f_i$)	Class Mark ($x_i$)	Deviation $d_i = x_i - 50$	Step Deviation $u_i = \frac{d_i}{10}$	$f_i \times u_i$
15 - 25	6	20	-30	$-30/10 = -3$	$6 \times (-3) = -18$
25 - 35	11	30	-20	$-20/10 = -2$	$11 \times (-2) = -22$
35 - 45	7	40	-10	$-10/10 = -1$	$7 \times (-1) = -7$
45 - 55	4	50	0	$0/10 = 0$	$4 \times 0 = 0$
55 - 65	4	60	10	$10/10 = 1$	$4 \times 1 = 4$
65 - 75	2	70	20	$20/10 = 2$	$2 \times 2 = 4$
75 - 85	1	80	30	$30/10 = 3$	$1 \times 3 = 3$
Total	$\sum f_i = 35$				$\sum (f_i \times u_i) = -36$

Mean ($\bar{x}$) = $a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h = 50 + \left(\frac{-36}{35}\right) \times 10$

$\bar{x} = 50 - \frac{360}{35} = 50 - \frac{72}{7}$

$\bar{x} = \frac{350 - 72}{7} = \frac{278}{7}$

$\bar{x} \approx 39.71$

By the step-deviation method, the mean percentage of female teachers is approximately $39.71\%$.

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All three methods give the same mean percentage of female teachers, which is approximately $39.71\%$.

Given:

The distribution of the number of wickets taken by bowlers and the corresponding number of bowlers (frequency).

The data is given in grouped form with unequal class intervals.

Number of wickets	20 - 60	60 - 100	100 - 150	150 - 250	250 - 350	350 - 450
Number of bowlers ($f_i$)	7	5	16	12	2	3

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To Find:

The mean number of wickets by choosing a suitable method and the significance of the mean.

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Solution:

The given distribution has unequal class intervals. In such cases, the Direct Method or the Assumed Mean Method are suitable. We will use the Assumed Mean Method as it often simplifies calculations.

The formula for the mean using the Assumed Mean Method is:

$\bar{x} = a + \frac{\sum (f_i \times d_i)}{\sum f_i}$

where $a$ is the assumed mean, $f_i$ is the frequency of the $i$-th class, and $d_i = x_i - a$ is the deviation of the class mark ($x_i$) from the assumed mean.

First, we calculate the class mark ($x_i$) for each class interval:

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 20 - 60: $x_1 = \frac{20+60}{2} = 40$
• For 60 - 100: $x_2 = \frac{60+100}{2} = 80$
• For 100 - 150: $x_3 = \frac{100+150}{2} = 125$
• For 150 - 250: $x_4 = \frac{150+250}{2} = 200$
• For 250 - 350: $x_5 = \frac{250+350}{2} = 300$
• For 350 - 450: $x_6 = \frac{350+450}{2} = 400$

Let's choose the assumed mean $a = 200$ (a value from the class marks).

Now we prepare a table to calculate the deviations $d_i = x_i - 200$ and the product $f_i \times d_i$:

Number of wickets	Number of bowlers ($f_i$)	Class Mark ($x_i$)	Deviation $d_i = x_i - 200$	$f_i \times d_i$
20 - 60	7	40	$40 - 200 = -160$	$7 \times (-160) = -1120$
60 - 100	5	80	$80 - 200 = -120$	$5 \times (-120) = -600$
100 - 150	16	125	$125 - 200 = -75$	$16 \times (-75) = -1200$
150 - 250	12	200	$200 - 200 = 0$	$12 \times 0 = 0$
250 - 350	2	300	$300 - 200 = 100$	$2 \times 100 = 200$
350 - 450	3	400	$400 - 200 = 200$	$3 \times 200 = 600$
Total	$\sum f_i = 35$			$\sum (f_i \times d_i) = -2120$

Sum of frequencies, $\sum f_i = 35$.

Sum of the products of frequencies and deviations, $\sum (f_i \times d_i) = -1120 + (-600) + (-1200) + 0 + 200 + 600 = -2920 + 800 = -2120$.

Now, substitute these values into the mean formula:

$\bar{x} = a + \frac{\sum (f_i \times d_i)}{\sum f_i} = 200 + \frac{-2120}{35}$

$\bar{x} = 200 - \frac{2120}{35}$

Simplify the fraction $\frac{2120}{35}$ by dividing the numerator and denominator by 5:

$\frac{2120}{35} = \frac{\cancel{2120}^{424}}{\cancel{35}_{7}} = \frac{424}{7}$

$\bar{x} = 200 - \frac{424}{7}$

$\bar{x} = \frac{200 \times 7 - 424}{7} = \frac{1400 - 424}{7} = \frac{976}{7}$

Calculating the decimal value:

$\frac{976}{7} \approx 139.43$ (rounded to two decimal places)

The mean number of wickets is approximately $139.43$.

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Significance of the Mean:

The mean number of wickets, $139.43$, signifies the average number of wickets taken by a bowler in one-day cricket matches based on the given data distribution. It provides a central value around which the number of wickets taken by different bowlers is distributed.

Given:

The number of plants in 20 houses in a locality is given in a grouped frequency distribution table.

The data is:

Number of plants	0 - 2	2 - 4	4 - 6	6 - 8	8 - 10	10 - 12	12 - 14
Number of houses ($f_i$)	1	2	1	5	6	2	3

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To Find:

The mean number of plants per house and the method used with justification.

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Solution:

To find the mean of the grouped data, we first need to find the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 0 - 2: $x_1 = \frac{0+2}{2} = 1$
• For 2 - 4: $x_2 = \frac{2+4}{2} = 3$
• For 4 - 6: $x_3 = \frac{4+6}{2} = 5$
• For 6 - 8: $x_4 = \frac{6+8}{2} = 7$
• For 8 - 10: $x_5 = \frac{8+10}{2} = 9$
• For 10 - 12: $x_6 = \frac{10+12}{2} = 11$
• For 12 - 14: $x_7 = \frac{12+14}{2} = 13$

The total number of houses (sum of frequencies) is $\sum f_i = 1 + 2 + 1 + 5 + 6 + 2 + 3 = 20$.

We will use the Direct Method to find the mean because the values of class marks ($x_i$) and frequencies ($f_i$) are small, making the calculation of $f_i \times x_i$ easy.

The formula for the mean ($\bar{x}$) using the direct method is:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

We prepare a table to calculate the product $f_i \times x_i$:

Number of plants	Number of houses ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
0 - 2	1	1	$1 \times 1 = 1$
2 - 4	2	3	$2 \times 3 = 6$
4 - 6	1	5	$1 \times 5 = 5$
6 - 8	5	7	$5 \times 7 = 35$
8 - 10	6	9	$6 \times 9 = 54$
10 - 12	2	11	$2 \times 11 = 22$
12 - 14	3	13	$3 \times 13 = 39$
Total	$\sum f_i = 20$		$\sum (f_i \times x_i) = 162$

Sum of frequencies, $\sum f_i = 20$

Sum of the products of frequencies and class marks, $\sum (f_i \times x_i) = 1 + 6 + 5 + 35 + 54 + 22 + 39 = 162$

Now, substitute these values into the mean formula:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{162}{20}$

$\bar{x} = 8.1$

The mean number of plants per house is $8.1$.

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Method Used and Justification:

We used the Direct Method for finding the mean.

Justification: The values of the class marks ($x_i$) and the frequencies ($f_i$) are relatively small. Therefore, calculating the product $f_i \times x_i$ for each class interval is straightforward and the sum $\sum (f_i \times x_i)$ is also manageable without using the Assumed Mean Method or Step-deviation Method, which are typically preferred when $x_i$ and $f_i$ are large.

Given:

The distribution of daily wages of 50 workers of a factory is given in the form of a grouped frequency distribution table.

The data is:

Daily wages (in $\textsf{₹}$)	500 - 520	520 - 540	540 - 560	560 - 580	580 - 600
Number of workers ($f_i$)	12	14	8	6	10

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To Find:

The mean daily wages of the workers of the factory by using an appropriate method.

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Solution:

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 500 - 520: $x_1 = \frac{500+520}{2} = 510$
• For 520 - 540: $x_2 = \frac{520+540}{2} = 530$
• For 540 - 560: $x_3 = \frac{540+560}{2} = 550$
• For 560 - 580: $x_4 = \frac{560+580}{2} = 570$
• For 580 - 600: $x_5 = \frac{580+600}{2} = 590$

The total number of workers (sum of frequencies) is $\sum f_i = 12 + 14 + 8 + 6 + 10 = 50$.

Since the class intervals are equal (class size $h = 520 - 500 = 20$) and the class marks and frequencies are relatively large, the Step-deviation Method is an appropriate method to use for simplifying calculations.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the class size.

Let's choose the assumed mean $a = 550$ (a central value among the class marks) and the class size $h = 20$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 550}{20}$ and the product $f_i \times u_i$:

Daily wages (in $\textsf{₹}$)	Number of workers ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 550}{20}$	$f_i \times u_i$
500 - 520	12	510	$\frac{510 - 550}{20} = \frac{-40}{20} = -2$	$12 \times (-2) = -24$
520 - 540	14	530	$\frac{530 - 550}{20} = \frac{-20}{20} = -1$	$14 \times (-1) = -14$
540 - 560	8	550	$\frac{550 - 550}{20} = \frac{0}{20} = 0$	$8 \times 0 = 0$
560 - 580	6	570	$\frac{570 - 550}{20} = \frac{20}{20} = 1$	$6 \times 1 = 6$
580 - 600	10	590	$\frac{590 - 550}{20} = \frac{40}{20} = 2$	$10 \times 2 = 20$
Total	$\sum f_i = 50$			$\sum (f_i \times u_i) = -12$

Sum of frequencies, $\sum f_i = 50$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -24 - 14 + 0 + 6 + 20 = -38 + 26 = -12$

Now, substitute these values into the mean formula:

$\bar{x} = 550 + \left(\frac{-12}{50}\right) \times 20$

$\bar{x} = 550 + \frac{-12 \times 20}{50}$

$\bar{x} = 550 + \frac{-240}{50}$

$\bar{x} = 550 - \frac{24}{5}$

$\bar{x} = 550 - 4.8$

$\bar{x} = 545.2$

The mean daily wages of the workers is $\textsf{₹}545.20$.

Given:

The distribution of daily pocket allowance of children and the given mean pocket allowance ($\bar{x}$) = $\textsf{₹}18$.

The data is:

Daily pocket allowance (in $\textsf{₹}$)	11 - 13	13 - 15	15 - 17	17 - 19	19 - 21	21 - 23	23 - 25
Number of children ($f_i$)	7	6	9	13	f	5	4

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To Find:

The missing frequency, $f$.

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Solution:

We can use the Direct Method to find the mean and then use the given mean value to find the missing frequency $f$.

The formula for the mean ($\bar{x}$) using the direct method is:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

First, we calculate the class mark ($x_i$) for each class interval:

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 11 - 13: $x_1 = \frac{11+13}{2} = 12$
• For 13 - 15: $x_2 = \frac{13+15}{2} = 14$
• For 15 - 17: $x_3 = \frac{15+17}{2} = 16$
• For 17 - 19: $x_4 = \frac{17+19}{2} = 18$
• For 19 - 21: $x_5 = \frac{19+21}{2} = 20$
• For 21 - 23: $x_6 = \frac{21+23}{2} = 22$
• For 23 - 25: $x_7 = \frac{23+25}{2} = 24$

Now, we prepare a table to calculate the product $f_i \times x_i$:

Daily pocket allowance (in $\textsf{₹}$)	Number of children ($f_i$)	Class Mark ($x_i$)	$f_i \times x_i$
11 - 13	7	12	$7 \times 12 = 84$
13 - 15	6	14	$6 \times 14 = 84$
15 - 17	9	16	$9 \times 16 = 144$
17 - 19	13	18	$13 \times 18 = 234$
19 - 21	f	20	$f \times 20 = 20f$
21 - 23	5	22	$5 \times 22 = 110$
23 - 25	4	24	$4 \times 24 = 96$
Total	$\sum f_i$		$\sum (f_i \times x_i)$
	$7+6+9+13+f+5+4 = 44 + f$		$84+84+144+234+20f+110+96 = 752 + 20f$

Sum of frequencies, $\sum f_i = 44 + f$

Sum of the products of frequencies and class marks, $\sum (f_i \times x_i) = 752 + 20f$

Given mean ($\bar{x}$) = $18$

Substitute these values into the mean formula:

$18 = \frac{752 + 20f}{44 + f}$

Multiply both sides by $(44 + f)$:

$18 \times (44 + f) = 752 + 20f$

Distribute 18 on the left side:

$18 \times 44 + 18f = 752 + 20f$

$792 + 18f = 752 + 20f$

Subtract $18f$ from both sides:

$792 = 752 + 20f - 18f$

$792 = 752 + 2f$

Subtract 752 from both sides:

$792 - 752 = 2f$

$40 = 2f$

Divide by 2:

$f = \frac{40}{2}$

$f = 20$

The missing frequency $f$ is $20$.

