Chapter 13 Statistics

Given:

The marks obtained ($x_i$) by 30 students and their corresponding frequencies ($f_i$).

The data is presented in the table:

To Find:

The mean of the marks obtained by the students.

Solution:

We can find the mean using the direct method. The formula for the mean ($\overline{x}$) for ungrouped data with frequencies is given by:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$

We will create a table to calculate $\sum f_i$ and $\sum f_i x_i$.

So, $\sum f_i = 30$ and $\sum f_i x_i = 1779$.

Now, calculate the mean:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1779}{30}$

$\overline{x} = \frac{177.9}{3}$

$\overline{x} = 59.3$

Answer:

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

Given:

The percentage distribution of female teachers in primary schools of rural areas.

The data is presented in the table:

To Find:

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

Solution:

First, we find the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

So, $\sum f_i = 35$.

Method 1: Direct Method

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

$\overline{x} = \frac{1390}{35}$

$\overline{x} = \frac{278}{7} \approx 39.714$

Method 2: Assumed Mean Method

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

Let's take the assumed mean $a$ as the class mark of the middle interval. Let $a = 50$.

$\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 50 + \frac{-360}{35}$

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

$\overline{x} = \frac{50 \times 7 - 72}{7} = \frac{350 - 72}{7} = \frac{278}{7} \approx 39.714$

Method 3: Step-deviation Method

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

The class size $h$ is the difference between the upper and lower limits of any class interval (assuming uniform class size), e.g., $25 - 15 = 10$. So, $h = 10$.

Let's again take the assumed mean $a = 50$.

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

$\overline{x} = 50 - \frac{36 \times 10}{35} = 50 - \frac{360}{35}$

$\overline{x} = 50 - \frac{72}{7} = \frac{350 - 72}{7} = \frac{278}{7} \approx 39.714$

Answer:

The mean percentage of female teachers by all three methods is $\frac{278}{7}$ or approximately $39.71\%$.

Given:

The distribution of the number of wickets taken by bowlers in one-day cricket matches.

The data is presented in the table:

To Find:

1. The mean number of wickets by choosing a suitable method.

2. The meaning of the calculated mean.

Solution:

First, we find the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

So, $\sum f_i = 45$.

The class intervals are of unequal width (40, 40, 50, 100, 100, 100). Therefore, the step-deviation method might not be the most suitable unless we make adjustments or use a large common factor, but the Assumed Mean Method is straightforward.

Let's use the Assumed Mean Method. The formula is $\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is the assumed mean and $d_i = x_i - a$.

Let's choose the assumed mean $a$ as the class mark of the interval with the highest frequency (100-150), so $a = 125$.

$\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 125 + \frac{1255}{45}$

Simplify the fraction $\frac{1255}{45}$ by dividing both by 5:

$\frac{1255}{45} = \frac{251}{9}$

$\overline{x} = 125 + \frac{251}{9}$

$\overline{x} = \frac{125 \times 9 + 251}{9} = \frac{1125 + 251}{9} = \frac{1376}{9}$

Performing the division:

$1376 \div 9 \approx 152.88...$

$\overline{x} \approx 152.89$

Meaning of the Mean:

The mean number of wickets ($\approx 152.89$) represents the average number of wickets taken by a bowler in this distribution of 45 bowlers in one-day cricket matches.

Answer:

The mean number of wickets is $\frac{1376}{9}$ or approximately $152.89$.

The mean signifies the average number of wickets taken per bowler in the given data set.

Given:

A survey regarding the number of plants in 20 houses.

The data is presented in the table:

To Find:

1. The mean number of plants per house.

2. The method used and the reason for choosing it.

Solution:

First, we find the class mark ($x_i$) for each interval: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

So, $\sum f_i = 20$.

The class marks ($x_i$) and frequencies ($f_i$) are relatively small numbers. Therefore, the Direct Method is suitable for calculating the mean as it involves simpler calculations without the need for assumed mean or deviations.

Method Used: Direct Method

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

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20}$

$\overline{x} = \frac{16.2}{2} = 8.1$

Reason for using the Direct Method:

The values of the class marks ($x_i$) and frequencies ($f_i$) are small. Using the direct method avoids the calculation of deviations ($d_i$) or step-deviations ($u_i$), making the calculations simpler and less prone to errors compared to the other methods for this specific data set.

Answer:

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

The method used is the Direct Method because the class marks and frequencies are small, making the calculation $\sum f_i x_i$ easy and efficient.

Given:

The distribution of daily wages of 50 workers of a factory.

The data is presented in the table:

To Find:

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

Solution:

First, we find the class mark ($x_i$) for each interval: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

Since the class intervals are of uniform width ($520 - 500 = 20$), and the class marks will be relatively large, the Step-deviation Method is an appropriate method to use for easier calculation.

The class size is $h = 20$.

We will use the formula for the mean ($\overline{x}$) using the step-deviation method: $\overline{x} = a + (\frac{\sum f_i u_i}{\sum f_i}) \times h$, where $a$ is the assumed mean and $u_i = \frac{x_i - a}{h}$.

Let's choose the assumed mean $a = 550$ (the class mark of the interval 540-560, which is in the middle).

Now, substitute the values into the formula:

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

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

$\overline{x} = 550 + (-0.24) \times 20$

$\overline{x} = 550 - 4.8$

$\overline{x} = 545.2$

Answer:

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

The method used is the Step-deviation Method. It is appropriate because the class intervals are uniform, and it simplifies calculations when the class marks are large numbers.

To find the missing frequency 'f', we will use the direct method for calculating the mean of a grouped frequency distribution.

First, we construct a frequency distribution table to calculate the class mark ($x_i$) for each interval and the product of frequency and class mark ($f_i x_i$).

Now, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($f_i x_i$).

$\sum f_i = 7 + 6 + 9 + 13 + f + 5 + 4$

$\sum f_i x_i = 84 + 84 + 144 + 234 + 20f + 110 + 96$

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

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$

We are given that the mean pocket allowance is $\textsf{₹} 18$. So, $\overline{x} = 18$.

