RD Sharma Solutions for Class 11 Maths Chapter 32 Statistics

RD Sharma Solutions Class 11 Maths Chapter 32 Avail Free PDF Updated for (2023 24) RD Sharma Solutions for Class 11 Maths Chapter 32 Statistics Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 32

RD Sharma Solutions Class 11 Maths Chapter 32 – Avail Free PDF Updated for (2023-24)

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics are provided here for students to score good marks in the board exams. In earlier classes, students have learnt methods of representing data graphically and in tabular form. Here, we shall see various methods of finding a representative value of the given data. Students are advised to refer to the PDF of RD Sharma Class 11 Maths Solutions in order to help them grasp the shortcut techniques and important formulas. For students to understand the concepts easily, the solutions are solved in a step-wise manner, as each step carries marks in the annual exam.

Chapter 32 – Statistics contains seven exercises, and RD Sharma Solutions offer accurate solutions to the questions present in each exercise. Students who aim to improve their academic performance can use RD Sharma as a key source of reference material to achieve their goals. Students can download the readily available PDF of RD Sharma Solutions from the links given below. Now, let us have a look at the concepts discussed in this chapter.

  • Measures of dispersion
  • Range
  • Mean deviation
    • Mean deviation for ungrouped data or individual observations.
    • Mean deviation of a discrete frequency distribution.
    • Mean deviation of a grouped or continuous frequency distribution.
  • Limitations of mean deviation
  • Variance and standard deviation
    • The variance of individual observations.
    • The variance of a discrete frequency distribution.
    • The variance of a grouped or continuous frequency distribution.
  • Analysis of frequency distribution

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics

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EXERCISE 32.1 PAGE NO: 32.6

1. Calculate the mean deviation about the median of the following observation:
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000

(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Solution:

(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000

To calculate the Median (M), let us arrange the numbers in ascending order.

The median is the middle number of all the observations.

2354, 2780, 3011, 3020, 3541, 4150, 5000

So, Median = 3020 and n = 7

By using the formula to calculate the Mean Deviation,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)
xi |di| = |xi – 3020|
3011 9
2780 240
3020 0
2354 666
3541 521
4150 1130
5000 1980
Total 4546
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/7 × 4546

= 649.42

∴ The Mean Deviation is 649.42.

(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44

To calculate the Median (M), let us arrange the numbers in ascending order.

The median is the middle number of all the observations.

34, 38, 42, 44, 46, 48, 54, 55, 63, 70

Here, the number of observations is even, then median = (46+48)/2 = 47

Median = 47 and n = 10

By using the formula to calculate the Mean Deviation,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)
xi |di| = |xi – 47|
38 9
70 23
48 1
34 13
42 5
55 8
63 16
46 1
54 7
44 3
Total 86
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 86

= 8.6

∴ The Mean Deviation is 8.6.

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

To calculate the Median (M), let us arrange the numbers in ascending order.

The median is the middle number of all the observations.

30, 34, 38, 40, 42, 44, 50, 51, 60, 66

Here, the number of observations is even, then the Median = (42+44)/2 = 43

Median = 43 and n = 10

By using the formula to calculate the Mean Deviation,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)
xi |di| = |xi – 43|
30 13
34 9
38 5
40 3
42 1
44 1
50 7
51 8
60 17
66 23
Total 87
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 87

= 8.7

∴ The Mean Deviation is 8.7.

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

To calculate the Median (M), let us arrange the numbers in ascending order.

The median is the middle number of all the observations.

22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Here, the number of observations is even, then the median = (28+29)/2 = 28.5

Median = 28.5 and n = 10

By using the formula to calculate the Mean Deviation,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)
xi |di| = |xi – 28.5|
22 6.5
24 4.5
30 1.5
27 1.5
29 0.5
31 2.5
25 3.5
28 0.5
41 12.5
42 13.5
Total 47
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 47

= 4.7

∴ The Mean Deviation is 4.7.

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

To calculate the Median (M), let us arrange the numbers in ascending order.

The median is the middle number of all the observations.

34, 38, 43, 44, 47, 48, 53, 55, 63, 70

Here, the number of observations is even, then median = (47+48)/2 = 47.5

Median = 47.5 and n = 10

By using the formula to calculate the Mean Deviation,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)
xi |di| = |xi – 47.5|
38 9.5
70 22.5
48 0.5
34 13.5
63 15.5
42 5.5
55 7.5
44 3.5
53 5.5
47 0.5
Total 84
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 84

= 8.4

∴ The Mean Deviation is 8.4.

