NCERT Solutions for Class 9 Maths Chapter 12 Statistics

Question 1: (ii) A survey conducted by an organization for the cause of illness and death among women between the ages 15 - 44 (in years) worldwide, found the following figures (in %) :

Which condition is the major cause of women’s ill health and death worldwide?

Answer:

From the graph, we can see reproductive health conditions are the major cause of women’s ill health and death worldwide. The female fatality rate is 31.8% due to reproductive health conditions.

Question 1: (iii) A survey conducted by an organization for the cause of illness and death among women between the ages 15 - 44 (in years) worldwide, found the following figures (in %):

Try to find out, with the help of your teacher, any two factors that play a major role in the cause in (ii) above being the major cause.

Answer:

Due to poor financial conditions and the failure of the government to provide necessary healthcare conditions to women, reproductive health conditions are the major cause of ill health and death of women worldwide.

Question 2: (i) The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

Represent the information above with a bar graph.

Answer:

The graphical representation of the given information is as follows.

Question 2: (ii) The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

In the classroom, discuss what conclusions can be arrived at from the graph

Answer:

From the graph, we can see that the number of girls per thousand boys is the least in urban society and the highest in the Scheduled Tribes.

910 in the case of urban society and 970 in that of Scheduled Tribes.

Question 3: (i) Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

Draw a bar graph to represent the polling results.

Answer:

The representation of the given data in the form of a bar graph is as follows.

Question 3: (ii) Given below are the seats won by different political parties in the polling outcome of state assembly elections:

Which political party won the maximum number of seats?

Answer:

Party A has won the maximum number of seats. Party A has won 75 seats.

Question 4: (i) The length of 40 leaves of a plant is measured correctly by one millimeter, and the obtained data is represented in the following table:

Draw a histogram to represent the given data. [Hint: First, make the class intervals continuous]

Answer:

As we can see from the given table, the data is discontinuous, and the difference between the upper limit of a class and the lower limit of the next class is 1; therefore, we change both of them by a value $\frac12$.

e.g 127 - 135 would become 126.5 - 235.5

The modified table, therefore, is

The representation of the above data through a histogram is as follows.

Question 4: (ii) The length of 40 leaves of a plant is measured correctly by one millimeter, and the obtained data is represented in the following table:

Is there any other suitable graphical representation for the same data?

Answer:

A frequency polygon could be another suitable graphical representation for the same data.

Question 4: (iii) The length of 40 leaves of a plant is measured correctly by one millimeter, and the obtained data is represented in the following table:

Is it correct to conclude that the maximum number of leaves is 153 mm long? Why?

Answer:

No, it is certainly not correct to conclude that the maximum number of leaves is 153 mm long because the given data does not tell us about the exact length of the leaves. It only tells us about the range in which their lengths lie. We can only conclude that the maximum number of leaves (12) have their lengths in the region 145 - 153.

Question 5: (i) The following table gives the lifetimes of 400 neon lamps:

Represent the given information with the help of a histogram.

Answer:

The representation of the given information in the form of a histogram is as follows.

Question 5: (ii) The following table gives the lifetimes of 400 neon lamps:

How many lamps have a lifetime of more than 700 hours?

Answer:

Lamps having a lifetime in the range of 700 - 800 = 74

Lamps having a lifetime in the range of 800 - 900 = 62

Lamps having a lifetime in the range 900 - 1000 = 48

Lamps having a lifetime of more than 700 hours = 74 + 62 + 48 = 184.

Question 6: The following table gives the distribution of students in two sections according to the marks obtained by them:

Represent the marks of the students of both sections on the same graph by two frequency polygons. From the two polygons, compare the performance of the two sections

Answer:

To make the frequency polygon, we first modify the table as follows

Class marks $= \frac{\text{Upper limit of class interval + Lower limit of class interval}}{2}$$\frac{\mathrm{}}{}$

To make the frequency polygon, we mark the marks on the x-axis and the number of students on the y-axis.

The representation of the given information in the form of a frequency polygon is as follows.

From the frequency polygon, we can see that the performance of section A is better.

Question 7: The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Represent the data of both teams on the same graph by frequency polygons. [ Hint: First, make the class intervals continuous.]

Answer:

The given data is not continuous; we therefore modify the limits of the class intervals as well to make the class intervals continuous.

To make the frequency polygon, we first modify the table as follows

Class marks $= \frac{\text{Upper limit of class interval + Lower limit of class interval}}{2}$$\frac{\mathrm{}}{}$

To make the frequency polygon, we mark the number of balls on the x-axis and the runs scored on the y-axis.

The representation of the given information in the form of a frequency polygon is as follows.

Question 8: A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.

Answer:

Since the class sizes vary to make the histogram, we have to calculate the weighted frequency for each rectangle as per its width

Weighted frequency$=\frac{\text{Minimum class size}}{\text{Class size of the interval}}\times$ Frequency
Length of the rectangle $=\frac{\text{Minimum width}}{\text{Width of the rectangle}}\times$ Frequency

Minimum class size = 2 - 1 = 1

The modified table showing the weighted frequency as per the size of the class intervals is as follows.

The histogram representing the information given in the above table is as follows.

Question 9: (i) 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Answer:

Since the class sizes vary, to make the histogram, we have to calculate the weighted frequency for each rectangle as per its width

Weighted frequency$=\frac{\text{Minimum class size}}{\text{Class size of the interval}}\times$ Frequency
Length of the rectangle $=\frac{\text{Minimum width}}{\text{Width of the rectangle}}\times$ Frequency

Minimum class size = 6 - 4 = 2

The modified table showing the weighted frequency as per the size of the class intervals is as follows.

The histogram representing the information given in the above table is as follows.

Question 9: (ii) 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Write the class interval in which the maximum number of surnames lie.

Answer:

The class interval in which the maximum number of surnames lie is 6 - 8

The weighted frequency of this class interval (taking 2 as the minimum class size) is 44.