Substitute the values from (i) and (ii) into the mean formula:

Now, we solve this equation for 'f'.

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

Distribute 18 on the left side:

Calculate $18 \times 44$:

$18 \times 44 = 792$

Substitute this value back into equation (iv):

Rearrange the equation to isolate 'f' terms on one side and constant terms on the other:

Perform the subtractions:

Divide both sides by 2 to find the value of 'f':

The missing frequency 'f' is 20.

To find the mean heartbeats per minute, we will use the Assumed Mean Method, as it is suitable for grouped data with relatively large class marks.

First, we construct a frequency distribution table including the class mark ($x_i$) and the deviation ($d_i = x_i - a$), where $a$ is the assumed mean. We also calculate the product of the frequency and deviation ($f_i d_i$).

We choose the assumed mean ($a$) as the class mark of the middle class interval (74 - 77), which is $a = \frac{74+77}{2} = \frac{151}{2} = 75.5$.

Now, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i d_i$).

$\sum f_i = 2 + 4 + 3 + 8 + 7 + 4 + 2$

$\sum f_i d_i = -18 - 24 - 9 + 0 + 21 + 24 + 18$

$\sum f_i d_i = (-18 - 24 - 9) + (21 + 24 + 18)$

$\sum f_i d_i = -51 + 63$

The formula for the mean ($\overline{x}$) using the Assumed Mean Method is:

Substitute the assumed mean ($a = 75.5$) and the values from (i) and (ii) into equation (iii):

Simplify the fraction $\frac{12}{30}$:

$\frac{12}{30} = \frac{\cancel{12}^{2}}{\cancel{30}_{5}} = \frac{2}{5} = 0.4$

Substitute this value back into the mean calculation:

The mean heartbeats per minute for these women is 75.9.

To find the mean number of mangoes per box, we will use the Step-Deviation Method. This method is suitable because the class size is uniform and the class marks are relatively large, making the calculation simpler.

First, we construct a frequency distribution table including the class mark ($x_i$), the deviation ($d_i = x_i - a$), where $a$ is the assumed mean, the step deviation ($u_i = d_i / h$), where $h$ is the class size, and the product of the frequency and step deviation ($f_i u_i$).

The class intervals are not continuous (50-52, 53-55, etc.). However, the class mark is the midpoint of the given interval, calculated as $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$. The difference between consecutive class marks is constant (e.g., $54 - 51 = 3$), which represents the class size $h = 3$.

We choose the assumed mean ($a$) as the class mark of the middle class interval (56 - 58), which is $a = \frac{56+58}{2} = \frac{114}{2} = 57$. The class size is $h = 3$.

Now, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 15 + 110 + 135 + 115 + 25$

$\sum f_i u_i = -30 - 110 + 0 + 115 + 50$

$\sum f_i u_i = (-30 - 110) + (115 + 50)$

$\sum f_i u_i = -140 + 165$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 57$), class size ($h = 3$), and the values from (i) and (ii) into equation (iii):

Simplify the fraction $\frac{25}{400}$:

$\frac{25}{400} = \frac{\cancel{25}^{1}}{\cancel{400}_{16}} = \frac{1}{16}$

Substitute this value back into the mean calculation:

Calculate $\frac{3}{16}$:

$\frac{3}{16} = 0.1875$

Now, add this to 57:

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

The method chosen was the Step-Deviation Method.

To find the mean daily expenditure on food, we will use the Step-Deviation Method. This method is suitable when the class size is uniform, as it simplifies calculations by working with smaller values.

First, we construct a frequency distribution table including the class mark ($x_i$), the deviation ($d_i = x_i - a$), where $a$ is the assumed mean, the step deviation ($u_i = d_i / h$), where $h$ is the class size, and the product of the frequency and step deviation ($f_i u_i$).

The class intervals are continuous, and the class size ($h$) is uniform:

We choose the assumed mean ($a$) as the class mark of the middle class interval (200 - 250), which is:

Now, we create the table:

Next, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 4 + 5 + 12 + 2 + 2$

$\sum f_i u_i = -8 + (-5) + 0 + 2 + 4$

$\sum f_i u_i = -13 + 6$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 225$), class size ($h = 50$), and the values from (iii) and (iv) into equation (v):

Simplify the expression:

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

The method used is the Step-Deviation Method.

To find the mean concentration of $\text{SO}_2$ in the air, we will use the Step-Deviation Method. This method is appropriate as the class intervals are continuous and have a uniform class size.

First, we construct a frequency distribution table to calculate the class mark ($x_i$), the deviation from an assumed mean ($d_i = x_i - a$), the step deviation ($u_i = d_i / h$), and the product of frequency and step deviation ($f_i u_i$).

The class intervals are continuous. The class size ($h$) is the difference between the upper and lower limits of any class interval:

We choose the assumed mean ($a$) as the class mark of one of the middle intervals. Let's choose the class interval 0.08 - 0.12. The class mark is:

Now, we create the table:

Now, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 4 + 9 + 9 + 2 + 4 + 2$

$\sum f_i u_i = -8 + (-9) + 0 + 2 + 8 + 6$

$\sum f_i u_i = (-8 - 9) + (0 + 2 + 8 + 6)$

$\sum f_i u_i = -17 + 16$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 0.10$), class size ($h = 0.04$), and the values from (iii) and (iv) into equation (v):

Simplify the expression:

Calculate $\frac{0.04}{30}$:

$\frac{0.04}{30} = \frac{4/100}{30} = \frac{4}{100 \times 30} = \frac{4}{3000} = \frac{1}{750}$

Now, perform the division $\frac{1}{750}$:

$\frac{1}{750} \approx 0.001333...$

Substitute this value back into the mean calculation:

Rounding to three decimal places (as the data is given up to two decimal places):

The mean concentration of $\text{SO}_2$ in the air is approximately 0.099 ppm.