2. Calculate the mean deviation from the mean for the following data:
(i) 4, 7, 8, 9, 10, 12, 13, 17

(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

Solution:

(i) 4, 7, 8, 9, 10, 12, 13, 17

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [4 + 7 + 8 + 9 + 10 + 12 + 13 + 17]/8

= 80/8

= 10

Number of observations, ‘n’ = 8

xi |di| = |xi – 10|
4 6
7 3
8 2
9 1
10 0
12 2
13 3
17 7
Total 24
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/8 × 24

= 3

∴ The Mean Deviation is 3.

(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [13 + 17 + 16 + 14 + 11 + 13 + 10 + 16 + 11 + 18 + 12 + 17]/12

= 168/12

= 14

Number of observations, ‘n’ = 12

xi |di| = |xi – 14|
13 1
17 3
16 2
14 0
11 3
13 1
10 4
16 2
11 3
18 4
12 2
17 3
Total 28
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/12 × 28

= 2.33

∴ The Mean Deviation is 2.33.

(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44]/10

= 500/10

= 50

Number of observations, ‘n’ = 10

xi |di| = |xi – 50|
38 12
70 20
48 2
40 10
42 8
55 5
63 13
46 4
54 4
44 6
Total 84
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 84

= 8.4

∴ The Mean Deviation is 8.4.

(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [36 + 72 + 46 + 42 + 60 + 45 + 53 + 46 + 51 + 49]/10

= 500/10

= 50

Number of observations, ‘n’ = 10

xi |di| = |xi – 50|
36 14
72 22
46 4
42 8
60 10
45 5
53 3
46 4
51 1
49 1
Total 72
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 72

= 7.2

∴ The Mean Deviation is 7.2.

(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59]/10

= 550/10

= 55

Number of observations, ‘n’ = 10

xi |di| = |xi – 55|
57 2
64 9
43 12
67 12
49 6
59 4
44 11
47 8
61 6
59 4
Total 74
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 74

= 7.4

∴ The Mean Deviation is 7.4.

3. Calculate the mean deviation of the following income groups of five and seven members from their medians:

I

Income in ₹

 II

Income in ₹
4000 3800
4200 4000
4400 4200
4600 4400
4800 4600
4800
5800

Solution:

Let us calculate the mean deviation for the first data set.

Since the data is arranged in ascending order,

4000, 4200, 4400, 4600, 4800

Median = 4400

Total observations = 5

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – M|

xi |di| = |xi – 4400|
4000 400
4200 200
4400 0
4600 200
4800 400
Total 1200
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/5 × 1200

= 240

Let us calculate the mean deviation for the second data set.

Since the data is arranged in ascending order,

3800, 4000, 4200, 4400, 4600, 4800, 5800

Median = 4400

Total observations = 7

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – M|

xi |di| = |xi – 4400|
3800 600
4000 400
4200 200
4400 0
4600 200
4800 400
5800 1400
Total 3200
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/7 × 3200

= 457.14

∴ The Mean Deviation of set 1 is 240, and set 2 is 457.14

4. The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i) Find the mean deviation from the median.
(ii) Find the mean deviation from the mean also.

Solution:

(i) Find the mean deviation from the median.

Let us arrange the data in ascending order,

15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – M|

The number of observations are even, then median = (40+52.3)/2 = 46.15

Median = 46.15

Number of observations, ‘n’ = 10

xi |di| = |xi – 46.15|
40.0 6.15
52.3 6.15
55.2 9.05
72.9 26.75
52.8 6.65
79.0 32.85
32.5 13.65
15.2 30.95
27.9 19.25
30.2 15.95
Total 167.4
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 167.4

= 16.74

∴ The Mean Deviation is 16.74.

(ii) Find the mean deviation from the mean also.

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [40.0 + 52.3 + 55.2 + 72.9 + 52.8 + 79.0 + 32.5 + 15.2 + 27.9 + 30.2]/10

= 458/10

= 45.8

Number of observations, ‘n’ = 10

xi |di| = |xi – 45.8|
40.0 5.8
52.3 6.5
55.2 9.4
72.9 27.1
52.8 7
79.0 33.2
32.5 13.3
15.2 30.6
27.9 17.9
30.2 15.6
Total 166.4
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 166.4

= 16.64

∴ The Mean Deviation is 16.64

5. In questions 1(iii), (iv), (v), find the number of observations lying between

\(\begin{array}{l}\overline{X} – M.D\end{array} \)
and
\(\begin{array}{l}\overline{X} + M.D\end{array} \)
, where M.D. is the mean deviation from the mean.