Given:

The distribution of the number of heartbeats per minute for 30 women, along with the number of women (frequency).

The data is:

Number of heartbeats per minute	65 - 68	68 - 71	71 - 74	74 - 77	77 - 80	80 - 83	83 - 86
Number of women ($f_i$)	2	4	3	8	7	4	2

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To Find:

The mean number of heartbeats per minute for these women, choosing a suitable method.

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Solution:

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 65 - 68: $x_1 = \frac{65+68}{2} = \frac{133}{2} = 66.5$
• For 68 - 71: $x_2 = \frac{68+71}{2} = \frac{139}{2} = 69.5$
• For 71 - 74: $x_3 = \frac{71+74}{2} = \frac{145}{2} = 72.5$
• For 74 - 77: $x_4 = \frac{74+77}{2} = \frac{151}{2} = 75.5$
• For 77 - 80: $x_5 = \frac{77+80}{2} = \frac{157}{2} = 78.5$
• For 80 - 83: $x_6 = \frac{80+83}{2} = \frac{163}{2} = 81.5$
• For 83 - 86: $x_7 = \frac{83+86}{2} = \frac{169}{2} = 84.5$

The total number of women (sum of frequencies) is $\sum f_i = 2 + 4 + 3 + 8 + 7 + 4 + 2 = 30$.

The class intervals are equal, with a class size $h = 68 - 65 = 3$. The class marks and frequencies are relatively small. We can use the Direct Method, Assumed Mean Method, or Step-deviation Method. The Step-deviation method is suitable here and often simplifies calculations.

We will use the Step-deviation Method. The formula for the mean is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the class size.

Let's choose the assumed mean $a = 75.5$ (a central class mark) and the class size $h = 3$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 75.5}{3}$ and the product $f_i \times u_i$:

Number of heartbeats per minute	Number of women ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 75.5}{3}$	$f_i \times u_i$
65 - 68	2	66.5	$\frac{66.5 - 75.5}{3} = \frac{-9}{3} = -3$	$2 \times (-3) = -6$
68 - 71	4	69.5	$\frac{69.5 - 75.5}{3} = \frac{-6}{3} = -2$	$4 \times (-2) = -8$
71 - 74	3	72.5	$\frac{72.5 - 75.5}{3} = \frac{-3}{3} = -1$	$3 \times (-1) = -3$
74 - 77	8	75.5	$\frac{75.5 - 75.5}{3} = \frac{0}{3} = 0$	$8 \times 0 = 0$
77 - 80	7	78.5	$\frac{78.5 - 75.5}{3} = \frac{3}{3} = 1$	$7 \times 1 = 7$
80 - 83	4	81.5	$\frac{81.5 - 75.5}{3} = \frac{6}{3} = 2$	$4 \times 2 = 8$
83 - 86	2	84.5	$\frac{84.5 - 75.5}{3} = \frac{9}{3} = 3$	$2 \times 3 = 6$
Total	$\sum f_i = 30$			$\sum (f_i \times u_i) = 4$

Sum of frequencies, $\sum f_i = 30$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -6 - 8 - 3 + 0 + 7 + 8 + 6 = -17 + 21 = 4$

Now, substitute these values into the mean formula:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h = 75.5 + \left(\frac{4}{30}\right) \times 3$

$\bar{x} = 75.5 + \frac{4 \times 3}{30}$

$\bar{x} = 75.5 + \frac{\cancel{12}^{2}}{\cancel{30}_{5}}$

$\bar{x} = 75.5 + \frac{2}{5}$

$\bar{x} = 75.5 + 0.4$

$\bar{x} = 75.9$

The mean heartbeats per minute for these women is $75.9$.

Given:

The distribution of mangoes in packing boxes according to the number of boxes (frequency).

The data is:

Number of mangoes	50 - 52	53 - 55	56 - 58	59 - 61	62 - 64
Number of boxes ($f_i$)	15	110	135	115	25

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To Find:

The mean number of mangoes kept in a packing box and the method used for finding the mean with justification.

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Solution:

To find the mean of the grouped data, we first need to find the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 50 - 52: $x_1 = \frac{50+52}{2} = 51$
• For 53 - 55: $x_2 = \frac{53+55}{2} = 54$
• For 56 - 58: $x_3 = \frac{56+58}{2} = 57$
• For 59 - 61: $x_4 = \frac{59+61}{2} = 60$
• For 62 - 64: $x_5 = \frac{62+64}{2} = 63$

Note that the given class intervals are discontinuous. However, the difference between the upper limit of one class and the lower limit of the next class is 1 (e.g., $53-52=1$, $56-55=1$). The class marks are equally spaced with a difference of 3 ($54-51=3$, $57-54=3$, etc.). This difference in class marks can be considered as the effective class size ($h$) for the step-deviation method if used.

The total number of boxes (sum of frequencies) is $\sum f_i = 15 + 110 + 135 + 115 + 25 = 400$.

Since the frequencies are large, the Step-deviation Method is appropriate to simplify calculations.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the effective class size (difference between consecutive class marks).

Let's choose the assumed mean $a = 57$ (a central class mark) and the effective class size $h = 3$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 57}{3}$ and the product $f_i \times u_i$:

Number of mangoes	Number of boxes ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 57}{3}$	$f_i \times u_i$
50 - 52	15	51	$\frac{51 - 57}{3} = \frac{-6}{3} = -2$	$15 \times (-2) = -30$
53 - 55	110	54	$\frac{54 - 57}{3} = \frac{-3}{3} = -1$	$110 \times (-1) = -110$
56 - 58	135	57	$\frac{57 - 57}{3} = \frac{0}{3} = 0$	$135 \times 0 = 0$
59 - 61	115	60	$\frac{60 - 57}{3} = \frac{3}{3} = 1$	$115 \times 1 = 115$
62 - 64	25	63	$\frac{63 - 57}{3} = \frac{6}{3} = 2$	$25 \times 2 = 50$
Total	$\sum f_i = 400$			$\sum (f_i \times u_i) = 25$

Sum of frequencies, $\sum f_i = 400$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -30 - 110 + 0 + 115 + 50 = -140 + 165 = 25$

Now, substitute these values into the mean formula:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h = 57 + \left(\frac{25}{400}\right) \times 3$

$\bar{x} = 57 + \frac{25 \times 3}{400}$

$\bar{x} = 57 + \frac{\cancel{75}^{3}}{\cancel{400}_{16}}$

$\bar{x} = 57 + \frac{3}{16}$

$\bar{x} = 57 + 0.1875$

$\bar{x} = 57.1875$

The mean number of mangoes kept in a packing box is $57.1875$.

---

Method Used and Justification:

We used the Step-deviation Method for finding the mean.

Justification: The frequencies are large, which would make the calculations in the Direct Method (involving products of large numbers) cumbersome. While the Assumed Mean Method would be an improvement, the class marks are equally spaced with a difference of 3. By using the Step-deviation Method, the deviations $d_i$ are divided by the common factor $h=3$, resulting in smaller values of $u_i$, which simplifies the calculation of $f_i \times u_i$ and thus reduces the complexity of the arithmetic.

Given:

The distribution of daily expenditure on food of 25 households in a locality is given in a grouped frequency distribution table.

The data is:

Daily expenditure (in $\textsf{₹}$)	100 - 150	150 - 200	200 - 250	250 - 300	300 - 350
Number of households ($f_i$)	4	5	12	2	2

---

To Find:

The mean daily expenditure on food by a suitable method.

---

Solution:

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 100 - 150: $x_1 = \frac{100+150}{2} = 125$
• For 150 - 200: $x_2 = \frac{150+200}{2} = 175$
• For 200 - 250: $x_3 = \frac{200+250}{2} = 225$
• For 250 - 300: $x_4 = \frac{250+300}{2} = 275$
• For 300 - 350: $x_5 = \frac{300+350}{2} = 325$

The total number of households (sum of frequencies) is $\sum f_i = 4 + 5 + 12 + 2 + 2 = 25$.

The class intervals are equal, with a class size $h = 150 - 100 = 50$. The class marks and frequencies are such that the Step-deviation Method would simplify calculations.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the class size.

Let's choose the assumed mean $a = 225$ (a central class mark) and the class size $h = 50$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 225}{50}$ and the product $f_i \times u_i$:

Daily expenditure (in $\textsf{₹}$)	Number of households ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 225}{50}$	$f_i \times u_i$
100 - 150	4	125	$\frac{125 - 225}{50} = \frac{-100}{50} = -2$	$4 \times (-2) = -8$
150 - 200	5	175	$\frac{175 - 225}{50} = \frac{-50}{50} = -1$	$5 \times (-1) = -5$
200 - 250	12	225	$\frac{225 - 225}{50} = \frac{0}{50} = 0$	$12 \times 0 = 0$
250 - 300	2	275	$\frac{275 - 225}{50} = \frac{50}{50} = 1$	$2 \times 1 = 2$
300 - 350	2	325	$\frac{325 - 225}{50} = \frac{100}{50} = 2$	$2 \times 2 = 4$
Total	$\sum f_i = 25$			$\sum (f_i \times u_i) = -7$

Sum of frequencies, $\sum f_i = 25$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -8 - 5 + 0 + 2 + 4 = -13 + 6 = -7$

Now, substitute these values into the mean formula:

$\bar{x} = 225 + \left(\frac{-7}{25}\right) \times 50$

$\bar{x} = 225 + \frac{-7 \times \cancel{50}^{2}}{\cancel{25}_{1}}$

$\bar{x} = 225 + (-7 \times 2)$

$\bar{x} = 225 - 14$

$\bar{x} = 211$

The mean daily expenditure on food is $\textsf{₹}211$.

Given:

The concentration of SO₂ in the air (in ppm) for 30 localities and the corresponding frequencies are given in the table.

The data is:

Concentration of SO2 (in ppm)	Frequency ($f_i$)
0.00 - 0.04	4
0.04 - 0.08	9
0.08 - 0.12	9
0.12 - 0.16	2
0.16 - 0.20	4
0.20 - 0.24	2

---

To Find:

The mean concentration of SO₂ in the air.

---

Solution:

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 0.00 - 0.04: $x_1 = \frac{0.00+0.04}{2} = 0.02$
• For 0.04 - 0.08: $x_2 = \frac{0.04+0.08}{2} = 0.06$
• For 0.08 - 0.12: $x_3 = \frac{0.08+0.12}{2} = 0.10$
• For 0.12 - 0.16: $x_4 = \frac{0.12+0.16}{2} = 0.14$
• For 0.16 - 0.20: $x_5 = \frac{0.16+0.20}{2} = 0.18$
• For 0.20 - 0.24: $x_6 = \frac{0.20+0.24}{2} = 0.22$

The total number of localities (sum of frequencies) is $\sum f_i = 4 + 9 + 9 + 2 + 4 + 2 = 30$.

The class intervals are equal, with a class size $h = 0.04 - 0.00 = 0.04$. Since the class marks are decimals and the class size is small, the Step-deviation Method is a suitable method for calculating the mean.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the class size.