To find the mean number of days a student was absent, we will use the Assumed Mean Method. This method is suitable as the class intervals are not uniform in size, which makes the Direct Method or Step-Deviation Method with a constant class size less convenient.

First, we construct a frequency distribution table to calculate the class mark ($x_i$) for each interval, the deviation ($d_i = x_i - a$), where $a$ is the assumed mean, and the product of the frequency and deviation ($f_i d_i$).

We choose the assumed mean ($a$) as the class mark of one of the middle class intervals. Let's choose the class interval 14 - 20. The class mark is:

Now, we create the table:

Next, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i d_i$).

$\sum f_i = 11 + 10 + 7 + 4 + 4 + 3 + 1$

$\sum f_i d_i = -154 + (-90) + (-35) + 0 + 28 + 48 + 22$

$\sum f_i d_i = (-154 - 90 - 35) + (28 + 48 + 22)$

$\sum f_i d_i = -279 + 98$

The formula for the mean ($\overline{x}$) using the Assumed Mean Method is:

Substitute the assumed mean ($a = 17$) and the values from (ii) and (iii) into equation (iv):

Simplify the expression:

So, $\frac{181}{40} = 4.525$.

Substitute this value back into the mean calculation:

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

To find the mean literacy rate, we will use the Step-Deviation Method. This method is suitable as the class intervals are continuous and have a uniform class size, which simplifies calculations.

First, we construct a frequency distribution table to calculate the class mark ($x_i$), the deviation from an assumed mean ($d_i = x_i - a$), the step deviation ($u_i = d_i / h$), where $h$ is the class size, and the product of frequency and step deviation ($f_i u_i$).

The class intervals are continuous. The class size ($h$) is the difference between the upper and lower limits of any class interval:

We choose the assumed mean ($a$) as the class mark of the middle class interval (65 - 75). The class mark is:

Now, we create the table:

Now, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 3 + 10 + 11 + 8 + 3$

$\sum f_i u_i = -6 + (-10) + 0 + 8 + 6$

$\sum f_i u_i = (-6 - 10) + (8 + 6)$

$\sum f_i u_i = -16 + 14$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 70$), class size ($h = 10$), and the values from (iii) and (iv) into equation (v):

Simplify the expression:

Simplify the fraction $\frac{20}{35}$:

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

Substitute this value back into the mean calculation:

Calculate the value of $\frac{4}{7}$.

$\frac{4}{7} \approx 0.5714$ (rounded to four decimal places)

Substitute this value back into the mean calculation:

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

The mean literacy rate is approximately 69.43 %.

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

Given Data:

The wickets taken by the bowler in 10 matches are: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3.

To Find:

The mode of the data.

Solution:

To find the mode, we count the frequency of each observation in the data set.

Arranging the data in ascending order can be helpful:

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

Now, let's count the occurrences of each distinct 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 observation that occurs most frequently is 2, as it appears 3 times.

The mode of the data is 2.

To find the mode of grouped data, we first identify the modal class and then use the formula for the mode.

Given Data:

The frequency distribution table for the number of family members in a household is given.

To Find:

The mode of the data.

Solution:

The modal class is the class with the highest frequency. In the given table, the highest frequency is 8, which corresponds to the class interval 3 - 5.

So, the modal class is 3 - 5.

Now, we identify the values required for the mode formula:

Lower limit of the modal class ($l$) = 3

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

Class size ($h$) = Upper limit of modal class - Lower limit of modal class = $5 - 3 = 2$

The formula for the mode of grouped data is:

Substitute the identified values into the formula:

Simplify the expression inside the parentheses:

Convert the fraction $\frac{2}{7}$ to a decimal:

$\frac{2}{7} \approx 0.2857$ (rounded to four decimal places)

Add this decimal to 3:

Rounding to two decimal places, the mode is 3.29.

The mode of the data is approximately 3.29.

To find the mode of the given grouped data, we first need to identify the modal class, which is the class interval with the highest frequency.

Given Data:

The frequency distribution table for the marks of 30 students is given:

To Find:

The mode of the data. Compare and interpret the mode and the mean.

Solution (Mode):

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

Therefore, the modal class is 40 - 55.

Now, we identify the values required for the mode formula:

Lower limit of the modal class ($l$) = 40

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)

Class size ($h$) = Upper limit of modal class - Lower limit of modal class = $55 - 40 = 15$

The formula for the mode of grouped data is:

Substitute the identified values into the formula:

Simplify the expression inside the parentheses:

Solution (Mean):

The mean ($\overline{x}$) can be calculated using the direct method since $\sum f_i$ and $\sum f_i x_i$ are already provided in the table.

The formula for the mean using the direct method is:

Substitute the given sum of $f_i x_i$ and sum of $f_i$ into the formula:

Simplify the expression:

Comparison and Interpretation:

Mode = 52

Mean = 62

The mode (52) indicates that the marks scored by the maximum number of students is 52. This value lies within the modal class (40 - 55).

The mean (62) represents the average marks scored by the students. It gives a central value for the overall performance of the students in the mathematics examination.

Comparing the two, the mean (62) is higher than the mode (52). This suggests that while the most frequent score is around 52, there are some students who scored higher marks, pulling the average score up towards the higher end of the distribution.

The mode of the data is 52.

The mean of the data is 62.

Interpretation: The most frequently occurring mark is 52, while the average mark is 62. The higher mean compared to the mode suggests a skew in the distribution towards higher marks.

To find the mode and the mean of the given data, we will use the appropriate methods for grouped data.

Given Data:

The frequency distribution table showing the ages of patients and their frequencies is provided.

To Find:

The mode and the mean of the data. Compare and interpret the two measures.

Solution (Mode):

To find the mode of grouped data, we first identify the modal class.

The modal class is the class with the highest frequency. From the table, the highest frequency is 23, which corresponds to the class interval 35 - 45.

Thus, the modal class is 35 - 45.