Solution:

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51]/10

= 455/10

= 45.5

Number of observations, ‘n’ = 10

xi |di| = |xi – 45.5|
34 11.5
66 20.5
30 15.5
38 7.5
44 1.5
50 4.5
40 5.5
60 14.5
42 3.5
51 5.5
Total 90
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 90

= 9

Now

\(\begin{array}{l}\overline{X} – M.D\end{array} \)
= 45.5 – 9 = 36.5
\(\begin{array}{l}\overline{X} + M.D\end{array} \)
= 45.5 + 9 = 54.5

So, there are a total of 6 observations between

\(\begin{array}{l}\overline{X} – M.D\end{array} \)
and 
\(\begin{array}{l}\overline{X} + M.D\end{array} \)

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42]/10

= 299/10

= 29.9

Number of observations, ‘n’ = 10

xi |di| = |xi – 29.9|
22 7.9
24 5.9
30 0.1
27 2.9
29 0.9
31 1.1
25 4.9
28 1.9
41 11.1
42 12.1
Total 48.8
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 48.8

= 4.88

Now

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 35

So, there are a total of 5 observations between RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 36
and RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 37

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

We know that,

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

Where, |di| = |xi – x|

So, let ‘x’ be the mean of the given observation.

x = [38 + 70 + 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47]/10

= 494/10

= 49.4

Number of observations, ‘n’ = 10

xi |di| = |xi – 49.4|
38 11.4
70 20.6
48 1.4
34 15.4
63 13.6
42 7.4
55 5.6
44 5.4
53 3.6
47 2.4
Total 86.8
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/10 × 86.8

= 8.68

Now

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 40

EXERCISE 32.2 PAGE NO: 32.11

1. Calculate the mean deviation from the median of the following frequency distribution:

Heights in inches 58 59 60 61 62 63 64 65 66
No. of students 15 20 32 35 35 22 20 10 8

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

We know, median is the middle term.

So, Median = 61

Let xi =Heights in inches

And, fi = Number of students

xi fi Cumulative Frequency |di| = |xi – M|

= |xi – 61|

 fi |di|
58 15 15 3 45
59 20 35 2 40
60 32 67 1 32
61 35 102 0 0
62 35 137 1 35
63 22 159 2 44
64 20 179 3 60
65 10 189 4 40
66 8 197 5 40
N = 197 Total = 336

N=197

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/197 × 336

= 1.70

∴ The mean deviation is 1.70.

2. The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:

Number of calls 0 1 2 3 4 5 6 7
Frequency 14 21 25 43 51 40 39 12

Compute the mean deviation about the median.

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

We know, median is the even term, (3+5)/2 = 4

So, Median = 8

Let xi =Number of calls

And, fi = Frequency

xi fi Cumulative Frequency |di| = |xi – M|

= |xi – 61|

 fi |di|
0 14 14 4 56
1 21 35 3 63
2 25 60 2 50
3 43 103 1 43
4 51 154 0 0
5 40 194 1 40
6 39 233 2 78
7 12 245 3 36
Total = 366
Total = 245

N = 245

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/245 × 336

= 1.49

∴ The mean deviation is 1.49.

3. Calculate the mean deviation about the median of the following frequency distribution:

xi 5 7 9 11 13 15 17
fi 2 4 6 8 10 12 8

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

We know N = 50

Median = (50)/2 = 25

So, the median Corresponding to 25 is 13.

xi fi Cumulative Frequency |di| = |xi – M|

= |xi – 61|

 fi |di|
5 2 2 8 16
7 4 6 6 24
9 6 12 4 24
11 8 20 2 16
13 10 30 0 0
15 12 42 2 24
17 8 50 4 32
Total = 50 Total = 136

N = 50

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/50 × 136

= 2.72

∴ The mean deviation is 2.72.