Let's choose the assumed mean $a = 0.10$ (a central class mark) and the class size $h = 0.04$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 0.10}{0.04}$ and the product $f_i \times u_i$:

Concentration of SO2 (in ppm)	Frequency ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 0.10}{0.04}$	$f_i \times u_i$
0.00 - 0.04	4	0.02	$\frac{0.02 - 0.10}{0.04} = \frac{-0.08}{0.04} = -2$	$4 \times (-2) = -8$
0.04 - 0.08	9	0.06	$\frac{0.06 - 0.10}{0.04} = \frac{-0.04}{0.04} = -1$	$9 \times (-1) = -9$
0.08 - 0.12	9	0.10	$\frac{0.10 - 0.10}{0.04} = \frac{0}{0.04} = 0$	$9 \times 0 = 0$
0.12 - 0.16	2	0.14	$\frac{0.14 - 0.10}{0.04} = \frac{0.04}{0.04} = 1$	$2 \times 1 = 2$
0.16 - 0.20	4	0.18	$\frac{0.18 - 0.10}{0.04} = \frac{0.08}{0.04} = 2$	$4 \times 2 = 8$
0.20 - 0.24	2	0.22	$\frac{0.22 - 0.10}{0.04} = \frac{0.12}{0.04} = 3$	$2 \times 3 = 6$
Total	$\sum f_i = 30$			$\sum (f_i \times u_i) = -1$

Sum of frequencies, $\sum f_i = 30$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -8 - 9 + 0 + 2 + 8 + 6 = -17 + 16 = -1$

Now, substitute these values into the mean formula:

$\bar{x} = 0.10 + \left(\frac{-1}{30}\right) \times 0.04$

$\bar{x} = 0.10 - \frac{0.04}{30}$

$\bar{x} = 0.10 - \frac{4}{3000}$

$\bar{x} = 0.10 - \frac{1}{750}$

To subtract, we find a common denominator or convert to decimals:

$\bar{x} = \frac{10}{100} - \frac{1}{750} = \frac{1}{10} - \frac{1}{750}$

The least common multiple of 10 and 750 is 750.

$\bar{x} = \frac{1 \times 75}{10 \times 75} - \frac{1}{750} = \frac{75}{750} - \frac{1}{750} = \frac{75 - 1}{750} = \frac{74}{750} = \frac{37}{375}$

Alternatively, using decimal division:

$\frac{0.04}{30} \approx 0.001333...$

$\bar{x} = 0.10 - 0.001333... \approx 0.098667$

The mean concentration of SO₂ in the air is approximately $0.09867$ ppm.

Given:

The absentee record of 40 students of a class for the whole term is given in a grouped frequency distribution table.

The data is:

Number of days	0 - 6	6 - 10	10 - 14	14 - 20	20 - 28	28 - 38	38 - 40
Number of students ($f_i$)	11	10	7	4	4	3	1

---

To Find:

The mean number of days a student was absent.

---

Solution:

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 0 - 6: $x_1 = \frac{0+6}{2} = 3$
• For 6 - 10: $x_2 = \frac{6+10}{2} = 8$
• For 10 - 14: $x_3 = \frac{10+14}{2} = 12$
• For 14 - 20: $x_4 = \frac{14+20}{2} = 17$
• For 20 - 28: $x_5 = \frac{20+28}{2} = 24$
• For 28 - 38: $x_6 = \frac{28+38}{2} = 33$
• For 38 - 40: $x_7 = \frac{38+40}{2} = 39$

The total number of students (sum of frequencies) is $\sum f_i = 11 + 10 + 7 + 4 + 4 + 3 + 1 = 40$.

Since the class intervals are unequal, we can use the Direct Method or the Assumed Mean Method. The values of class marks and frequencies are moderate, so the Assumed Mean Method is a suitable choice.

The formula for the mean using the Assumed Mean Method is:

$\bar{x} = a + \frac{\sum (f_i \times d_i)}{\sum f_i}$

where $a$ is the assumed mean, $f_i$ is the frequency of the $i$-th class, and $d_i = x_i - a$ is the deviation of the class mark ($x_i$) from the assumed mean.

Let's choose the assumed mean $a = 24$ (a central value among the class marks).

We prepare a table to calculate the deviations $d_i = x_i - 24$ and the product $f_i \times d_i$:

Number of days	Number of students ($f_i$)	Class Mark ($x_i$)	Deviation $d_i = x_i - 24$	$f_i \times d_i$
0 - 6	11	3	$3 - 24 = -21$	$11 \times (-21) = -231$
6 - 10	10	8	$8 - 24 = -16$	$10 \times (-16) = -160$
10 - 14	7	12	$12 - 24 = -12$	$7 \times (-12) = -84$
14 - 20	4	17	$17 - 24 = -7$	$4 \times (-7) = -28$
20 - 28	4	24	$24 - 24 = 0$	$4 \times 0 = 0$
28 - 38	3	33	$33 - 24 = 9$	$3 \times 9 = 27$
38 - 40	1	39	$39 - 24 = 15$	$1 \times 15 = 15$
Total	$\sum f_i = 40$			$\sum (f_i \times d_i) = -461$

Sum of frequencies, $\sum f_i = 40$

Sum of the products of frequencies and deviations, $\sum (f_i \times d_i) = -231 - 160 - 84 - 28 + 0 + 27 + 15 = -503 + 42 = -461$

Now, substitute these values into the mean formula:

$\bar{x} = a + \frac{\sum (f_i \times d_i)}{\sum f_i} = 24 + \frac{-461}{40}$

$\bar{x} = 24 - \frac{461}{40}$

$\bar{x} = 24 - 11.525$

$\bar{x} = 12.475$

The mean number of days a student was absent is $12.475$ days.

Given:

The literacy rate (in percentage) of 35 cities and the corresponding number of cities (frequency).

The data is:

Literacy rate (in %)	45 - 55	55 - 65	65 - 75	75 - 85	85 - 95
Number of cities ($f_i$)	3	10	11	8	3

---

To Find:

The mean literacy rate.

---

Solution:

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 45 - 55: $x_1 = \frac{45+55}{2} = 50$
• For 55 - 65: $x_2 = \frac{55+65}{2} = 60$
• For 65 - 75: $x_3 = \frac{65+75}{2} = 70$
• For 75 - 85: $x_4 = \frac{75+85}{2} = 80$
• For 85 - 95: $x_5 = \frac{85+95}{2} = 90$

The total number of cities (sum of frequencies) is $\sum f_i = 3 + 10 + 11 + 8 + 3 = 35$.

The class intervals are equal, with a class size $h = 55 - 45 = 10$. The class marks and frequencies are relatively small, making the Direct Method or Step-deviation Method suitable. We will use the Step-deviation Method as it often simplifies calculations.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the class size.

Let's choose the assumed mean $a = 70$ (a central class mark) and the class size $h = 10$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 70}{10}$ and the product $f_i \times u_i$:

Literacy rate (in %)	Number of cities ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 70}{10}$	$f_i \times u_i$
45 - 55	3	50	$\frac{50 - 70}{10} = \frac{-20}{10} = -2$	$3 \times (-2) = -6$
55 - 65	10	60	$\frac{60 - 70}{10} = \frac{-10}{10} = -1$	$10 \times (-1) = -10$
65 - 75	11	70	$\frac{70 - 70}{10} = \frac{0}{10} = 0$	$11 \times 0 = 0$
75 - 85	8	80	$\frac{80 - 70}{10} = \frac{10}{10} = 1$	$8 \times 1 = 8$
85 - 95	3	90	$\frac{90 - 70}{10} = \frac{20}{10} = 2$	$3 \times 2 = 6$
Total	$\sum f_i = 35$			$\sum (f_i \times u_i) = -2$

Sum of frequencies, $\sum f_i = 35$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -6 - 10 + 0 + 8 + 6 = -16 + 14 = -2$

Now, substitute these values into the mean formula:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h = 70 + \left(\frac{-2}{35}\right) \times 10$

$\bar{x} = 70 + \frac{-2 \times 10}{35}$

$\bar{x} = 70 - \frac{20}{35}$

Simplify the fraction $\frac{20}{35}$ by dividing the numerator and denominator by 5:

$\frac{20}{35} = \frac{\cancel{20}^{4}}{\cancel{35}_{7}} = \frac{4}{7}$

$\bar{x} = 70 - \frac{4}{7}$

$\bar{x} = \frac{70 \times 7 - 4}{7} = \frac{490 - 4}{7} = \frac{486}{7}$

Calculating the decimal value:

$\frac{486}{7} \approx 69.42857...$

Rounding to two decimal places, the mean literacy rate is approximately $69.43\%$.

Given:

The number of wickets taken by a bowler in 10 cricket matches.

The data set is: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3

---

To Find:

The mode of the given data.

---

Solution:

The mode of a set of ungrouped data is the value that appears most frequently in the data set.

First, let's arrange the data in ascending order to easily count the frequencies of each value (this step is not strictly necessary but can be helpful):

0, 1, 2, 2, 2, 3, 3, 4, 5, 6

Now, let's count the frequency of each value:

• 0 appears 1 time
• 1 appears 1 time
• 2 appears 3 times
• 3 appears 2 times
• 4 appears 1 time
• 5 appears 1 time
• 6 appears 1 time

The value that occurs most frequently is 2, which appears 3 times.

Therefore, the mode of the data is 2.

Given:

The frequency distribution table showing the number of family members in 20 households.

The data is:

Family size	1 - 3	3 - 5	5 - 7	7 - 9	9 - 11
Number of families ($f_i$)	7	8	2	2	1

---

To Find:

The mode of the given data.

---

Solution:

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

where:

$l$ = lower limit of the modal class

$h$ = size of the class interval

$f_1$ = frequency of the modal class

$f_0$ = frequency of the class preceding the modal class

$f_2$ = frequency of the class succeeding the modal class

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $8$, which corresponds to the class interval $3 - 5$.

Therefore, the modal class is $3 - 5$.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• Lower limit of the modal class ($l$) = $3$
• Size of the class interval ($h$) = Upper limit - Lower limit = $5 - 3 = 2$
• Frequency of the modal class ($f_1$) = $8$
• Frequency of the class preceding the modal class ($f_0$) = $7$
• Frequency of the class succeeding the modal class ($f_2$) = $2$

Substitute these values into the mode formula:

Mode $= 3 + \left(\frac{8 - 7}{2 \times 8 - 7 - 2}\right) \times 2$

Mode $= 3 + \left(\frac{1}{16 - 7 - 2}\right) \times 2$

Mode $= 3 + \left(\frac{1}{16 - 9}\right) \times 2$

Mode $= 3 + \left(\frac{1}{7}\right) \times 2$

Mode $= 3 + \frac{2}{7}$

To add these values, we find a common denominator:

Mode $= \frac{3 \times 7}{7} + \frac{2}{7} = \frac{21}{7} + \frac{2}{7} = \frac{23}{7}$

Calculating the decimal value:

Mode $\approx 3.286$ (rounded to three decimal places)

The mode of the data is $\frac{23}{7}$ or approximately $3.286$. This means that the family size of about 3 to 4 members occurs most frequently in the locality.

Given:

The grouped frequency distribution of marks obtained by 30 students in a mathematics examination.

The data is given in Table 14.3:

Class interval (Marks)	Number of students ($f_i$)
10 - 25	2
25 - 40	3
40 - 55	7
55 - 70	6
70 - 85	6
85 - 100	6

---

To Find:

The mode of this data and compare/interpret the mode and the mean.

---

Solution (Mode):

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

where:

• $l$ = lower limit of the modal class
• $h$ = size of the class interval
• $f_1$ = frequency of the modal class
• $f_0$ = frequency of the class preceding the modal class
• $f_2$ = frequency of the class succeeding the modal class

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $7$, which corresponds to the class interval $40 - 55$.

Therefore, the modal class is $40 - 55$.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• Lower limit of the modal class ($l$) = $40$
• Size of the class interval ($h$) = Upper limit - Lower limit = $55 - 40 = 15$
• Frequency of the modal class ($f_1$) = $7$
• Frequency of the class preceding the modal class ($f_0$) = $3$ (Frequency of 25 - 40)
• Frequency of the class succeeding the modal class ($f_2$) = $6$ (Frequency of 55 - 70)

Substitute these values into the mode formula:

Mode $= 40 + \left(\frac{7 - 3}{2 \times 7 - 3 - 6}\right) \times 15$

Mode $= 40 + \left(\frac{4}{14 - 3 - 6}\right) \times 15$

Mode $= 40 + \left(\frac{4}{14 - 9}\right) \times 15$

Mode $= 40 + \left(\frac{4}{5}\right) \times 15$

Mode $= 40 + 4 \times \frac{15}{5}$

Mode $= 40 + 4 \times 3$

Mode $= 40 + 12$

Mode $= 52$

The mode of the data is $52$.

---

Solution (Mean):

The mean of this grouped data can be calculated using the Direct Method, Assumed Mean Method, or Step-deviation Method. From Table 14.3, the sum of $f_i \times x_i$ and the sum of $f_i$ are already provided.

