NCERT Solution Class 9th Maths Chapter – 12 Statistics Example

September 28, 2023

NCERT Solution Class 9th Maths Chapter – 12 Statistics

Textbook	NCERT
Class	9th
Subject	Mathematics
Chapter	12^th
Chapter Name	Statistics
Category	Class 9th Math Solutions
Medium	English
Source	Last Doubt

NCERT Solution Class 9th Maths Chapter – 12 Statistics

Chapter – 12

Statistics

Exercises

Example 1 – In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained: Observe the bar graph given above and answer the following questions: (i) How many students were born in the month of November? (ii) In which month were the maximum number of students born? NCERT Solution Class 10th Maths Chapter – 12 Statistics Exercises

Solution – Note that the variable here is the ‘month of birth’, and the value of the variable is the ‘Number of students born’. (i) 4 students were born in the month of November. (ii) The Maximum number of students were born in the month of August. Let us now recall how a bar graph is constructed by considering the following example.

Example 2 : A family with a monthly income of ` 20,000 had planned the following expenditures per month under various heads: Draw a bar graph for the data above.

Heads	Expenditure (in thousand rupees)
Grocery	4
Rent	5
Education of children	5
Medicine	2
Fuel	2
Entertainment	1
Miscellaneous	1

Solution – We draw the bar graph of this data in the following steps. Note that the unit in the second column is thousand rupees. So, ‘4’ against ‘grocery’ means `4000. 1. We represent the Heads (variable) on the horizontal axis choosing any scale, since the width of the bar is not important. But for clarity, we take equal widths for all bars and maintain equal gaps in between. Let one Head be represented by one unit. 2. We represent the expenditure (value) on the vertical axis. Since the maximum expenditure is `5000, we can choose the scale as 1 unit = `1000. 3. To represent our first Head, i.e., grocery, we draw a rectangular bar with width 1 unit and height 4 units. 4. Similarly, other Heads are represented leaving a gap of 1 unit in between two consecutive bars.

Example 3 : A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows: 0 – 20, 20 – 30, . . ., 60 – 70, 70 – 100. Then she formed the following table:Solution

Marks	Number of students
0 – 20	7
20 – 30	10
30 – 40	10
40 – 50	20
50 – 60	20
60 – 70	15
70 – above	8
Total	90

Example 4 : Consider the marks, out of 100, obtained by 51 students of a class in a test, given in Table 12.5.

Marks	Number of students
0 – 10	5
10 – 20	10
20 – 30	4
30 – 40	6
40 – 50	7
50 – 60	3
60 – 70	2
70 – 80	2
80 – 90	3
90 – 100	9
Total	51

Draw a frequency polygon corresponding to this frequency distribution table.Solution – Let us first draw a histogram for this data and mark the mid-points of the tops of the rectangles as B, C, D, E, F, G, H, I, J, K, respectively. Here, the first class is 0-10. So, to find the class preceeding 0-10, we extend the horizontal axis in the negative direction and find the mid-point of the imaginary class-interval (–10) – 0. The first end point, i.e., B is joined to this mid-point with zero frequency on the negative direction of the horizontal axis. The point where this line segment meets the vertical axis is marked as A. Let L be the mid-point of the class succeeding the last class of the given data. Then OABCDEFGHIJKL is the frequency polygon, which is shown in Fig. 12.7

Example 5 : In a city, the weekly observations made in a study on the cost of living index are given in the following table.

Cost of living index	Number of weeks
140 – 150	5
150 – 160	10
160 – 170	20
170 – 180	9
180 – 190	6
190 – 200	2
Total	52

Draw a frequency polygon for the data above (without constructing a histogram).Solution – Since we want to draw a frequency polygon without a histogram, let us find the class-marks of the classes given above, that is of 140 – 150, 150 – 160,…. For 140 – 150, the upper limit = 150, and the lower limit = 140 So, the class-mark = 150 + 140/2 = 290/2 = 145

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