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

Where:

• $l$ is the lower limit of the modal class = 35
• $f_1$ is the frequency of the modal class = 23
• $f_0$ is the frequency of the class preceding the modal class = 21 (frequency of 25 - 35)
• $f_2$ is the frequency of the class succeeding the modal class = 14 (frequency of 45 - 55)
• $h$ is the class size = $45 - 35 = 10$

Substitute these values into the formula (i):

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

Simplify the expression:

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

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

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

So, $\frac{20}{11} \approx 1.8181...$

$\text{Mode} \approx 35 + 1.8181$

$\text{Mode} \approx 36.8181$

Rounding to one decimal place as per the expected answer, the mode is 36.8 years.

Solution (Mean):

We will use the Step-Deviation Method to find the mean, as the class size is uniform.

First, we construct a frequency distribution table including the class mark ($x_i$), deviation ($d_i = x_i - a$), step deviation ($u_i = d_i / h$), and the product $f_i u_i$.

The class size ($h$) = $15 - 5 = 10$.

We choose the assumed mean ($a$) as the class mark of the interval 35 - 45, which is $a = \frac{35+45}{2} = 40$.

Now, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 80$

$\sum f_i u_i = -18 - 22 - 21 + 0 + 14 + 10 = -61 + 24 = -37$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 40$), class size ($h = 10$), and the calculated sums into equation (ii):

$\overline{x} = 40 + 10 \left(\frac{-37}{80}\right)$

Simplify the expression:

$\overline{x} = 40 + \frac{10 \times (-37)}{80}$

$\overline{x} = 40 + \frac{-370}{80}$

$\overline{x} = 40 - \frac{37}{8}$

So, $\frac{37}{8} = 4.625$.

$\overline{x} = 40 - 4.625$

$\overline{x} = 35.375$

Rounding or approximating to two decimal places as per the expected answer, the mean is 35.37 years.

Comparison and Interpretation:

Mode $\approx 36.8$ years

Mean $\approx 35.37$ years

The mode (36.8 years) represents the age around which the maximum number of patients are admitted to the hospital. This is the most frequent age (or age group) among the admitted patients. Approximately 36.8 years is the age that occurred most often among the patients.

The mean (35.37 years) represents the average age of patients admitted to the hospital during the year. It gives a central value for the age distribution of the patients. On average, a patient admitted to the hospital is 35.37 years old.

Comparing the two measures, the mode (36.8) is slightly higher than the mean (35.37). This indicates that while the most frequent age group is 35-45, the distribution of ages is slightly skewed towards younger ages, pulling the mean downwards compared to the mode. Both measures give a central value for the ages, but the mode indicates the most common age, while the mean indicates the average age.

The mode of the data is approximately 36.8 years.

The mean of the data is approximately 35.37 years.

Interpretation: The maximum number of patients admitted in the hospital are of the age 36.8 years (approx.), while on an average the age of a patient admitted to the hospital is 35.37 years.

To determine the modal lifetimes of the components, we need to find the mode of the given grouped data.

Given Data:

The frequency distribution table for the lifetimes of 225 electrical components is provided.

To Find:

The modal lifetimes of the components (i.e., the mode of the data).

Solution:

To find the mode of grouped data, we first identify the modal class.

The modal class is the class with the highest frequency. From the table, the highest frequency is 61, which corresponds to the class interval 60 - 80.

Thus, the modal class is 60 - 80.

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

Where:

• $l$ is the lower limit of the modal class = 60
• $f_1$ is the frequency of the modal class = 61
• $f_0$ is the frequency of the class preceding the modal class = 52 (frequency of 40 - 60)
• $f_2$ is the frequency of the class succeeding the modal class = 38 (frequency of 80 - 100)
• $h$ is the class size = $80 - 60 = 20$

Substitute these values into the formula (i):

Simplify the expression:

Simplify the fraction $\frac{180}{32}$ by dividing the numerator and denominator by their greatest common divisor, which is 4:

$\frac{180}{32} = \frac{\cancel{180}^{45}}{\cancel{32}_{8}} = \frac{45}{8}$

Substitute the simplified fraction back into the mode calculation:

So, $\frac{45}{8} = 5.625$.

Substitute this value back into the mode calculation:

The modal lifetimes of the components is 65.625 hours.

To find the modal and mean monthly household expenditure, we will use the appropriate methods for grouped data.

Given Data:

The frequency distribution table for the monthly household expenditure of 200 families is provided.

To Find:

The modal monthly expenditure and the mean monthly expenditure.

Solution (Mode):

To find the mode of grouped data, we first identify the modal class.

The modal class is the class with the highest frequency. From the table, the highest frequency is 40, which corresponds to the class interval 1500 - 2000.

Thus, the modal class is 1500 - 2000.

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

Where:

• $l$ is the lower limit of the modal class = 1500
• $f_1$ is the frequency of the modal class = 40
• $f_0$ is the frequency of the class preceding the modal class = 24 (frequency of 1000 - 1500)
• $f_2$ is the frequency of the class succeeding the modal class = 33 (frequency of 2000 - 2500)
• $h$ is the class size = $2000 - 1500 = 500$

Substitute these values into the formula (i):

Simplify the expression:

Convert the fraction $\frac{8000}{23}$ to a decimal:

$\frac{8000}{23} \approx 347.82608...$

Substitute this value back into the mode calculation:

Rounding to two decimal places, the modal monthly expenditure is $\textsf{₹} 1847.83$.

Solution (Mean):

We will use the Step-Deviation Method to find the mean, as the class size is uniform and the expenditure values are relatively large.

First, we construct a frequency distribution table including the class mark ($x_i$), deviation ($d_i = x_i - a$), step deviation ($u_i = d_i / h$), and the product $f_i u_i$.

The class size ($h$) = $1500 - 1000 = 500$.

We choose the assumed mean ($a$) as the class mark of one of the middle intervals. Let's choose the class interval 2500 - 3000. The class mark is:

Now, we create the table:

Next, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 24 + 40 + 33 + 28 + 30 + 22 + 16 + 7$

$\sum f_i u_i = -72 + (-80) + (-33) + 0 + 30 + 44 + 48 + 28$

$\sum f_i u_i = (-72 - 80 - 33) + (30 + 44 + 48 + 28)$

$\sum f_i u_i = -185 + 150$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 2750$), class size ($h = 500$), and the values from (ix) and (x) into equation (xi):

Simplify the expression:

Calculate $\frac{175}{2} = 87.5$.