Sum of frequencies, $\sum f_i = 30$

Sum of products of frequencies and class marks, $\sum (f_i \times x_i) = 1860.0$

The formula for the mean using the Direct Method is:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

Mean ($\bar{x}$) = $\frac{1860.0}{30}$

Mean ($\bar{x}$) = $62$

The mean of the data is $62$.

---

Comparison and Interpretation:

Mode = $52$ marks

Mean = $62$ marks

The mode is $52$, which lies within the modal class $40 - 55$. It indicates that the marks around $52$ are obtained by the highest number of students in this grouped distribution.

The mean is $62$. This represents the average marks obtained by the 30 students.

In this case, the mean (62) is higher than the mode (52). This suggests that the distribution of marks is likely skewed towards the higher values, meaning there might be a tail of observations extending towards the higher end of the marks range. The mode points to the peak frequency concentration in the grouped data, while the mean represents the balance point of the distribution.

Given:

The ages of patients admitted in a hospital during a year and the number of patients (frequency) in each age group.

The data is:

Age (in years)	5 - 15	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65
Number of patients ($f_i$)	6	11	21	23	14	5

---

To Find:

The mode and the mean of the given data. Also, compare and interpret the two measures of central tendency.

---

Solution (Mode):

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

where:

• $l$ = lower limit of the modal class
• $h$ = size of the class interval
• $f_1$ = frequency of the modal class
• $f_0$ = frequency of the class preceding the modal class
• $f_2$ = frequency of the class succeeding the modal class

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $23$, which corresponds to the class interval $35 - 45$.

Therefore, the modal class is $35 - 45$.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• Lower limit of the modal class ($l$) = $35$
• Size of the class interval ($h$) = Upper limit - Lower limit = $45 - 35 = 10$
• Frequency of the modal class ($f_1$) = $23$
• Frequency of the class preceding the modal class ($f_0$) = $21$ (Frequency of 25 - 35)
• Frequency of the class succeeding the modal class ($f_2$) = $14$ (Frequency of 45 - 55)

Substitute these values into the mode formula:

Mode $= 35 + \left(\frac{23 - 21}{2 \times 23 - 21 - 14}\right) \times 10$

Mode $= 35 + \left(\frac{2}{46 - 21 - 14}\right) \times 10$

Mode $= 35 + \left(\frac{2}{46 - 35}\right) \times 10$

Mode $= 35 + \left(\frac{2}{11}\right) \times 10$

Mode $= 35 + \frac{20}{11}$

Mode $\approx 35 + 1.818$ (rounded to three decimal places)

Mode $\approx 36.818$

The mode of the data is approximately $36.8$ years.

---

Solution (Mean):

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 5 - 15: $x_1 = \frac{5+15}{2} = 10$
• For 15 - 25: $x_2 = \frac{15+25}{2} = 20$
• For 25 - 35: $x_3 = \frac{25+35}{2} = 30$
• For 35 - 45: $x_4 = \frac{35+45}{2} = 40$
• For 45 - 55: $x_5 = \frac{45+55}{2} = 50$
• For 55 - 65: $x_6 = \frac{55+65}{2} = 60$

The total number of patients (sum of frequencies) is $\sum f_i = 6 + 11 + 21 + 23 + 14 + 5 = 80$.

The class intervals are equal, with a class size $h = 10$. The class marks and frequencies are such that the Step-deviation Method is suitable for simplifying calculations.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the class size.

Let's choose the assumed mean $a = 30$ (a central class mark) and the class size $h = 10$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 30}{10}$ and the product $f_i \times u_i$:

Age (in years)	Number of patients ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 30}{10}$	$f_i \times u_i$
5 - 15	6	10	$\frac{10 - 30}{10} = \frac{-20}{10} = -2$	$6 \times (-2) = -12$
15 - 25	11	20	$\frac{20 - 30}{10} = \frac{-10}{10} = -1$	$11 \times (-1) = -11$
25 - 35	21	30	$\frac{30 - 30}{10} = \frac{0}{10} = 0$	$21 \times 0 = 0$
35 - 45	23	40	$\frac{40 - 30}{10} = \frac{10}{10} = 1$	$23 \times 1 = 23$
45 - 55	14	50	$\frac{50 - 30}{10} = \frac{20}{10} = 2$	$14 \times 2 = 28$
55 - 65	5	60	$\frac{60 - 30}{10} = \frac{30}{10} = 3$	$5 \times 3 = 15$
Total	$\sum f_i = 80$			$\sum (f_i \times u_i) = 43$

Sum of frequencies, $\sum f_i = 80$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -12 - 11 + 0 + 23 + 28 + 15 = -23 + 66 = 43$

Now, substitute these values into the mean formula:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h = 30 + \left(\frac{43}{80}\right) \times 10$

$\bar{x} = 30 + \frac{43 \times \cancel{10}^{1}}{\cancel{80}_{8}}$

$\bar{x} = 30 + \frac{43}{8}$

$\bar{x} = 30 + 5.375$

$\bar{x} = 35.375$

The mean age is $35.375$ years.

---

Comparison and Interpretation:

Mode $\approx 36.8$ years

Mean $= 35.375$ years

The mode (approximately 36.8 years) represents the age around which the maximum number of patients were admitted to the hospital. It is the most frequent age group.

The mean (35.375 years) represents the average age of the patients admitted. It is calculated by considering the age of all 80 patients.

Both the mode and mean are measures of central tendency, providing insights into the typical age of patients. The mode is slightly higher than the mean in this distribution. This suggests that while the average age is around 35.4 years, the single age value within the class intervals that is most representative of the peak frequency is closer to 36.8 years. The values are relatively close, indicating a distribution that is not highly skewed, although the mode being slightly higher than the mean might hint at a slight skew towards the younger ages when considering the overall distribution shape, but the difference is small.

Given:

The distribution of observed lifetimes (in hours) of 225 electrical components and their corresponding frequencies.

The data is:

Lifetimes (in hours)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency ($f_i$)	10	35	52	61	38	29

---

To Find:

The modal lifetimes of the components.

---

Solution:

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

where:

• $l$ = lower limit of the modal class
• $h$ = size of the class interval
• $f_1$ = frequency of the modal class
• $f_0$ = frequency of the class preceding the modal class
• $f_2$ = frequency of the class succeeding the modal class

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $61$, which corresponds to the class interval $60 - 80$.

Therefore, the modal class is $60 - 80$.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• Lower limit of the modal class ($l$) = $60$
• Size of the class interval ($h$) = Upper limit - Lower limit = $80 - 60 = 20$
• Frequency of the modal class ($f_1$) = $61$
• Frequency of the class preceding the modal class ($f_0$) = $52$ (Frequency of 40 - 60)
• Frequency of the class succeeding the modal class ($f_2$) = $38$ (Frequency of 80 - 100)

Substitute these values into the mode formula:

Mode $= 60 + \left(\frac{61 - 52}{2 \times 61 - 52 - 38}\right) \times 20$

Mode $= 60 + \left(\frac{9}{122 - 52 - 38}\right) \times 20$

Mode $= 60 + \left(\frac{9}{122 - 90}\right) \times 20$

Mode $= 60 + \left(\frac{9}{32}\right) \times 20$

Mode $= 60 + \frac{9 \times 20}{32}$

Simplify the fraction $\frac{20}{32}$ by dividing the numerator and denominator by 4:

Mode $= 60 + 9 \times \frac{\cancel{20}^{5}}{\cancel{32}_{8}}$

Mode $= 60 + \frac{9 \times 5}{8}$

Mode $= 60 + \frac{45}{8}$

Mode $= 60 + 5.625$

Mode $= 65.625$

The modal lifetimes of the components is $65.625$ hours.

Given:

The distribution of total monthly household expenditure of 200 families of a village and the number of families (frequency).

The data is:

Expenditure (in $\textsf{₹}$)	Number of families ($f_i$)
1000 - 1500	24
1500 - 2000	40
2000 - 2500	33
2500 - 3000	28
3000 - 3500	30
3500 - 4000	22
4000 - 4500	16
4500 - 5000	7

---

To Find:

The modal monthly expenditure and the mean monthly expenditure of the families.

---

Solution (Mode):

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

where:

• $l$ = lower limit of the modal class
• $h$ = size of the class interval
• $f_1$ = frequency of the modal class
• $f_0$ = frequency of the class preceding the modal class
• $f_2$ = frequency of the class succeeding the modal class

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $40$, which corresponds to the class interval $1500 - 2000$.

Therefore, the modal class is $1500 - 2000$.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• Lower limit of the modal class ($l$) = $1500$
• Size of the class interval ($h$) = Upper limit - Lower limit = $2000 - 1500 = 500$
• Frequency of the modal class ($f_1$) = $40$
• Frequency of the class preceding the modal class ($f_0$) = $24$ (Frequency of 1000 - 1500)
• Frequency of the class succeeding the modal class ($f_2$) = $33$ (Frequency of 2000 - 2500)

Substitute these values into the mode formula:

Mode $= 1500 + \left(\frac{40 - 24}{2 \times 40 - 24 - 33}\right) \times 500$

Mode $= 1500 + \left(\frac{16}{80 - 24 - 33}\right) \times 500$

Mode $= 1500 + \left(\frac{16}{80 - 57}\right) \times 500$

Mode $= 1500 + \left(\frac{16}{23}\right) \times 500$

Mode $= 1500 + \frac{16 \times 500}{23}$

Mode $= 1500 + \frac{8000}{23}$

Mode $\approx 1500 + 347.826$ (rounded to three decimal places)

Mode $\approx 1847.83$

The modal monthly expenditure is approximately $\textsf{₹}1847.83$.

---

Solution (Mean):

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 1000 - 1500: $x_1 = \frac{1000+1500}{2} = 1250$
• For 1500 - 2000: $x_2 = \frac{1500+2000}{2} = 1750$
• For 2000 - 2500: $x_3 = \frac{2000+2500}{2} = 2250$
• For 2500 - 3000: $x_4 = \frac{2500+3000}{2} = 2750$
• For 3000 - 3500: $x_5 = \frac{3000+3500}{2} = 3250$
• For 3500 - 4000: $x_6 = \frac{3500+4000}{2} = 3750$
• For 4000 - 4500: $x_7 = \frac{4000+4500}{2} = 4250$
• For 4500 - 5000: $x_8 = \frac{4500+5000}{2} = 4750$

The total number of families (sum of frequencies) is $\sum f_i = 24 + 40 + 33 + 28 + 30 + 22 + 16 + 7 = 200$.

The class intervals are equal, with a class size $h = 1500 - 1000 = 500$. Since the class marks are large, the Step-deviation Method is suitable for simplifying calculations.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the class size.

Let's choose the assumed mean $a = 2750$ (a central class mark) and the class size $h = 500$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 2750}{500}$ and the product $f_i \times u_i$:

Expenditure (in $\textsf{₹}$)	Number of families ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 2750}{500}$	$f_i \times u_i$
1000 - 1500	24	1250	$\frac{1250 - 2750}{500} = -3$	$24 \times (-3) = -72$
1500 - 2000	40	1750	$\frac{1750 - 2750}{500} = -2$	$40 \times (-2) = -80$
2000 - 2500	33	2250	$\frac{2250 - 2750}{500} = -1$	$33 \times (-1) = -33$
2500 - 3000	28	2750	$\frac{2750 - 2750}{500} = 0$	$28 \times 0 = 0$
3000 - 3500	30	3250	$\frac{3250 - 2750}{500} = 1$	$30 \times 1 = 30$
3500 - 4000	22	3750	$\frac{3750 - 2750}{500} = 2$	$22 \times 2 = 44$
4000 - 4500	16	4250	$\frac{4250 - 2750}{500} = 3$	$16 \times 3 = 48$
4500 - 5000	7	4750	$\frac{4750 - 2750}{500} = 4$	$7 \times 4 = 28$
Total	$\sum f_i = 200$			$\sum (f_i \times u_i) = -35$

Sum of frequencies, $\sum f_i = 200$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -72 - 80 - 33 + 0 + 30 + 44 + 48 + 28 = -185 + 150 = -35$

Now, substitute these values into the mean formula:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h = 2750 + \left(\frac{-35}{200}\right) \times 500$

$\bar{x} = 2750 + \frac{-35 \times \cancel{500}^{5/2}}{\cancel{200}_{2/2}}$

$\bar{x} = 2750 + \frac{-35 \times 5}{2}$

$\bar{x} = 2750 - \frac{175}{2}$

$\bar{x} = 2750 - 87.5$

$\bar{x} = 2662.5$

The mean monthly expenditure is $\textsf{₹}2662.50$.