4. Find the mean deviation from the mean for the following data:

(i)

xi 5 7 9 10 12 15
fi 8 6 2 2 2 6

Solution:

To find the mean deviation from the mean, firstly, let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 44

xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di|
5 8 40 4 32
7 6 42 2 12
9 2 18 0 0
10 2 20 1 2
12 2 24 3 6
15 6 90 6 36
Total = 26 Total = 234 Total = 88

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 45

= 234/26

= 9

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 46

= 88/26

= 3.3

∴ The mean deviation is 3.3

(ii)

xi 5 10 15 20 25
fi 7 4 6 3 5

Solution:

To find the mean deviation from the mean, firstly, let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 47

xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di|
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
Total = 25 Total = 350 Total = 158

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 48

= 350/25

= 14

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 49

= 158/25

= 6.32

∴ The mean deviation is 6.32

(iii)

xi 10 30 50 70 90
fi 4 24 28 16 8

Solution:

To find the mean deviation from the mean, firstly, let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 50

xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di|
10 4 40 40 160
30 24 720 20 480
50 28 1400 0 0
70 16 1120 20 320
90 8 720 40 320
Total = 80 Total = 4000 Total = 1280

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 51

= 4000/80

= 50

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 52

= 1280/80

= 16

∴ The mean deviation is 16

5. Find the mean deviation from the median for the following data:

(i)

xi 15 21 27 30
fi 3 5 6 7

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

We know N = 21

Median = (21)/2 = 10.5

So, the median corresponding to 10.5 is 27.

xi fi Cumulative Frequency |di| = |xi – M| fi |di|
15 3 3 15 45
21 5 8 9 45
27 6 14 3 18
30 7 21 0 0
Total = 21 Total = 46 Total = 108

N = 21

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/21 × 108

= 5.14

∴ The mean deviation is 5.14

(ii)

xi 74 89 42 54 91 94 35
fi 20 12 2 4 5 3 4

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

We know N = 50

Median = (50)/2 = 25

So, the median corresponding to 25 is 74.

xi fi Cumulative Frequency |di| = |xi – M| fi |di|
74 20 4 39 156
89 12 6 32 64
42 2 10 20 80
54 4 30 0 0
91 5 42 15 180
94 3 47 17 85
35 4 50 20 60
Total = 50 Total = 189 Total = 625

N = 50

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/50 × 625

= 12.5

∴ The mean deviation is 12.5

(iii)

Marks obtained 10 11 12 14 15
No. of students 2 3 8 3 4

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

We know N = 20

Median = (20)/2 = 10

So, the median corresponding to 10 is 12.

xi fi Cumulative Frequency |di| = |xi – M| fi |di|
10 2 2 2 4
11 3 5 1 3
12 8 13 0 0
14 3 16 2 6
15 4 20 3 12
Total = 20 Total = 25

N = 20

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/20 × 25

= 1.25

∴ The mean deviation is 1.25

EXERCISE 32.3 PAGE NO: 32.16

1. Compute the mean deviation from the median of the following distribution:

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 10 20 5 10

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

The median is the middle term of Xi.

Here, the middle term is 25.

So, Median = 25

Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di|
0-10 5 5 5 20 100
10-20 15 10 15 10 100
20-30 25 20 35 0 0
30-40 35 5 91 10 50
40-50 45 10 101 20 200
Total = 50 Total = 450
\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/50 × 450

= 9

∴ The mean deviation is 9

2. Find the mean deviation from the mean for the following data:

(i)

Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Frequencies 4 8 9 10 7 5 4 3

Solution:

To find the mean deviation from the mean, firstly, let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 57

= 17900/50

= 358

Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di|
0-100 50 4 200 308 1232
100-200 150 8 1200 208 1664
200-300 250 9 2250 108 972
300-400 350 10 3500 8 80
400-500 450 7 3150 92 644
500-600 550 5 2750 192 960
600-700 650 4 2600 292 1168
700-800 750 3 2250 392 1176
Total = 50 Total = 17900 Total = 7896

N = 50

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/50 × 7896

= 157.92

∴ The mean deviation is 157.92

(ii)

Classes 95-105 105-115 115-125 125-135 135-145 145-155
Frequencies 9 13 16 26 30 12

Solution:

To find the mean deviation from the mean, firstly, let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 59

= 13630/106

= 128.58

Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di|
95-105 100 9 900 28.58 257.22
105-115 110 13 1430 18.58 241.54
115-125 120 16 1920 8.58 137.28
125-135 130 26 3380 1.42 36.92
135-145 140 30 4200 11.42 342.6
145-155 150 12 1800 21.42 257.04
N = 106 Total = 13630 Total = 1272.6

N = 106

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/106 × 1272.6

= 12.005

∴ The mean deviation is 12.005

3. Compute mean deviation from the mean of the following distribution:

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of students 8 10 15 25 20 18 9 5

Solution:

To find the mean deviation from the mean, firstly, let us calculate the mean.