So, the mean monthly expenditure is $\textsf{₹} 2662.50$.

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

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

To find the mode and the mean of the given data, we will use the appropriate methods for grouped data.

Given Data:

The frequency distribution table showing the state-wise teacher-student ratio is provided.

To Find:

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

Solution (Mode):

To find the mode of grouped data, we first identify the modal class.

The modal class is the class with the highest frequency. From the table, the highest frequency is 10, which corresponds to the class interval 30 - 35.

Thus, the modal class is 30 - 35.

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

Where:

• $l$ is the lower limit of the modal class = 30
• $f_1$ is the frequency of the modal class = 10
• $f_0$ is the frequency of the class preceding the modal class = 9 (frequency of 25 - 30)
• $f_2$ is the frequency of the class succeeding the modal class = 3 (frequency of 35 - 40)
• $h$ is the class size = $35 - 30 = 5$

Substitute these values into the formula (i):

Simplify the expression:

Convert the fraction $\frac{5}{8}$ to a decimal:

$\frac{5}{8} = 0.625$

Substitute this value back into the mode calculation:

Solution (Mean):

We will use the Step-Deviation Method to find the mean, as the class size is uniform.

First, we construct a frequency distribution table including the class mark ($x_i$), deviation ($d_i = x_i - a$), step deviation ($u_i = d_i / h$), and the product $f_i u_i$.

The class size ($h$) = $20 - 15 = 5$.

We choose the assumed mean ($a$) as the class mark of the class interval 30 - 35. The class mark is:

Now, we create the table:

Next, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 3 + 8 + 9 + 10 + 3 + 0 + 0 + 2$

$\sum f_i u_i = -9 + (-16) + (-9) + 0 + 3 + 0 + 0 + 8$

$\sum f_i u_i = (-9 - 16 - 9) + (3 + 8)$

$\sum f_i u_i = -34 + 11$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 32.5$), class size ($h = 5$), and the values from (viii) and (ix) into equation (x):

Simplify the expression:

Convert the fraction $\frac{23}{7}$ to a decimal:

$\frac{23}{7} \approx 3.2857$ (rounded to four decimal places)

Substitute this value back into the mean calculation:

Rounding to two decimal places, the mean is 29.21.

Comparison and Interpretation:

Mode $= 30.625$

Mean $\approx 29.2143$

The mode (30.625) indicates that the most frequent teacher-student ratio among the states/UTs is approximately 30.6 students per teacher. This value falls within the modal class (30 - 35), which has the highest number of states/UTs.

The mean (approximately 29.21) represents the average teacher-student ratio across all 35 states/UTs. It provides a central value for the distribution of the ratio.

Comparing the two measures, the mode (30.625) is slightly higher than the mean (29.2143). This suggests that the distribution of the teacher-student ratio is slightly skewed towards lower values (negatively skewed), as the presence of states with very high ratios (like 50 - 55) is balanced by a greater concentration of states with ratios in the 20s and early 30s.

The mode of the data is 30.625.

The mean of the data is approximately 29.21.

Interpretation: The most common teacher-student ratio is around 30.6 students per teacher. The average teacher-student ratio across the states/UTs is approximately 29.2 students per teacher. The mean being slightly lower than the mode suggests that there is a slight tendency towards lower teacher-student ratios overall, despite the most frequently observed ratio being a bit higher.

To find the mode of the given distribution, we need to find the modal class and then use the formula for the mode of grouped data.

Given Data:

The frequency distribution table for the runs scored by batsmen is provided.

To Find:

The mode of the data (modal runs scored).

Solution:

The modal class is the class interval with the highest frequency. From the table, the highest frequency is 18, which corresponds to the class interval 4000 - 5000.

Thus, the modal class is 4000 - 5000.

Now, we identify the values required for the mode formula:

• $l$ is the lower limit of the modal class = 4000
• $f_1$ is the frequency of the modal class = 18
• $f_0$ is the frequency of the class preceding the modal class = 4 (frequency of 3000 - 4000)
• $f_2$ is the frequency of the class succeeding the modal class = 9 (frequency of 5000 - 6000)
• $h$ is the class size = $5000 - 4000 = 1000$

The formula for the mode of grouped data is:

Substitute these values into the formula (i):

Simplify the expression:

Convert the fraction $\frac{14000}{23}$ to a decimal:

$\frac{14000}{23} \approx 608.69565...$

Substitute this value back into the mode calculation:

Rounding to two decimal places, the modal runs scored is 4608.70.

The mode of the data (modal runs scored) is approximately 4608.70.

To find the mode of the given data, we need to find the modal class and then use the formula for the mode of grouped data.

Given Data:

The frequency distribution table showing the number of cars passing through a spot is provided.

To Find:

The mode of the data.

Solution:

The modal class is the class interval with the highest frequency. From the table, the highest frequency is 20, which corresponds to the class interval 40 - 50.

Thus, the modal class is 40 - 50.

Now, we identify the values required for the mode formula:

• $l$ is the lower limit of the modal class = 40
• $f_1$ is the frequency of the modal class = 20
• $f_0$ is the frequency of the class preceding the modal class = 12 (frequency of 30 - 40)
• $f_2$ is the frequency of the class succeeding the modal class = 11 (frequency of 50 - 60)
• $h$ is the class size = $50 - 40 = 10$

The formula for the mode of grouped data is:

Substitute these values into the formula (i):

Simplify the expression inside the parentheses:

Convert the fraction $\frac{80}{17}$ to a decimal:

$\frac{80}{17} \approx 4.70588...$

Substitute this value back into the mode calculation:

Rounding to two decimal places, the mode is 44.71.

The mode of the data is approximately 44.71.