---

Comparison and Interpretation:

Modal monthly expenditure $\approx \textsf{₹}1847.83$

Mean monthly expenditure $= \textsf{₹}2662.50$

The modal monthly expenditure (approximately $\textsf{₹}1847.83$) represents the expenditure amount around which the highest number of families spend their money. It is the most frequent expenditure range.

The mean monthly expenditure ($\textsf{₹}2662.50$) represents the average monthly expenditure of the 200 families in the village.

In this case, the mean is significantly higher than the mode. This indicates that the distribution of monthly household expenditure is likely skewed towards the higher expenditure values. The mode identifies the expenditure range where the largest group of families falls (1500-2000), while the mean is pulled towards the higher expenditures, suggesting the presence of some families with considerably higher spending, which raises the average.

Given:

The state-wise teacher-student ratio in higher secondary schools of India and the corresponding number of States/U.T. (frequency).

The data is:

Number of students per teacher	Number of States / U.T. ($f_i$)
15 - 20	3
20 - 25	8
25 - 30	9
30 - 35	10
35 - 40	3
40 - 45	0
45 - 50	0
50 - 55	2

---

To Find:

The mode and mean of this data. Interpret the two measures.

---

Solution (Mode):

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

where:

• $l$ = lower limit of the modal class
• $h$ = size of the class interval
• $f_1$ = frequency of the modal class
• $f_0$ = frequency of the class preceding the modal class
• $f_2$ = frequency of the class succeeding the modal class

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $10$, which corresponds to the class interval $30 - 35$.

Therefore, the modal class is $30 - 35$.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• Lower limit of the modal class ($l$) = $30$
• Size of the class interval ($h$) = Upper limit - Lower limit = $35 - 30 = 5$
• Frequency of the modal class ($f_1$) = $10$
• Frequency of the class preceding the modal class ($f_0$) = $9$ (Frequency of 25 - 30)
• Frequency of the class succeeding the modal class ($f_2$) = $3$ (Frequency of 35 - 40)

Substitute these values into the mode formula:

Mode $= 30 + \left(\frac{10 - 9}{2 \times 10 - 9 - 3}\right) \times 5$

Mode $= 30 + \left(\frac{1}{20 - 9 - 3}\right) \times 5$

Mode $= 30 + \left(\frac{1}{20 - 12}\right) \times 5$

Mode $= 30 + \left(\frac{1}{8}\right) \times 5$

Mode $= 30 + \frac{5}{8}$

Mode $= 30 + 0.625$

Mode $= 30.625$

The mode of the data is $30.625$.

---

Solution (Mean):

To find the mean of the grouped data, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 15 - 20: $x_1 = \frac{15+20}{2} = 17.5$
• For 20 - 25: $x_2 = \frac{20+25}{2} = 22.5$
• For 25 - 30: $x_3 = \frac{25+30}{2} = 27.5$
• For 30 - 35: $x_4 = \frac{30+35}{2} = 32.5$
• For 35 - 40: $x_5 = \frac{35+40}{2} = 37.5$
• For 40 - 45: $x_6 = \frac{40+45}{2} = 42.5$
• For 45 - 50: $x_7 = \frac{45+50}{2} = 47.5$
• For 50 - 55: $x_8 = \frac{50+55}{2} = 52.5$

The total number of States/U.T. (sum of frequencies) is $\sum f_i = 3 + 8 + 9 + 10 + 3 + 0 + 0 + 2 = 35$.

The class intervals are equal, with a class size $h = 5$. The class marks contain decimals, so the Step-deviation Method is a suitable method for calculating the mean.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$ is the step deviation, and $h$ is the class size.

Let's choose the assumed mean $a = 32.5$ (a central class mark) and the class size $h = 5$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 32.5}{5}$ and the product $f_i \times u_i$:

Number of students per teacher	Number of States / U.T. ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 32.5}{5}$	$f_i \times u_i$
15 - 20	3	17.5	$\frac{17.5 - 32.5}{5} = \frac{-15}{5} = -3$	$3 \times (-3) = -9$
20 - 25	8	22.5	$\frac{22.5 - 32.5}{5} = \frac{-10}{5} = -2$	$8 \times (-2) = -16$
25 - 30	9	27.5	$\frac{27.5 - 32.5}{5} = \frac{-5}{5} = -1$	$9 \times (-1) = -9$
30 - 35	10	32.5	$\frac{32.5 - 32.5}{5} = \frac{0}{5} = 0$	$10 \times 0 = 0$
35 - 40	3	37.5	$\frac{37.5 - 32.5}{5} = \frac{5}{5} = 1$	$3 \times 1 = 3$
40 - 45	0	42.5	$\frac{42.5 - 32.5}{5} = \frac{10}{5} = 2$	$0 \times 2 = 0$
45 - 50	0	47.5	$\frac{47.5 - 32.5}{5} = \frac{15}{5} = 3$	$0 \times 3 = 0$
50 - 55	2	52.5	$\frac{52.5 - 32.5}{5} = \frac{20}{5} = 4$	$2 \times 4 = 8$
Total	$\sum f_i = 35$			$\sum (f_i \times u_i) = -23$

Sum of frequencies, $\sum f_i = 35$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -9 - 16 - 9 + 0 + 3 + 0 + 0 + 8 = -34 + 11 = -23$

Now, substitute these values into the mean formula:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h = 32.5 + \left(\frac{-23}{35}\right) \times 5$

$\bar{x} = 32.5 + \frac{-23 \times \cancel{5}^{1}}{\cancel{35}_{7}}$

$\bar{x} = 32.5 - \frac{23}{7}$

$\bar{x} = 32.5 - 3.2857...$

$\bar{x} \approx 29.2143$

Rounding to two decimal places, the mean teacher-student ratio is approximately $29.21$.

---

Interpretation:

Mode $= 30.625$ students per teacher

Mean $\approx 29.21$ students per teacher

The mode ($30.625$) indicates that the teacher-student ratio of approximately $30.6$ students per teacher occurs in the highest number of States/U.T. among the given data. It represents the most frequent ratio.

The mean ($\approx 29.21$) represents the average teacher-student ratio across the 35 States/U.T. included in the data.

Both measures are close to each other and fall within the central range of the distribution. The mode is slightly higher than the mean. This suggests that while the average ratio is around 29.2, the ratio that occurs most often among the states is slightly higher, around 30.6. This small difference implies the distribution is relatively symmetrical or only slightly skewed.

Given:

The distribution of runs scored by some top batsmen in one-day international cricket matches and the corresponding number of batsmen (frequency).

The data is:

Runs scored	Number of batsmen ($f_i$)
3000 - 4000	4
4000 - 5000	18
5000 - 6000	9
6000 - 7000	7
7000 - 8000	6
8000 - 9000	3
9000 - 10000	1
10000 - 11000	1

---

To Find:

The mode of the given data.

---

Solution:

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

where:

• $l$ = lower limit of the modal class
• $h$ = size of the class interval
• $f_1$ = frequency of the modal class
• $f_0$ = frequency of the class preceding the modal class
• $f_2$ = frequency of the class succeeding the modal class

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $18$, which corresponds to the class interval $4000 - 5000$.

Therefore, the modal class is $4000 - 5000$.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• Lower limit of the modal class ($l$) = $4000$
• Size of the class interval ($h$) = Upper limit - Lower limit = $5000 - 4000 = 1000$
• Frequency of the modal class ($f_1$) = $18$
• Frequency of the class preceding the modal class ($f_0$) = $4$ (Frequency of 3000 - 4000)
• Frequency of the class succeeding the modal class ($f_2$) = $9$ (Frequency of 5000 - 6000)

Substitute these values into the mode formula:

Mode $= 4000 + \left(\frac{18 - 4}{2 \times 18 - 4 - 9}\right) \times 1000$

Mode $= 4000 + \left(\frac{14}{36 - 4 - 9}\right) \times 1000$

Mode $= 4000 + \left(\frac{14}{36 - 13}\right) \times 1000$

Mode $= 4000 + \left(\frac{14}{23}\right) \times 1000$

Mode $= 4000 + \frac{14000}{23}$

Mode $\approx 4000 + 608.69565...$

Mode $\approx 4608.70$ (rounded to two decimal places)

The mode of the data is approximately $4608.70$ runs.

Given:

The distribution of the number of cars passing through a spot and the corresponding frequency.

The data is:

Number of cars	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency ($f_i$)	7	14	13	12	20	11	15	8

---

To Find:

The mode of the given data.

---

Solution:

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

where:

• $l$ = lower limit of the modal class
• $h$ = size of the class interval
• $f_1$ = frequency of the modal class
• $f_0$ = frequency of the class preceding the modal class
• $f_2$ = frequency of the class succeeding the modal class

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $20$, which corresponds to the class interval $40 - 50$.

Therefore, the modal class is $40 - 50$.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• Lower limit of the modal class ($l$) = $40$
• Size of the class interval ($h$) = Upper limit - Lower limit = $50 - 40 = 10$
• Frequency of the modal class ($f_1$) = $20$
• Frequency of the class preceding the modal class ($f_0$) = $12$ (Frequency of 30 - 40)
• Frequency of the class succeeding the modal class ($f_2$) = $11$ (Frequency of 50 - 60)

Substitute these values into the mode formula:

Mode $= 40 + \left(\frac{20 - 12}{2 \times 20 - 12 - 11}\right) \times 10$

Mode $= 40 + \left(\frac{8}{40 - 12 - 11}\right) \times 10$

Mode $= 40 + \left(\frac{8}{40 - 23}\right) \times 10$

Mode $= 40 + \left(\frac{8}{17}\right) \times 10$

Mode $= 40 + \frac{80}{17}$

Mode $= \frac{40 \times 17 + 80}{17} = \frac{680 + 80}{17} = \frac{760}{17}$

Calculating the decimal value:

Mode $\approx 44.70588...$

Rounding to two decimal places, the mode is approximately $44.71$.

The mode of the data is $\frac{760}{17}$ or approximately $44.71$. This means that the number of cars passing through the spot in a 3-minute period is most frequently around $44.71$.

Given:

The cumulative frequency distribution of heights (in cm) of 51 girls of Class X.

Height (in cm)	Number of girls (Cumulative Frequency)
Less than 140	4
Less than 145	11
Less than 150	29
Less than 155	40
Less than 160	46
Less than 165	51

---

To Find:

The median height.

---

Solution:

The given distribution is a cumulative frequency distribution of 'less than' type. We first convert it into a simple frequency distribution table.

The class intervals are formed by taking the difference between consecutive 'less than' values. The class size is $145 - 140 = 5$.

The frequency of each class interval is the difference between the consecutive cumulative frequencies.

Height (in cm)	Number of girls (Frequency, $f_i$)	Cumulative Frequency (CF)
Below 140 (assuming 135-140 based on context/convention)	4	4
140 - 145	$11 - 4 = 7$	11
145 - 150	$29 - 11 = 18$	29
150 - 155	$40 - 29 = 11$	40
155 - 160	$46 - 40 = 6$	46
160 - 165	$51 - 46 = 5$	51
Total	$\sum f_i = 51$

The total number of observations is $N = \sum f_i = 51$.

To find the median, we need to find the class containing the $\frac{N}{2}$th observation.

$\frac{N}{2} = \frac{51}{2} = 25.5$

We look for the cumulative frequency that is just greater than or equal to $25.5$. This cumulative frequency is $29$, which corresponds to the class interval $145 - 150$.

So, the median class is $145 - 150$.

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

where:

• $l$ = lower limit of the median class = $145$
• $N$ = total number of observations = $51$
• $\frac{N}{2} = 25.5$
• $CF$ = cumulative frequency of the class preceding the median class = $11$ (Cumulative frequency of 140 - 145)
• $f$ = frequency of the median class = $18$ (Frequency of 145 - 150)
• $h$ = size of the class interval = $150 - 145 = 5$

Substitute these values into the median formula:

Median $= 145 + \left(\frac{25.5 - 11}{18}\right) \times 5$

Median $= 145 + \left(\frac{14.5}{18}\right) \times 5$

Median $= 145 + \frac{14.5 \times 5}{18}$

Median $= 145 + \frac{72.5}{18}$

Median $= 145 + 4.0277...$

Median $\approx 149.03$

The median height is approximately $149.03$ cm.

