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NCERT Class 11 Maths Textbook Chapter 15 With Answer Book PDF Free Download
Chapter 15: Statistics
15.1 Introduction
We know that statistics deals with data collected for specific purposes. We can make decisions about the data by analyzing and interpreting it.
In earlier classes, we have studied methods of representing data graphically and in tabular form.
This representation reveals certain salient features or characteristics of the data. We have also studied the methods of finding a representative value for the given data.
This value is called the measure of central tendency. Recall mean (arithmetic mean), median and mode are three measures of central tendency.
A measure of central tendency gives us a rough idea of where data points are centered.
But, in order to make a better interpretation of the data, we should also have an idea of how the data are scattered or how much they are bunched around a measure of central tendency.
Consider now the runs scored by two batsmen in their last ten matches as follows:
Batsman A : 30, 91, 0, 64, 42, 80, 30, 5, 117, 71
Batsman B : 53, 46, 48, 50, 53, 53, 58, 60, 57, 52
Clearly, the mean and median of the data are Batsman A Batsman B
Mean 53 53
Median 53 53
Recall that, we calculate the mean of a data (denoted by x ) by dividing the sum
of the observations by the number of observations, i.e
| Author | NCERT |
| Language | English |
| No. of Pages | 36 |
| PDF Size | 2.5 MB |
| Category | Mathematics |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 11 Maths Chapter 15 Statistics
Find the mean deviation about the mean for the data in Exercises 1 and 2.
1. 4, 7, 8, 9, 10, 12, 13, 17
Solution:-
First we have to find (x̅) of the given data
So, the respective values of the deviations from mean,
i.e., xi – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 9 = 1, 10 – 10 = 0,
10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7
6, 3, 2, 1, 0, -2, -3, -7
Now absolute values of the deviations,
6, 3, 2, 1, 0, 2, 3, 7
MD = sum of deviations/ number of observations
= 24/8
= 3
So, the mean deviation for the given data is 3.
2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:-
First we have to find (x̅) of the given data
So, the respective values of the deviations from mean,
i.e., xi – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,
50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6
-12, 20, -2, -10, -8, 5, 13, -4, 4, -6
Now absolute values of the deviations,
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
MD = sum of deviations/ number of observations
= 84/10
= 8.4
So, the mean deviation for the given data is 8.4.
Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Solution:-
First we have to arrange the given observations into ascending order,
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
The number of observations is 12
Then,
Median = ((12/2)th observation + ((12/2)+ 1)th observation)/2
(12/2)th observation = 6th = 13
(12/2)+ 1)th observation = 6 + 1
= 7th = 14
Median = (13 + 14)/2
= 27/2
= 13.5
So, the absolute values of the respective deviations from the median, i.e., |xi – M| are
3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
Mean Deviation,
= (1/12) × 28
= 2.33
So, the mean deviation about the median for the given data is 2.33.
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Solution:-
First we have to arrange the given observations into ascending order,
36, 42, 45, 46, 46, 49, 51, 53, 60, 72.
The number of observations is 10
Then,
Median = ((10/2)th observation + ((10/2)+ 1)th observation)/2
(10/2)th observation = 5th = 46
(10/2)+ 1)th observation = 5 + 1
= 6th = 49
Median = (46 + 49)/2
= 95
= 47.5
So, the absolute values of the respective deviations from the median, i.e., |xi – M| are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Mean Deviation,
= (1/10) × 70
= 7
So, the mean deviation about the median for the given data is 7.
Find the mean deviation about the mean for the data in Exercises 5 and 6.
5.
| xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
| Xi | fi | fixi | |xi – x̅| | fi |xi – x̅| |
| 5 | 7 | 35 | 9 | 63 |
| 10 | 4 | 40 | 4 | 16 |
| 15 | 6 | 90 | 1 | 6 |
| 20 | 3 | 60 | 6 | 18 |
| 25 | 5 | 125 | 11 | 55 |
| 25 | 350 | 158 |
The sum of calculated data,
The absolute values of the deviations from the mean, i.e., |xi – x̅|, as shown in the table.
NCERT Class 11 Maths Textbook Chapter 15 With Answer Book PDF Free Download