The given data is a cumulative frequency distribution of the 'less than' type. To find the median, we first convert this into a regular frequency distribution table.

Given Data:

Cumulative frequency distribution table for the heights of 51 girls.

To Find:

The median height.

Solution:

First, we construct the frequency distribution table with class intervals and their frequencies ($f_i$). We also include the cumulative frequency ($cf$) column for reference.

The upper limits of the class intervals are given. The class size ($h$) is uniform: $145 - 140 = 5$, $150 - 145 = 5$, etc. So, $h = 5$.

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

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

We need to find the value corresponding to $\frac{N}{2}$.

The median class is the class whose cumulative frequency is greater than or equal to 25.5 for the first time. From the cumulative frequency column, 29 is the first value greater than or equal to 25.5, and it corresponds to the class interval 145 - 150.

So, the median class is 145 - 150.

Now, we identify the values required for the median formula:

• $l$ is the lower limit of the median class = 145
• $N$ is the total number of observations = 51
• $cf$ is the cumulative frequency of the class preceding the median class = 11 (cumulative frequency of 140 - 145)
• $f$ is the frequency of the median class = 18 (frequency of 145 - 150)
• $h$ is the class size = 5

The formula for the median of grouped data is:

Substitute the identified values into the formula (ii):

Simplify the expression inside the parentheses:

Convert the fraction $\frac{72.5}{18}$ to a decimal:

$\frac{72.5}{18} = \frac{725}{180} = \frac{\cancel{725}^{145}}{\cancel{180}_{36}} = \frac{145}{36}$

So, $\frac{72.5}{18} \approx 4.0277...$

Substitute this value back into the median calculation:

Rounding to two decimal places, the median height is 149.03 cm.

The median height is approximately 149.03 cm.

To find the missing frequencies x and y, we will first construct the cumulative frequency table and use the given information about the total frequency and the median.

Given:

Frequency distribution table with missing frequencies x and y.

Total frequency ($N$) = 100

Median = 525

To Find:

The values of x and y.

Solution:

First, let's create a table with cumulative frequencies.

We are given that the total frequency is 100. The sum of all frequencies in the table must be equal to the total frequency.

We are given that the median is 525. Since 525 lies in the class interval 500 - 600, this is the median class.

For the median class (500 - 600):

• Lower limit ($l$) = 500
• Frequency ($f$) = 20
• Cumulative frequency of the class preceding the median class ($cf$) = $36 + x$ (cumulative frequency of 400 - 500)
• Class size ($h$) = $600 - 500 = 100$

The total frequency is $N = 100$, so $\frac{N}{2} = \frac{100}{2} = 50$.

The formula for the median of grouped data is:

Substitute the known values into the median formula:

Solve the equation for x:

Now substitute the value of x (9) into equation (iii) to find y:

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

To find the median, mean, and mode of the given data, we will use the appropriate methods for grouped frequency distributions.

Given Data:

The frequency distribution table showing the monthly consumption of electricity of 68 consumers is provided.

To Find:

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

Solution (Mode):

To find the mode of grouped data, we first identify the modal class.

The modal class is the class with the highest frequency. From the table, the highest frequency is 20, which corresponds to the class interval 125 - 145.

Thus, the modal class is 125 - 145.

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

Where:

• $l$ is the lower limit of the modal class = 125
• $f_1$ is the frequency of the modal class = 20
• $f_0$ is the frequency of the class preceding the modal class = 13 (frequency of 105 - 125)
• $f_2$ is the frequency of the class succeeding the modal class = 14 (frequency of 145 - 165)
• $h$ is the class size = $145 - 125 = 20$

Substitute these values into the formula (i):

Simplify the expression:

Convert the fraction $\frac{140}{13}$ to a decimal:

$\frac{140}{13} \approx 10.7692...$

Substitute this value back into the mode calculation:

Rounding to two decimal places, the mode is 135.77 units.

Solution (Median):

To find the median, we first need to calculate the cumulative frequencies.

The total number of observations is $N = \sum f_i = 68$.

We need to find the value corresponding to $\frac{N}{2}$.

The median class is the class whose cumulative frequency is greater than or equal to 34 for the first time. From the cumulative frequency column, 42 is the first value greater than or equal to 34, and it corresponds to the class interval 125 - 145.

So, the median class is 125 - 145.

Now, we identify the values required for the median formula:

• $l$ is the lower limit of the median class = 125
• $N$ is the total number of observations = 68
• $cf$ is the cumulative frequency of the class preceding the median class = 22 (cumulative frequency of 105 - 125)
• $f$ is the frequency of the median class = 20 (frequency of 125 - 145)
• $h$ is the class size = $145 - 125 = 20$

The formula for the median of grouped data is:

Substitute the identified values into the formula (viii):

Simplify the expression inside the parentheses:

The median is 137 units.

Solution (Mean):

We will use the Step-Deviation Method to find the mean, as the class size is uniform ($h = 20$).

First, we construct a frequency distribution table including the class mark ($x_i$), deviation ($d_i = x_i - a$), step deviation ($u_i = d_i / h$), and the product $f_i u_i$.

We choose the assumed mean ($a$) as the class mark of the median class (125 - 145), which is $a = \frac{125+145}{2} = \frac{270}{2} = 135$.

Now, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 4 + 5 + 13 + 20 + 14 + 8 + 4 = 68$ (Given)

$\sum f_i u_i = -12 + (-10) + (-13) + 0 + 14 + 16 + 12$

$\sum f_i u_i = (-12 - 10 - 13) + (14 + 16 + 12)$

$\sum f_i u_i = -35 + 42$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 135$), class size ($h = 20$), and the values into equation (xiv):

Simplify the expression:

Simplify the fraction $\frac{140}{68}$:

$\frac{140}{68} = \frac{\cancel{140}^{35}}{\cancel{68}_{17}} = \frac{35}{17}$

Substitute the simplified fraction back into the mean calculation:

Convert the fraction $\frac{35}{17}$ to a decimal:

$\frac{35}{17} \approx 2.05882...$

Substitute this value back into the mean calculation:

Rounding to two decimal places, the mean is 137.06 units.