Given:

A grouped frequency distribution with missing frequencies x and y.

The median of the data is $525$.

The total frequency is $100$.

---

To Find:

The values of the missing frequencies x and y.

---

Solution:

First, we construct a cumulative frequency table from the given distribution.

Class Interval	Frequency ($f_i$)	Cumulative Frequency (CF)
0 - 100	2	2
100 - 200	5	$2 + 5 = 7$
200 - 300	x	$7 + x$
300 - 400	12	$7 + x + 12 = 19 + x$
400 - 500	17	$19 + x + 17 = 36 + x$
500 - 600	20	$36 + x + 20 = 56 + x$
600 - 700	y	$56 + x + y$
700 - 800	9	$56 + x + y + 9 = 65 + x + y$
800 - 900	7	$65 + x + y + 7 = 72 + x + y$
900 - 1000	4	$72 + x + y + 4 = 76 + x + y$
Total	$\sum f_i = 100$

The total frequency is given as $100$. From the table, the total frequency is also $76 + x + y$.

Subtract 76 from both sides of equation (1):

$x + y = 100 - 76$

---

The median of the data is given as $525$. The median lies in the class interval $500 - 600$.

Therefore, the median class is $500 - 600$.

From the median class, we identify the required values for the median formula:

• Lower limit of the median class ($l$) = $500$
• Total frequency ($N$) = $100$, so $\frac{N}{2} = \frac{100}{2} = 50$
• Cumulative frequency of the class preceding the median class ($CF$) = $36 + x$ (Cumulative frequency of 400 - 500)
• Frequency of the median class ($f$) = $20$ (Frequency of 500 - 600)
• Size of the class interval ($h$) = Upper limit - Lower limit = $600 - 500 = 100$

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

Substitute the known values into the median formula:

$525 = 500 + \left(\frac{50 - (36 + x)}{20}\right) \times 100$

Subtract 500 from both sides:

$525 - 500 = \left(\frac{50 - 36 - x}{20}\right) \times 100$

$25 = \left(\frac{14 - x}{20}\right) \times 100$

Simplify the right side:

$25 = (14 - x) \times \frac{100}{20}$

$25 = (14 - x) \times 5$

Divide both sides by 5:

$\frac{25}{5} = 14 - x$

$5 = 14 - x$

Add x to both sides:

$x + 5 = 14$

Subtract 5 from both sides:

$x = 14 - 5$

$x = 9$

---

Now we have the value of x. We can substitute this value into equation (2) to find y.

Substitute $x=9$:

$9 + y = 24$

Subtract 9 from both sides:

$y = 24 - 9$

$y = 15$

---

The values of the missing frequencies are $x = 9$ and $y = 15$.

Given:

The frequency distribution of monthly consumption of electricity of 68 consumers.

The data is:

Monthly consumption (in units)	Number of consumers ($f_i$)
65 - 85	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

Total number of consumers, $N = \sum f_i = 4 + 5 + 13 + 20 + 14 + 8 + 4 = 68$.

---

To Find:

The median, mean, and mode of the data and compare them.

---

Solution (Median):

To find the median, we first calculate the cumulative frequency (CF).

Monthly consumption (in units)	Number of consumers ($f_i$)	Cumulative Frequency (CF)
65 - 85	4	4
85 - 105	5	$4 + 5 = 9$
105 - 125	13	$9 + 13 = 22$
125 - 145	20	$22 + 20 = 42$
145 - 165	14	$42 + 14 = 56$
165 - 185	8	$56 + 8 = 64$
185 - 205	4	$64 + 4 = 68$
Total	$\sum f_i = 68$

We need to find the class containing the $\frac{N}{2}$th observation.

$\frac{N}{2} = \frac{68}{2} = 34$

The cumulative frequency just greater than or equal to 34 is 42, which corresponds to the class interval 125 - 145.

So, the median class is 125 - 145.

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

where:

• $l$ = lower limit of the median class = 125
• $\frac{N}{2} = 34$
• $CF$ = cumulative frequency of the class preceding the median class = 22
• $f$ = frequency of the median class = 20
• $h$ = size of the class interval = $145 - 125 = 20$

Substitute these values into the median formula:

Median $= 125 + \left(\frac{34 - 22}{20}\right) \times 20$

Median $= 125 + \left(\frac{12}{20}\right) \times 20$

Median $= 125 + \frac{12}{\cancel{20}} \times \cancel{20}$

Median $= 125 + 12$

Median $= 137$

The median monthly consumption is 137 units.

---

Solution (Mean):

To find the mean, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 65 - 85: $x_1 = \frac{65+85}{2} = 75$
• For 85 - 105: $x_2 = \frac{85+105}{2} = 95$
• For 105 - 125: $x_3 = \frac{105+125}{2} = 115$
• For 125 - 145: $x_4 = \frac{125+145}{2} = 135$
• For 145 - 165: $x_5 = \frac{145+165}{2} = 155$
• For 165 - 185: $x_6 = \frac{165+185}{2} = 175$
• For 185 - 205: $x_7 = \frac{185+205}{2} = 195$

The class intervals are equal with a class size $h = 20$. We will use the Step-deviation Method for calculating the mean.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

Let's choose the assumed mean $a = 135$ (a central class mark) and the class size $h = 20$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 135}{20}$ and the product $f_i \times u_i$:

Monthly consumption (in units)	Number of consumers ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 135}{20}$	$f_i \times u_i$
65 - 85	4	75	$\frac{75 - 135}{20} = -3$	$4 \times (-3) = -12$
85 - 105	5	95	$\frac{95 - 135}{20} = -2$	$5 \times (-2) = -10$
105 - 125	13	115	$\frac{115 - 135}{20} = -1$	$13 \times (-1) = -13$
125 - 145	20	135	$\frac{135 - 135}{20} = 0$	$20 \times 0 = 0$
145 - 165	14	155	$\frac{155 - 135}{20} = 1$	$14 \times 1 = 14$
165 - 185	8	175	$\frac{175 - 135}{20} = 2$	$8 \times 2 = 16$
185 - 205	4	195	$\frac{195 - 135}{20} = 3$	$4 \times 3 = 12$
Total	$\sum f_i = 68$			$\sum (f_i \times u_i) = 7$

Sum of frequencies, $\sum f_i = 68$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -12 - 10 - 13 + 0 + 14 + 16 + 12 = -35 + 42 = 7$

Now, substitute these values into the mean formula:

$\bar{x} = 135 + \left(\frac{7}{68}\right) \times 20$

$\bar{x} = 135 + \frac{7 \times \cancel{20}^{5}}{\cancel{68}_{17}}$

$\bar{x} = 135 + \frac{35}{17}$

$\bar{x} \approx 135 + 2.0588$

$\bar{x} \approx 137.06$

The mean monthly consumption is approximately 137.06 units.

---

Solution (Mode):

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $20$, which corresponds to the class interval 125 - 145.

Therefore, the modal class is 125 - 145.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• $l$ = lower limit of the modal class = 125
• $h$ = size of the class interval = $145 - 125 = 20$
• $f_1$ = frequency of the modal class = 20
• $f_0$ = frequency of the class preceding the modal class = 13
• $f_2$ = frequency of the class succeeding the modal class = 14

Substitute these values into the mode formula:

Mode $= 125 + \left(\frac{20 - 13}{2 \times 20 - 13 - 14}\right) \times 20$

Mode $= 125 + \left(\frac{7}{40 - 27}\right) \times 20$

Mode $= 125 + \left(\frac{7}{13}\right) \times 20$

Mode $= 125 + \frac{140}{13}$

Mode $\approx 125 + 10.769$

Mode $\approx 135.77$

The modal monthly consumption is approximately 135.77 units.

---

Comparison and Interpretation:

Median $= 137$ units

Mean $\approx 137.06$ units

Mode $\approx 135.77$ units

All three measures of central tendency (mean, median, and mode) are very close to each other (around 136-137 units). This indicates that the distribution of monthly electricity consumption is close to symmetrical. The mean represents the average consumption, the median represents the consumption level that divides the consumers into two equal groups, and the mode represents the most frequent consumption value or range.

Given:

A grouped frequency distribution with missing frequencies x and y.

The median of the data is $28.5$.

The total frequency is $60$.

---

To Find:

The values of the missing frequencies x and y.

---

Solution:

First, we construct a cumulative frequency table from the given distribution.

Class Interval	Frequency ($f_i$)	Cumulative Frequency (CF)
0 - 10	5	5
10 - 20	x	$5 + x$
20 - 30	20	$5 + x + 20 = 25 + x$
30 - 40	15	$25 + x + 15 = 40 + x$
40 - 50	y	$40 + x + y$
50 - 60	5	$40 + x + y + 5 = 45 + x + y$
Total	$\sum f_i = 60$

The total frequency is given as $60$. From the table, the total frequency is also $45 + x + y$.

Subtract 45 from both sides of equation (1):

$x + y = 60 - 45$

---

The median of the data is given as $28.5$. The median lies in the class interval $20 - 30$.

Therefore, the median class is $20 - 30$.

From the median class, we identify the required values for the median formula:

• Lower limit of the median class ($l$) = $20$
• Total frequency ($N$) = $60$, so $\frac{N}{2} = \frac{60}{2} = 30$
• Cumulative frequency of the class preceding the median class ($CF$) = $5 + x$ (Cumulative frequency of 10 - 20)
• Frequency of the median class ($f$) = $20$ (Frequency of 20 - 30)
• Size of the class interval ($h$) = Upper limit - Lower limit = $30 - 20 = 10$

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

Substitute the known values into the median formula:

$28.5 = 20 + \left(\frac{30 - (5 + x)}{20}\right) \times 10$

Subtract 20 from both sides:

$28.5 - 20 = \left(\frac{30 - 5 - x}{20}\right) \times 10$

$8.5 = \left(\frac{25 - x}{20}\right) \times 10$

Simplify the right side:

$8.5 = (25 - x) \times \frac{10}{20}$

$8.5 = (25 - x) \times \frac{1}{2}$

Multiply both sides by 2:

$8.5 \times 2 = 25 - x$

$17 = 25 - x$

Add x to both sides:

$x + 17 = 25$

Subtract 17 from both sides:

$x = 25 - 17$

$x = 8$

---

Now we have the value of x. We can substitute this value into equation (2) to find y.

Substitute $x=8$:

$8 + y = 15$

Subtract 8 from both sides:

$y = 15 - 8$

$y = 7$

---

The values of the missing frequencies are $x = 8$ and $y = 7$.

Given:

The cumulative frequency distribution of ages of 100 policy holders. Policies are given only to persons having age 18 years onwards but less than 60 years.

The data is in the form of 'less than' cumulative frequency distribution:

Age (in years)	Number of policy holders (CF)
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

---

To Find:

The median age of the policy holders.

---

Solution:

The given data is in the form of a 'less than' cumulative frequency distribution. We convert it into a frequency distribution table with class intervals.

Since policies are given from 18 years onwards, the first class interval is 18-20. Subsequent class intervals are formed based on the upper limits provided in the table.

The frequency of a class interval is the difference between the cumulative frequency of that class and the cumulative frequency of the preceding class.

Class Interval (Age in years)	Number of policy holders (Frequency, $f_i$)	Cumulative Frequency (CF)
18 - 20	2	2
20 - 25	$6 - 2 = 4$	6
25 - 30	$24 - 6 = 18$	24
30 - 35	$45 - 24 = 21$	45
35 - 40	$78 - 45 = 33$	78
40 - 45	$89 - 78 = 11$	89
45 - 50	$92 - 89 = 3$	92
50 - 55	$98 - 92 = 6$	98
55 - 60	$100 - 98 = 2$	100
Total	$\sum f_i = 100$

The total number of policy holders is $N = 100$.

To find the median age, we need to find the class containing the $\frac{N}{2}$th observation.

$\frac{N}{2} = \frac{100}{2} = 50$

We look for the cumulative frequency that is just greater than or equal to $50$. From the table, the CF 45 corresponds to the class 30 - 35, and the CF 78 corresponds to the class 35 - 40. Since 50 is greater than 45, the 50th observation lies in the class 35 - 40.

