Chapter 12 Statistics

Given:

A bar graph showing the number of students born in different months of a year for a section of Class IX.

To Find:

(i) Number of students born in the month of November.

(ii) The month in which the maximum number of students were born.

Solution:

By observing the given bar graph, we can conclude the following:

(i) Number of students born in November:

On the horizontal axis (x-axis), locate the month 'November'. The height of the bar corresponding to November is 4 units on the vertical axis.

(ii) Month with maximum births:

The maximum number of students born will correspond to the tallest bar in the graph. By looking at the graph, the bar for the month of 'August' is the tallest, reaching a height of 6 units.

Given:

Monthly income of the family = $\textsf{₹} 20,000$.

Expenditure data (in thousand rupees) under various heads: Grocery (4), Rent (5), Education (5), Medicine (2), Fuel (2), Entertainment (1), and Miscellaneous (1).

To Draw:

A bar graph representing the monthly expenditure.

Solution:

To represent the data in a bar graph, we follow these steps:

1. We represent the 'Heads' on the horizontal axis and the 'Expenditure' on the vertical axis.

2. We choose a suitable scale for the vertical axis. Let 1 unit length = $\textsf{₹} 1000$ (or 1 thousand rupees).

3. We draw bars of equal width and equal gaps, with heights proportional to the expenditure.

Given:

Frequency distribution table of marks with varying class interval sizes.

To Draw:

A histogram for the given distribution.

Solution:

Since the class intervals are of unequal widths, we must adjust the frequencies so that the area of the rectangles is proportional to the frequency. We calculate the adjusted frequency using the formula:

The minimum class width here is 10 (from the intervals like 20 - 30).

Given:

Marks of 51 students grouped in class intervals of 10.

To Draw:

A frequency polygon.

Solution:

To draw a frequency polygon, we first calculate the Class Mark for each interval.

We also include two imaginary classes with frequency 0: one before the first class ($-10 - 0$) and one after the last class ($100 - 110$).

Given:

Cost of living index data for 52 weeks.

To Draw:

A frequency polygon.

Solution:

We calculate the class marks for the given intervals. We add an imaginary class before (130 - 140) and after (200 - 210) with 0 frequency to close the polygon.

Plotting the points $(135, 0), (145, 5), (155, 10), $$ \ (165, 20), $$ \ (175, 9), $$ \ (185, 6), $$ \ (195, 2), $$ \ (205, 0)$ and joining them with straight lines gives the required frequency polygon.

Given:

The percentage of female fatality rates worldwide for women aged 15–44 due to various causes.

To Find/Do:

(i) Graphical representation of the data.

(ii) Identify the major cause of ill health and death.

(iii) Factors playing a major role in the major cause.

Solution:

(i) Graphical Representation:

The given data can be represented using a Bar Graph. We represent 'Causes' on the x-axis and 'Female Fatality Rate (%)' on the y-axis. Since the data is categorical, we draw bars of equal width with equal gaps between them.

(ii) Major Cause:

From the table and graph, it is observed that Reproductive health conditions have the highest fatality rate of $31.8\%$. Thus, it is the major cause of women's ill health and death worldwide.

(iii) Factors for Major Cause:

Two major factors contributing to reproductive health issues being the primary cause are:

1. Lack of awareness and education: In many regions, women are not educated about reproductive health and hygiene.

2. Inadequate medical facilities: Poor access to quality healthcare, prenatal care, and safe delivery services increases the risk of fatality.

Given:

Number of girls per thousand boys in different sections of Indian society.

To Do:

(i) Represent the information by a bar graph.

(ii) Discuss conclusions from the graph.

Solution:

(i) Bar Graph:

We represent 'Section' on the x-axis and 'Number of girls per thousand boys' on the y-axis. To highlight the differences clearly, we can use a "kink" or "zigzag line" on the y-axis starting from 900.

(ii) Conclusions:

1. The Scheduled Tribe (ST) section has the highest number of girls per thousand boys ($970$).

2. The Urban section has the lowest number of girls per thousand boys ($910$), indicating a higher gender disparity in cities compared to rural areas ($930$).