Comparison and Interpretation:

Mode $\approx 135.77$ units

Median $= 137$ units

Mean $\approx 137.06$ units

The mode (135.77 units) indicates the electricity consumption value that occurs most frequently among the consumers. It falls within the modal class (125 - 145), which has the highest number of consumers.

The median (137 units) divides the distribution into two equal halves; 50% of the consumers consume less than 137 units, and 50% consume more than 137 units.

The mean (137.06 units) represents the average monthly electricity consumption per consumer.

Comparing the three measures (Mode $\approx$ 135.77, Median = 137, Mean $\approx$ 137.06), we observe that they are very close to each other. This suggests that the distribution of monthly electricity consumption is almost symmetric. The peak consumption is around 135.77 units, the midpoint consumption is 137 units, and the average consumption is about 137.06 units. The small difference between the mean and median (Mean > Median) indicates a very slight positive skew in the data, meaning there might be a few consumers with relatively higher consumption pulling the mean slightly above the median and mode.

The mode of the data is approximately 135.77 units.

The median of the data is 137 units.

The mean of the data is approximately 137.06 units.

Interpretation: The three measures of central tendency (mean, median, and mode) are very close, suggesting that the distribution of electricity consumption is nearly symmetric. The most frequent consumption value is around 135.77 units, the median consumption is 137 units, and the average consumption is about 137.06 units.

To find the missing frequencies x and y, we will first construct the cumulative frequency table and use the given information about the total frequency and the median.

Given:

Frequency distribution table with missing frequencies x and y.

Total frequency ($N$) = 60

Median = 28.5

To Find:

The values of x and y.

Solution:

First, let's create a table with cumulative frequencies ($cf$).

We are given that the total frequency is 60. The sum of all frequencies in the table must be equal to the total frequency, or the last value in the cumulative frequency column must be 60.

We are given that the median is 28.5. Since 28.5 lies in the class interval 20 - 30, this is the median class.

For the median class (20 - 30):

• Lower limit ($l$) = 20
• Frequency ($f$) = 20 (frequency of the median class)
• Cumulative frequency of the class preceding the median class ($cf$) = $5 + x$ (cumulative frequency of 10 - 20)
• Class size ($h$) = $30 - 20 = 10$

The total frequency is $N = 60$, so $\frac{N}{2} = \frac{60}{2} = 30$.

The formula for the median of grouped data is:

Substitute the known values into the median formula:

Solve the equation for x:

Now substitute the value of x (8) into equation (ii) to find y:

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

The given data is a cumulative frequency distribution of the 'below' type. To find the median age, we first convert this into a regular frequency distribution table.

Given Data:

Cumulative frequency distribution table for the ages of 100 policy holders.

Total number of policy holders = 100.

Policies are given to persons aged 18 years onwards but less than 60 years.

To Find:

The median age.

Solution:

First, we construct the frequency distribution table with class intervals and their frequencies ($f_i$). The class intervals are formed based on the 'below' values, and the frequency of each class is the difference between consecutive cumulative frequencies.

Assuming the policies start effectively from age 18 and considering the 'Below 20' as the first group from 18 up to 20 (or the lower limit of the previous interval), and the class width appears to be consistent at 5 years for most intervals (25-20, 30-25, etc.), we form the following classes:

The total number of observations is $N = \sum f_i = 100$.

We need to find the value corresponding to $\frac{N}{2}$.

The median class is the class whose cumulative frequency is greater than or equal to 50 for the first time. From the cumulative frequency column in our constructed table, 78 is the first value greater than or equal to 50, and it corresponds to the class interval 35 - 40.

So, the median class is 35 - 40.

Now, we identify the values required for the median formula:

• $l$ is the lower limit of the median class = 35
• $N$ is the total number of observations = 100
• $cf$ is the cumulative frequency of the class preceding the median class = 45 (cumulative frequency of 30 - 35)
• $f$ is the frequency of the median class = 33 (frequency of 35 - 40)
• $h$ is the class size = $40 - 35 = 5$

The formula for the median of grouped data is:

Substitute the identified values into the formula (ii):

Simplify the expression inside the parentheses:

Convert the fraction $\frac{25}{33}$ to a decimal:

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

Substitute this value back into the median calculation:

Rounding to two decimal places, the median age is 35.76 years.

The median age is approximately 35.76 years.

The given class intervals are discontinuous. To find the median, we first need to convert them into continuous classes. This is done by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

Given Data:

The frequency distribution table for the lengths of 40 leaves.

To Find:

The median length of the leaves.

Solution:

We convert the given distribution into continuous classes and calculate the cumulative frequencies ($cf$).

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

We need to find the value corresponding to $\frac{N}{2}$.

The median class is the class whose cumulative frequency is greater than or equal to 20 for the first time. From the cumulative frequency column, 29 is the first value greater than or equal to 20, and it corresponds to the class interval 144.5 - 153.5.

So, the median class is 144.5 - 153.5.

Now, we identify the values required for the median formula:

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

The formula for the median of grouped data is:

Substitute the identified values into the formula (ii):

Simplify the expression inside the parentheses:

Convert the fraction $\frac{9}{4}$ to a decimal:

$\frac{9}{4} = 2.25$

Substitute this value back into the median calculation:

The median length of the leaves is 146.75 mm.

To find the median life time of a lamp, we first construct the cumulative frequency distribution table.

Given Data:

The frequency distribution table for the life time of 400 neon lamps.

To Find:

The median life time of a lamp.

Solution:

We calculate the cumulative frequencies ($cf$).

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

We need to find the value corresponding to $\frac{N}{2}$.

The median class is the class whose cumulative frequency is greater than or equal to 200 for the first time. From the cumulative frequency column, 216 is the first value greater than or equal to 200, and it corresponds to the class interval 3000 - 3500.

So, the median class is 3000 - 3500.