So, the median class is 35 - 40.

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

where:

• $l$ = lower limit of the median class = $35$
• $N$ = total number of observations = $100$
• $\frac{N}{2} = 50$
• $CF$ = cumulative frequency of the class preceding the median class = $45$ (CF of 30 - 35)
• $f$ = frequency of the median class = $33$ (Frequency of 35 - 40)
• $h$ = size of the median class interval = $40 - 35 = 5$

Substitute these values into the median formula:

Median $= 35 + \left(\frac{50 - 45}{33}\right) \times 5$

Median $= 35 + \left(\frac{5}{33}\right) \times 5$

Median $= 35 + \frac{25}{33}$

Calculating the decimal value of $\frac{25}{33}$:

$\frac{25}{33} \approx 0.757575...$

Median $\approx 35 + 0.7576$ (rounding to four decimal places)

Median $\approx 35.7576$ years.

The median age of the policy holders is approximately $35.76$ years.

Given:

The frequency distribution of the lengths of 40 leaves of a plant.

The data is given in a table with discontinuous class intervals.

---

To Find:

The median length of the leaves.

---

Solution:

The given class intervals are discontinuous (e.g., the upper limit of the first class is 126, and the lower limit of the next class is 127). To find the median of grouped data, the classes must be continuous.

We convert the given discontinuous class intervals into continuous class intervals by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class.

We also calculate the cumulative frequency (CF).

Length (in mm) (Continuous Classes)	Number of leaves (Frequency, $f_i$)	Cumulative Frequency (CF)
117.5 - 126.5	3	3
126.5 - 135.5	5	$3 + 5 = 8$
135.5 - 144.5	9	$8 + 9 = 17$
144.5 - 153.5	12	$17 + 12 = 29$
153.5 - 162.5	5	$29 + 5 = 34$
162.5 - 171.5	4	$34 + 4 = 38$
171.5 - 180.5	2	$38 + 2 = 40$
Total	$\sum f_i = 40$

The total number of leaves is $N = \sum f_i = 40$.

To find the median, we need to find the class containing the $\frac{N}{2}$th observation.

$\frac{N}{2} = \frac{40}{2} = 20$

We look for the cumulative frequency that is just greater than or equal to $20$. From the table, the CF 17 corresponds to the class 135.5 - 144.5, and the CF 29 corresponds to the class 144.5 - 153.5. Since 20 is greater than 17, the 20th observation lies in the class 144.5 - 153.5.

So, the median class is 144.5 - 153.5.

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

where:

• $l$ = lower limit of the median class = $144.5$
• $N$ = total number of observations = $40$
• $\frac{N}{2} = 20$
• $CF$ = cumulative frequency of the class preceding the median class = $17$ (CF of 135.5 - 144.5)
• $f$ = frequency of the median class = $12$ (Frequency of 144.5 - 153.5)
• $h$ = size of the median class interval = $153.5 - 144.5 = 9$

Substitute these values into the median formula:

Median $= 144.5 + \left(\frac{20 - 17}{12}\right) \times 9$

Median $= 144.5 + \left(\frac{3}{12}\right) \times 9$

Median $= 144.5 + \frac{\cancel{3}^{1}}{\cancel{12}_{4}} \times 9$

Median $= 144.5 + \frac{1 \times 9}{4}$

Median $= 144.5 + \frac{9}{4}$

Median $= 144.5 + 2.25$

Median $= 146.75$

The median length of the leaves is $146.75$ mm.

Given:

The distribution of the life time (in hours) of 400 neon lamps and the corresponding number of lamps (frequency).

The data is:

Life time (in hours)	Number of lamps ($f_i$)
1500 - 2000	14
2000 - 2500	56
2500 - 3000	60
3000 - 3500	86
3500 - 4000	74
4000 - 4500	62
4500 - 5000	48

---

To Find:

The median life time of a lamp.

---

Solution:

To find the median, we first construct the cumulative frequency (CF) table.

Life time (in hours)	Number of lamps ($f_i$)	Cumulative Frequency (CF)
1500 - 2000	14	14
2000 - 2500	56	$14 + 56 = 70$
2500 - 3000	60	$70 + 60 = 130$
3000 - 3500	86	$130 + 86 = 216$
3500 - 4000	74	$216 + 74 = 290$
4000 - 4500	62	$290 + 62 = 352$
4500 - 5000	48	$352 + 48 = 400$
Total	$\sum f_i = 400$

The total number of observations is $N = \sum f_i = 400$.

To find the median, we need to find the class containing the $\frac{N}{2}$th observation.

$\frac{N}{2} = \frac{400}{2} = 200$

We look for the cumulative frequency that is just greater than or equal to $200$. From the table, the cumulative frequency just greater than 200 is 216, which corresponds to the class interval 3000 - 3500.

So, the median class is 3000 - 3500.

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

where:

• $l$ = lower limit of the median class = $3000$
• $N$ = total number of observations = $400$
• $\frac{N}{2} = 200$
• $CF$ = cumulative frequency of the class preceding the median class = $130$ (CF of 2500 - 3000)
• $f$ = frequency of the median class = $86$ (Frequency of 3000 - 3500)
• $h$ = size of the median class interval = $3500 - 3000 = 500$

Substitute these values into the median formula:

Median $= 3000 + \left(\frac{200 - 130}{86}\right) \times 500$

Median $= 3000 + \left(\frac{70}{86}\right) \times 500$

Median $= 3000 + \frac{70 \times 500}{86}$

Median $= 3000 + \frac{35000}{86}$

Simplify the fraction $\frac{35000}{86}$ by dividing the numerator and denominator by 2:

Median $= 3000 + \frac{\cancel{35000}^{17500}}{\cancel{86}_{43}}$

Median $= 3000 + \frac{17500}{43}$

Calculating the decimal value:

$\frac{17500}{43} \approx 406.9767$

Median $\approx 3000 + 406.9767$

Median $\approx 3406.98$ (rounded to two decimal places)

The median life time of a lamp is approximately $3406.98$ hours.

Given:

The frequency distribution of the number of letters in English alphabets in 100 surnames.

The data is:

Number of letters	1 - 4	4 - 7	7 - 10	10 - 13	13 - 16	16 - 19
Number of surnames ($f_i$)	6	30	40	16	4	4

Total number of surnames, $N = \sum f_i = 6 + 30 + 40 + 16 + 4 + 4 = 100$.

---

To Find:

The median, mean, and mode of the data.

---

Solution (Median):

To find the median, we first construct the cumulative frequency (CF) table.

Number of letters	Number of surnames ($f_i$)	Cumulative Frequency (CF)
1 - 4	6	6
4 - 7	30	$6 + 30 = 36$
7 - 10	40	$36 + 40 = 76$
10 - 13	16	$76 + 16 = 92$
13 - 16	4	$92 + 4 = 96$
16 - 19	4	$96 + 4 = 100$
Total	$\sum f_i = 100$

We need to find the class containing the $\frac{N}{2}$th observation.

$\frac{N}{2} = \frac{100}{2} = 50$

The cumulative frequency just greater than or equal to 50 is 76, which corresponds to the class interval 7 - 10.

So, the median class is 7 - 10.

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

where:

• $l$ = lower limit of the median class = 7
• $\frac{N}{2} = 50$
• $CF$ = cumulative frequency of the class preceding the median class = 36
• $f$ = frequency of the median class = 40
• $h$ = size of the class interval = $10 - 7 = 3$

Substitute these values into the median formula:

Median $= 7 + \left(\frac{50 - 36}{40}\right) \times 3$

Median $= 7 + \left(\frac{14}{40}\right) \times 3$

Median $= 7 + \frac{\cancel{14}^{7}}{\cancel{40}_{20}} \times 3$

Median $= 7 + \frac{7 \times 3}{20}$

Median $= 7 + \frac{21}{20}$

Median $= 7 + 1.05$

Median $= 8.05$

The median number of letters in the surnames is 8.05.

---

Solution (Mean):

To find the mean, we first calculate the class mark ($x_i$) for each interval.

$x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• For 1 - 4: $x_1 = \frac{1+4}{2} = 2.5$
• For 4 - 7: $x_2 = \frac{4+7}{2} = 5.5$
• For 7 - 10: $x_3 = \frac{7+10}{2} = 8.5$
• For 10 - 13: $x_4 = \frac{10+13}{2} = 11.5$
• For 13 - 16: $x_5 = \frac{13+16}{2} = 14.5$
• For 16 - 19: $x_6 = \frac{16+19}{2} = 17.5$

The class intervals have a constant size $h = 3$ ($4-1=3$, $7-4=3$, etc.). We will use the Step-deviation Method for calculating the mean.

The formula for the mean using the Step-deviation Method is:

$\bar{x} = a + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h$

Let's choose the assumed mean $a = 8.5$ (a central class mark) and the class size $h = 3$.

We prepare a table to calculate the step deviations $u_i = \frac{x_i - 8.5}{3}$ and the product $f_i \times u_i$:

Number of letters	Number of surnames ($f_i$)	Class Mark ($x_i$)	$u_i = \frac{x_i - 8.5}{3}$	$f_i \times u_i$
1 - 4	6	2.5	$\frac{2.5 - 8.5}{3} = -2$	$6 \times (-2) = -12$
4 - 7	30	5.5	$\frac{5.5 - 8.5}{3} = -1$	$30 \times (-1) = -30$
7 - 10	40	8.5	$\frac{8.5 - 8.5}{3} = 0$	$40 \times 0 = 0$
10 - 13	16	11.5	$\frac{11.5 - 8.5}{3} = 1$	$16 \times 1 = 16$
13 - 16	4	14.5	$\frac{14.5 - 8.5}{3} = 2$	$4 \times 2 = 8$
16 - 19	4	17.5	$\frac{17.5 - 8.5}{3} = 3$	$4 \times 3 = 12$
Total	$\sum f_i = 100$			$\sum (f_i \times u_i) = -6$

Sum of frequencies, $\sum f_i = 100$

Sum of the products of frequencies and step deviations, $\sum (f_i \times u_i) = -12 - 30 + 0 + 16 + 8 + 12 = -42 + 36 = -6$

Now, substitute these values into the mean formula:

$\bar{x} = 8.5 + \left(\frac{-6}{100}\right) \times 3$

$\bar{x} = 8.5 - \frac{18}{100}$

$\bar{x} = 8.5 - 0.18$

$\bar{x} = 8.32$

The mean number of letters in the surnames is 8.32.

---

Solution (Mode):

To find the mode of grouped data, we use the formula:

Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

First, we identify the modal class. The modal class is the class with the highest frequency.

From the given table, the highest frequency is $40$, which corresponds to the class interval 7 - 10.

Therefore, the modal class is 7 - 10.

Now, we find the values of $l, h, f_1, f_0,$ and $f_2$ for the modal class:

• $l$ = lower limit of the modal class = 7
• $h$ = size of the class interval = $10 - 7 = 3$
• $f_1$ = frequency of the modal class = 40
• $f_0$ = frequency of the class preceding the modal class = 30
• $f_2$ = frequency of the class succeeding the modal class = 16

Substitute these values into the mode formula:

Mode $= 7 + \left(\frac{40 - 30}{2 \times 40 - 30 - 16}\right) \times 3$

Mode $= 7 + \left(\frac{10}{80 - 46}\right) \times 3$

Mode $= 7 + \left(\frac{10}{34}\right) \times 3$

Mode $= 7 + \frac{10 \times 3}{34}$

Mode $= 7 + \frac{30}{34}$

Mode $= 7 + \frac{15}{17}$

Mode $\approx 7 + 0.8824$ (rounded to four decimal places)

Mode $\approx 7.8824$

The modal size of the surnames is approximately 7.88 letters.

Given:

The distribution of weights of 30 students of a class and the corresponding number of students (frequency).

The data is:

Weight (in kg)	40 - 45	45 - 50	50 - 55	55 - 60	60 - 65	65 - 70	70 - 75
Number of students ($f_i$)	2	3	8	6	6	3	2

Total number of students, $N = \sum f_i = 2 + 3 + 8 + 6 + 6 + 3 + 2 = 30$.

---

To Find:

The median weight of the students.

---

Solution:

To find the median, we first construct the cumulative frequency (CF) table.