3. Backward districts have a higher sex ratio ($950$) than non-backward districts ($920$).

Given:

Seats won by different political parties: $A=75, B=55, $$ C=37, $$ D=29, $$ E=10, F=37$.

To Do:

(i) Draw a bar graph.

(ii) Identify the party with maximum seats.

Solution:

(i) Bar Graph:

The 'Political Party' is plotted on the x-axis and 'Seats Won' on the y-axis. We use a scale of $1 \text{ unit} = 10 \text{ seats}$.

(ii) Result:

Political party A won the maximum number of seats, which is $75$.

Given:

The following table represents the lengths of $40$ leaves of a plant measured to the nearest millimetre:

To Find / To Do:

(i) Represent the data using a histogram after making the class intervals continuous.

(ii) Identify any other suitable graphical representation for the same data.

(iii) Determine whether the conclusion "the maximum number of leaves are $153$ mm long" is correct and provide a reason.

Solution:

(i) Histogram Representation

To draw a histogram, the class intervals must be continuous. However, the given intervals are discontinuous (there is a gap of $1$ unit between the upper limit of one class and the lower limit of the next, e.g., $126$ and $127$).

To make the classes continuous, we calculate the adjustment factor:

We subtract $0.5$ from each lower limit and add $0.5$ to each upper limit to obtain the Modified Continuous Classes:

Now, we can represent this on a graph:

1. On the x-axis, we represent the length of leaves (in mm). Since the first class starts at $117.5$, we use a kink (zigzag line) near the origin to show that the scale does not start from zero.

2. On the y-axis, we represent the number of leaves (frequency).

3. We draw rectangles with bases equal to the class size and heights proportional to the corresponding frequencies.

(ii) Other Suitable Graphical Representation

Apart from a histogram, another suitable graphical representation for this data is a Frequency Polygon.

To draw a frequency polygon, we plot the class marks (mid-points of the class intervals) on the x-axis and the frequencies on the y-axis, then join the points with straight lines.

(iii) Evaluation of the Conclusion

The conclusion that "the maximum number of leaves are $153$ mm long" is incorrect.

Reason:

The given data is grouped frequency distribution. From the table, we only know that the maximum number of leaves ($12$) fall within the range of $144.5$ mm to $153.5$ mm. It does not imply that all $12$ leaves are exactly $153$ mm long. In fact, their individual lengths could be any value within that specific interval.

Alternate Solution (Frequency Polygon Class Marks):

For Part (ii), if we were to draw a frequency polygon, we would calculate the class marks as follows:

For the first interval:

By following this for all classes, we get the points needed for the polygon.

Given:

Lifetime of 400 neon lamps.

To Do:

(i) Represent using a histogram.

(ii) Number of lamps with lifetime > 700 hours.

Solution:

(i) Histogram:

The intervals are already continuous. We plot 'Lifetime' on the x-axis and 'Number of lamps' on the y-axis.

(ii) Lamps with life > 700 hours:

So, 184 lamps have a lifetime of more than 700 hours.

To Do:

Represent both sections on the same graph using frequency polygons.

Solution:

First, we find the class marks $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

Comparison: Section A performed better because the majority of its students scored in higher ranges compared to Section B, whose polygon is skewed towards lower marks.

Given:

The runs scored by two teams, Team A and Team B, on the first $60$ balls in a cricket match are provided in a grouped frequency distribution table.

To Represent:

The data of both teams on the same graph using frequency polygons.

Solution:

To draw frequency polygons, we first need to make the class intervals continuous and then find the class marks for each interval.

Step 1: Making Class Intervals Continuous

The difference between the upper limit of one class and the lower limit of the next class is $7 - 6 = 1$.

The adjustment factor is calculated as:

Subtracting $0.5$ from the lower limits and adding $0.5$ to the upper limits, we get continuous intervals like $0.5 - 6.5$, $6.5 - 12.5$, and so on.

Step 2: Calculating Class Marks

The class mark ($x_i$) is the mid-point of each class interval, calculated as:

For the first interval $(0.5 - 6.5)$:

Step 3: Closing the Polygon

To close the polygon, we imagine a class before the first class and after the last class with zero frequency. The class size is $6$.

The class mark before $3.5$ is $3.5 - 6 = -2.5$.