Now, we identify the values required for the median formula:

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

The formula for the median of grouped data is:

Substitute the identified values into the formula (ii):

Simplify the expression inside the parentheses:

Simplify the fraction $\frac{35000}{86}$ by dividing the numerator and denominator by their greatest common divisor, which is 2:

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

Substitute the simplified fraction back into the median calculation:

Convert the fraction $\frac{17500}{43}$ to a decimal:

$\frac{17500}{43} \approx 406.97674...$

Substitute this value back into the median calculation:

Rounding to two decimal places, the median life time is 3406.98 hours.

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

To find the median, mean, and mode of the given data, we will use the appropriate methods for grouped frequency distributions.

Given Data:

The frequency distribution table showing the number of letters in surnames and their frequencies is provided. The total number of surnames is 100.

To Find:

The median number of letters, the mean number of letters, and the modal size of the surnames.

Solution (Mode):

To find the mode of grouped data, we first identify the modal class, which is the class interval with the highest frequency.

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

Thus, the modal class is 7 - 10.

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

Where:

• $l$ is the lower limit of the modal class = 7
• $f_1$ is the frequency of the modal class = 40
• $f_0$ is the frequency of the class preceding the modal class = 30 (frequency of 4 - 7)
• $f_2$ is the frequency of the class succeeding the modal class = 16 (frequency of 10 - 13)
• $h$ is the class size. The class marks are calculated as $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$. For 1-4, $x_i = 2.5$. For 4-7, $x_i = 5.5$. The difference between consecutive class marks is $5.5 - 2.5 = 3$. This is the effective class size $h$. $h = 3$.

Substitute these values into the formula (i):

Simplify the expression:

Convert the fraction $\frac{15}{17}$ to a decimal:

$\frac{15}{17} \approx 0.88235$

Substitute this value back into the mode calculation:

Rounding to two decimal places, the modal size is 7.88 letters.

Solution (Median):

To find the median, we first need to calculate the cumulative frequencies ($cf$).

The total number of observations is $N = \sum f_i = 100$.

We need to find the value corresponding to $\frac{N}{2}$.

The median class is the class whose cumulative frequency is greater than or equal to 50 for the first time. From the cumulative frequency column, 76 is the first value greater than or equal to 50, and it corresponds to the class interval 7 - 10.

So, the median class is 7 - 10.

Now, we identify the values required for the median formula:

• $l$ is the lower limit of the median class = 7
• $N$ is the total number of observations = 100
• $\frac{N}{2} = 50$
• $cf$ is the cumulative frequency of the class preceding the median class = 36 (cumulative frequency of 4 - 7)
• $f$ is the frequency of the median class = 40 (frequency of 7 - 10)
• $h$ is the class size = 3

The formula for the median of grouped data is:

Substitute the identified values into the formula (ix):

Simplify the expression inside the parentheses:

Convert the fraction $\frac{21}{20}$ to a decimal:

$\frac{21}{20} = 1.05$

Substitute this value back into the median calculation:

The median number of letters is 8.05.

Solution (Mean):

We will use the Step-Deviation Method to find the mean, as the class size is uniform ($h = 3$) and the numbers are relatively small.

First, we construct a frequency distribution table including the class mark ($x_i$), deviation ($d_i = x_i - a$), step deviation ($u_i = d_i / h$), and the product $f_i u_i$.

The class size ($h$) = 3.

The class marks ($x_i$) are: $\frac{1+4}{2}=2.5$, $\frac{4+7}{2}=5.5$, $\frac{7+10}{2}=8.5$, $\frac{10+13}{2}=11.5$, $\frac{13+16}{2}=14.5$, $\frac{16+19}{2}=17.5$.

We choose the assumed mean ($a$) as the class mark of the class interval 7 - 10, which is $a = 8.5$.

Now, we calculate the sum of frequencies ($\sum f_i$) and the sum of the products ($\sum f_i u_i$).

$\sum f_i = 6 + 30 + 40 + 16 + 4 + 4 = 100$ (Given)

$\sum f_i u_i = -12 + (-30) + 0 + 16 + 8 + 12$

$\sum f_i u_i = (-12 - 30) + (16 + 8 + 12)$

$\sum f_i u_i = -42 + 36$

The formula for the mean ($\overline{x}$) using the Step-Deviation Method is:

Substitute the assumed mean ($a = 8.5$), class size ($h = 3$), and the values into equation (xvii):

Simplify the expression:

The mean number of letters is 8.32.

The median number of letters in the surnames is 8.05.

The mean number of letters in the surnames is 8.32.

The modal size of the surnames is approximately 7.88.

To find the median weight of the students, we first construct the cumulative frequency distribution table.

Given Data:

The frequency distribution table for the weights of 30 students.

To Find:

The median weight of the students.

Solution:

We calculate the cumulative frequencies ($cf$).

The total number of students (observations) is $N = \sum f_i = 30$.

We need to find the value corresponding to $\frac{N}{2}$.

The median class is the class whose cumulative frequency is greater than or equal to 15 for the first time. From the cumulative frequency column, 19 is the first value greater than or equal to 15, and it corresponds to the class interval 55 - 60.

So, the median class is 55 - 60.

Now, we identify the values required for the median formula:

• $l$ is the lower limit of the median class = 55
• $N$ is the total number of observations = 30
• $\frac{N}{2} = 15$
• $cf$ is the cumulative frequency of the class preceding the median class = 13 (cumulative frequency of 50 - 55)
• $f$ is the frequency of the median class = 6 (frequency of 55 - 60)
• $h$ is the class size = $60 - 55 = 5$

The formula for the median of grouped data is:

Substitute the identified values into the formula (ii):

Simplify the expression inside the parentheses:

Convert the fraction $\frac{5}{3}$ to a decimal:

$\frac{5}{3} \approx 1.6667$ (rounded to four decimal places)

Substitute this value back into the median calculation:

Rounding to two decimal places, the median weight is 56.67 kg.

The median weight of the students is approximately 56.67 kg.