Weight (in kg)	Number of students ($f_i$)	Cumulative Frequency (CF)
40 - 45	2	2
45 - 50	3	$2 + 3 = 5$
50 - 55	8	$5 + 8 = 13$
55 - 60	6	$13 + 6 = 19$
60 - 65	6	$19 + 6 = 25$
65 - 70	3	$25 + 3 = 28$
70 - 75	2	$28 + 2 = 30$
Total	$\sum f_i = 30$

We need to find the class containing the $\frac{N}{2}$th observation.

$\frac{N}{2} = \frac{30}{2} = 15$

The cumulative frequency just greater than or equal to 15 is 19, which corresponds to the class interval 55 - 60.

So, the median class is 55 - 60.

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

where:

• $l$ = lower limit of the median class = 55
• $\frac{N}{2} = 15$
• $CF$ = cumulative frequency of the class preceding the median class = 13 (CF of 50 - 55)
• $f$ = frequency of the median class = 6 (Frequency of 55 - 60)
• $h$ = size of the median class interval = $60 - 55 = 5$

Substitute these values into the median formula:

Median $= 55 + \left(\frac{15 - 13}{6}\right) \times 5$

Median $= 55 + \left(\frac{2}{6}\right) \times 5$

Median $= 55 + \frac{\cancel{2}^{1}}{\cancel{6}_{3}} \times 5$

Median $= 55 + \frac{1 \times 5}{3}$

Median $= 55 + \frac{5}{3}$

Median $= 55 + 1.666...$

Median $\approx 56.67$

The median weight of the students is approximately $56.67$ kg.

Given:

The 'more than or equal to' cumulative frequency distribution of annual profits of 30 shops.

Profit (in lakhs of $\textsf{₹}$) More than or equal to	Number of shops (cumulative frequency)
More than or equal to 5	30
More than or equal to 10	28
More than or equal to 15	16
More than or equal to 20	14
More than or equal to 25	10
More than or equal to 30	7
More than or equal to 35	3

Total number of shops, $N = 30$.

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To Find:

The median profit by drawing both ogives.

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Solution:

We are given the 'more than or equal to' cumulative frequency distribution. We first convert this into a simple frequency distribution and a 'less than' cumulative frequency distribution.

The class intervals are formed using the 'More than or equal to' values as the lower limits. The difference between consecutive lower limits is 5, which is the class size. The last implicit class would be 35-40 (or more, but the frequency for >= 40 is 0).

The frequency of a class interval (L-U) is the number of observations $\ge L$ minus the number of observations $\ge U$.

The 'less than' cumulative frequency for an upper limit U is the total frequency minus the 'more than or equal to' frequency for that upper limit.

Profit (in lakhs of $\textsf{₹}$) Class Interval	Number of shops (Frequency, $f_i$)	Cumulative Frequency (Less than type, CF)
5 - 10	$30 - 28 = 2$	2
10 - 15	$28 - 16 = 12$	$2 + 12 = 14$
15 - 20	$16 - 14 = 2$	$14 + 2 = 16$
20 - 25	$14 - 10 = 4$	$16 + 4 = 20$
25 - 30	$10 - 7 = 3$	$20 + 3 = 23$
30 - 35	$7 - 3 = 4$	$23 + 4 = 27$
35 - 40	$3 - 0 = 3$	$27 + 3 = 30$
Total	$\sum f_i = 30$

The total number of observations is $N = 30$. The median position is $\frac{N}{2} = \frac{30}{2} = 15$.

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Drawing the Ogives:

1. 'Less than' Ogive:

Plot the upper limits of the class intervals on the x-axis and the corresponding cumulative frequencies on the y-axis. The points to be plotted are:

(10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27), (40, 30)

Start the ogive at the lower limit of the first class with a cumulative frequency of 0. Since the first class is 5-10, the first point can be (5, 0).

Join these points with a smooth curve.

2. 'More than' Ogive:

Plot the lower limits of the class intervals on the x-axis and the corresponding 'more than or equal to' cumulative frequencies on the y-axis. The points to be plotted are:

(5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7), (35, 3)

Join these points with a smooth curve.

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Obtaining the Median from Ogives:

The median is the x-coordinate of the point where the 'less than' ogive and the 'more than' ogive intersect.

Alternatively, the median can also be found by locating $\frac{N}{2} = 15$ on the y-axis. Draw a horizontal line from this point to intersect the ogive(s). If using only one ogive (say, 'less than' ogive), drop a perpendicular from the intersection point to the x-axis. The point where it meets the x-axis is the median. If using both ogives, the intersection point of the two ogives will have a y-coordinate equal to $\frac{N}{2}$. The x-coordinate of this intersection point is the median.

By drawing the ogives and finding their intersection point, you will find the x-coordinate to be 17.5.

Therefore, the median profit is $\textsf{₹}17.5$ lakhs.

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Verification using Median Formula:

From the frequency distribution table, the median class is 15 - 20 ($\frac{N}{2}=15$ lies in this class's CF range 14-16).

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

• $l$ = 15
• $\frac{N}{2} = 15$
• $CF$ (of preceding class 10-15) = 14
• $f$ (of median class 15-20) = 2
• $h$ = $20 - 15 = 5$

Median $= 15 + \left(\frac{15 - 14}{2}\right) \times 5$

Median $= 15 + \left(\frac{1}{2}\right) \times 5$

Median $= 15 + 0.5 \times 5$

Median $= 15 + 2.5$

Median $= 17.5$

The median profit is $\textsf{₹}17.5$ lakhs, which matches the value obtained from the intersection of the ogives.

Given:

The frequency distribution of the daily income of 50 workers of a factory.

Daily income (in $\textsf{₹}$)	Number of workers ($f_i$)
100 - 120	12
120 - 140	14
140 - 160	8
160 - 180	6
180 - 200	10
Total	$\sum f_i = 50$

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To Find:

The 'less than' type cumulative frequency distribution and draw its ogive.

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Solution:

To convert the given frequency distribution into a 'less than' type cumulative frequency distribution, we consider the upper limit of each class interval and the cumulative frequency up to that upper limit.

Daily income (in $\textsf{₹}$) (Less than)	Number of workers (Cumulative Frequency)
Less than 120	12
Less than 140	$12 + 14 = 26$
Less than 160	$26 + 8 = 34$
Less than 180	$34 + 6 = 40$
Less than 200	$40 + 10 = 50$

The 'less than' cumulative frequency distribution is shown in the table above.

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Drawing the Ogive:

To draw the 'less than' ogive, we plot the points with the upper limits of the class intervals on the x-axis and their corresponding cumulative frequencies on the y-axis.

The points to be plotted are:

(120, 12), (140, 26), (160, 34), (180, 40), (200, 50)

Since the frequency of the first class (100-120) is 12, it means that for values less than 100, the cumulative frequency is 0. So, we start the ogive from the point (100, 0) on the x-axis.

We plot the points (100, 0), (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50) on a graph paper.

We take the upper limits (100, 120, 140, 160, 180, 200) on the x-axis and the cumulative frequencies (0, 12, 26, 34, 40, 50) on the y-axis.

Join these points with a smooth curve. This curve is the 'less than' ogive.

(Note: A graphical representation cannot be provided in this text format. The description above details how to plot the ogive based on the calculated cumulative frequencies).

Given:

The weights of 35 students recorded during a medical check-up, presented as a 'less than' cumulative frequency distribution.

Weight (in kg)	Number of students (Cumulative Frequency)
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

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To Find:

The median weight by drawing a 'less than' ogive and verifying the result using the median formula.

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Solution:

The given distribution is a 'less than' type cumulative frequency distribution. The upper limits of the class intervals are given in the first column, and the cumulative frequencies are given in the second column.

To draw the 'less than' ogive, we plot the points with the upper limits of the class intervals on the x-axis and the corresponding cumulative frequencies on the y-axis.

The points to be plotted are:

(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)

Plot these points on a graph paper. Take the upper limits (38, 40, 42, 44, 46, 48, 50, 52) on the x-axis and the cumulative frequencies (0, 3, 5, 9, 14, 28, 32, 35) on the y-axis. Join these points with a smooth curve. This curve is the 'less than' ogive.

(Note: A graphical representation cannot be provided in this text format. The description above details how to plot the ogive based on the given data).

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Obtaining the Median from the Ogive:

The total number of students is $N = 35$.

The median corresponds to the $\frac{N}{2}$th observation.

$\frac{N}{2} = \frac{35}{2} = 17.5$

To find the median from the 'less than' ogive, locate the value $17.5$ on the y-axis. Draw a horizontal line from $y = 17.5$ to intersect the ogive. From the point of intersection on the ogive, draw a vertical line down to the x-axis. The value on the x-axis where this vertical line meets is the median. By carefully drawing the ogive, this value will be $46.5$.

Thus, the median weight obtained from the graph is approximately $46.5$ kg.

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Verification using the Median Formula:

To verify the result using the formula, we first convert the 'less than' cumulative frequency distribution into a standard frequency distribution table.

Weight (in kg) (Class Interval)	Number of students (Frequency, $f_i$)	Cumulative Frequency (CF)
38 - 40	$3 - 0 = 3$	3
40 - 42	$5 - 3 = 2$	5
42 - 44	$9 - 5 = 4$	9
44 - 46	$14 - 9 = 5$	14
46 - 48	$28 - 14 = 14$	28
48 - 50	$32 - 28 = 4$	32
50 - 52	$35 - 32 = 3$	35
Total	$\sum f_i = 35$

The total number of observations is $N = 35$.

We need to find the class containing the $\frac{N}{2}$th observation, which is $17.5$.

The cumulative frequency just greater than or equal to $17.5$ is $28$, which corresponds to the class interval $46 - 48$.

So, the median class is $46 - 48$.

Now, we use the formula for the median of grouped data:

Median $= l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h$

where:

• $l$ = lower limit of the median class = $46$
• $\frac{N}{2} = 17.5$
• $CF$ = cumulative frequency of the class preceding the median class = $14$ (CF of 44 - 46)
• $f$ = frequency of the median class = $14$ (Frequency of 46 - 48)
• $h$ = size of the median class interval = $48 - 46 = 2$

Substitute these values into the median formula:

Median $= 46 + \left(\frac{17.5 - 14}{14}\right) \times 2$

Median $= 46 + \left(\frac{3.5}{14}\right) \times 2$

Median $= 46 + \frac{3.5 \times \cancel{2}^{1}}{\cancel{14}_{7}}$

Median $= 46 + \frac{3.5}{7}$

Median $= 46 + 0.5$

Median $= 46.5$

The median weight calculated using the formula is $46.5$ kg.

The median obtained from the graph (approximately 46.5 kg) and the median calculated using the formula (46.5 kg) are the same. This verifies the result.

Given:

The frequency distribution of production yield per hectare of wheat of 100 farms.

Production yield (in kg/ha)	Number of farms ($f_i$)
50 - 55	2
55 - 60	8
60 - 65	12
65 - 70	24
70 - 75	38
75 - 80	16
Total	$\sum f_i = 100$

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To Find:

The 'more than type' cumulative frequency distribution and draw its ogive.

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Solution:

To convert the given frequency distribution into a 'more than type' cumulative frequency distribution, we consider the lower limit of each class interval and the cumulative frequency of observations greater than or equal to that lower limit.

The total number of farms is $N = 100$.

Production yield (in kg/ha) (More than or equal to)	Number of farms (Cumulative Frequency)
More than or equal to 50	100
More than or equal to 55	$100 - 2 = 98$
More than or equal to 60	$98 - 8 = 90$
More than or equal to 65	$90 - 12 = 78$
More than or equal to 70	$78 - 24 = 54$
More than or equal to 75	$54 - 38 = 16$
More than or equal to 80	$16 - 16 = 0$

The 'more than type' cumulative frequency distribution is shown in the table above.

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Drawing the Ogive:

To draw the 'more than' ogive, we plot the points with the lower limits of the class intervals on the x-axis and their corresponding 'more than or equal to' cumulative frequencies on the y-axis.

The points to be plotted are:

(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16), (80, 0)

Plot these points on a graph paper. Take the lower limits (50, 55, 60, 65, 70, 75, 80) on the x-axis and the cumulative frequencies (100, 98, 90, 78, 54, 16, 0) on the y-axis.

Join these points with a smooth curve. This curve is the 'more than' ogive.

(Note: A graphical representation cannot be provided in this text format. The description above details how to plot the ogive based on the calculated cumulative frequencies).