The class mark after $57.5$ is $57.5 + 6 = 63.5$.

Table for Frequency Polygon:

Step 4: Plotting the Graph

1. Plot the Class Marks on the horizontal x-axis.

2. Plot the Runs Scored on the vertical y-axis.

3. For Team A, plot the points: $(-2.5, 0)$, $(3.5, 2)$, $(9.5, 1)$, $(15.5, 8)$, $(21.5, 9)$, $(27.5, 4)$, $(33.5, 5)$, $(39.5, 6)$, $(45.5, 10)$, $(51.5, 6)$, $(57.5, 2)$, and $(63.5, 0)$. Join them with straight lines.

4. For Team B, plot the points: $(-2.5, 0)$, $(3.5, 5)$, $(9.5, 6)$, $(15.5, 2)$, $(21.5, 10)$, $(27.5, 5)$, $(33.5, 6)$, $(39.5, 3)$, $(45.5, 4)$, $(51.5, 8)$, $(57.5, 10)$, and $(63.5, 0)$ on the same axis. Join them with a different line style (e.g., dotted or different colour).

Alternate Solution:

Alternatively, the frequency polygon can also be drawn by first constructing a histogram for the continuous data and then joining the mid-points of the top of the rectangles. However, the class mark method used above is more direct for representing two datasets on the same graph.

Given:

The following table shows the number of children of various age groups playing in a park:

To Draw:

A histogram to represent the given data.

Solution:

In the given data, the class intervals are continuous, but the class widths are unequal. To draw a histogram with unequal class widths, we need to adjust the heights of the rectangles (frequencies) so that the area of each rectangle is proportional to the frequency.

We choose the minimum class width, which is $1$ (for classes $1-2$ and $2-3$).

The formula for the adjusted frequency (height of the rectangle) is:

Step 1: Calculate the adjusted frequencies

Step 2: Drawing the Histogram

1. Represent the Age (in years) on the horizontal x-axis.

2. Represent the Adjusted Frequency (Number of children per year interval) on the vertical y-axis.

3. Draw rectangles where the width corresponds to the age group and the height corresponds to the adjusted frequency calculated above.

Alternate Explanation:

In a standard histogram, the area of the bar represents the frequency. When widths are different, simply plotting the raw frequency as height would mislead the viewer into thinking there are more children in wider groups than there actually are. By using the Adjusted Frequency, we ensure that the area $ \text{Width} \times \text{Height} $ correctly represents the actual number of children in that age group.

For example, for the age group $10 - 15$:

Given:

A frequency distribution of the number of letters in $100$ randomly picked surnames from a local telephone directory.

To Find:

(i) Draw a histogram to depict the given information.

(ii) Identify the class interval in which the maximum number of surnames lie.

Solution:

Part (i): Drawing the Histogram

In the given data, the class intervals are continuous, but the class widths are unequal. To represent such data in a histogram, the heights of the rectangles must be adjusted so that the area of each bar is proportional to the frequency.

First, we determine the minimum class width among all the intervals.

Class widths are calculated as: $4-1=3$, $6-4=2$, $8-6=2$, $12-8=4$, and $20-12=8$.

The minimum class width ($h$) is $2$.

The Adjusted Frequency (Height of the rectangle) is calculated using the following formula:

Now, let us calculate the adjusted frequencies for each class:

Steps to draw the histogram:

1. Represent the Number of letters on the horizontal x-axis with a suitable scale.

2. Represent the Adjusted Frequency on the vertical y-axis.

3. Draw rectangles with bases corresponding to the class intervals and heights equal to the adjusted frequencies calculated above.

Part (ii): Class interval with maximum surnames

To find the class interval in which the maximum number of surnames lie, we look at the original frequency column in the table.

The maximum frequency is $44$.

The class interval corresponding to the frequency $44$ is $6 - 8$.

Therefore, the maximum number of surnames lie in the class interval $6 - 8$.

Alternate Explanation:

In histograms with unequal widths, the frequency density is what is plotted on the y-axis. By multiplying the frequency-to-width ratio with the minimum class width ($2$), we standardise the heights so that a rectangle with a width of $2$ units shows its actual frequency as its height, while wider rectangles are proportionally flattened to maintain the correct area representation.