By using the formula,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 61

= 5390/110

= 49

Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di|
10-20 15 8 120 34 272
20-30 25 10 250 24 240
30-40 35 15 525 14 210
40-50 45 25 1125 4 100
50-60 55 20 1100 6 120
60-70 65 18 1170 16 288
70-80 75 9 675 26 234
80-90 85 5 425 36 180
N = 110 Total = 5390 Total = 1644

N = 110

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/110 × 1644

= 14.94

∴ The mean deviation is 14.94

4. The age distribution of 100 life-insurance policyholders is as follows:

Age (on nearest birthday) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5
No. of persons 5 16 12 26 14 12 6 5

Calculate the mean deviation from the median age.

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

N = 96

So, N/2 = 96/2 = 48

The cumulative frequency just greater than 48 is 59, and the corresponding value of x is 38.25

So, Median = 38.25

Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di|
17-19.5 18.25 5 5 20 100
20-25.5 22.75 16 21 15.5 248
36-35.5 30.75 12 33 7.5 90
36-40.5 38.25 26 59 0 0
41-50.5 45.75 14 73 7.5 105
51-55.5 53.25 12 85 15 180
56-60.5 58.25 6 91 20 120
61-70.5 65.75 5 96 27.5 137.5
Total = 96 Total = 980.5

N = 96

\(\begin{array}{l}MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\end{array} \)

= 1/96 × 980.5

= 10.21

∴ The mean deviation is 10.21

5. Find the mean deviation from the mean and from a median of the following distribution:

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 5 8 15 16 6

Solution:

To find the mean deviation from the median, firstly, let us calculate the median.

N = 50

So, N/2 = 50/2 = 25

The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28.

So, Median = 28

By using the formula to calculate the mean,

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 64

= 1350/50

= 27

Class Interval xi fi Cumulative Frequency |di| = |xi – Median| fi |di| FiXi |Xi – Mean| Fi |Xi – Mean|
0-10 5 5 5 23 115 25 22 110
10-20 15 8 13 13 104 120 12 96
20-30 25 15 28 3 45 375 2 30
30-40 35 16 44 7 112 560 8 128
40-50 45 6 50 17 102 270 18 108
N = 50 Total = 478 Total = 1350 Total = 472

The mean deviation from median = 478/50 = 9.56

And, the mean deviation from median = 472/50 = 9.44

∴ The Mean Deviation from the median is 9.56 and from the mean is 9.44.

EXERCISE 32.4 PAGE NO: 32.28

1. Find the mean, variance and standard deviation for the following data:
(i) 2, 4, 5, 6, 8, 17

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 65

(ii) 6, 7, 10, 12, 13, 4, 8, 12

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 66

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 67

2. The variance of 20 observations is 4. If each observation is multiplied by 2, find the variance of the resulting observations.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 68

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 69

3. The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 70

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 71

4. The mean of 5 observations is 4.4, and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 72

5. The mean and standard deviation of 6 observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 73

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 74

∴ The mean of new observation is 24, and the standard deviation of new observation is 12.

6. The mean and variance of 8 observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 75

EXERCISE 32.5 PAGE NO: 32.37

1. Find the standard deviation for the following distribution:

x: 4.5 14.5 24.5 34.5 44.5 54.5 64.5
f: 1 5 12 22 17 9 4

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 76

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 77

= 100 [1.857 – 0.0987]

= 100 [1.7583]

Var (X) = 175.83

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 78

2. Table below shows the frequency f with which ‘x’ alpha particles were radiated from a diskette

x: 0 1 2 3 4 5 6 7 8 9 10 11 12
f: 51 203 383 525 532 408 273 139 43 27 10 4 2

Calculate the mean and variance.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 79

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 80

3. Find the mean and standard deviation for the following data:
(i)

Year render: 10 20 30 40 50 60
No. of persons (cumulative) 15 32 51 78 97 109

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 81

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 82

(ii)

Marks: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Frequency: 1 6 6 8 8 2 2 3 0 2 1 0 0 0 1

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 83

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 84

4. Find the standard deviation for the following data:

(i)

x: 3 8 13 18 23
f: 7 10 15 10 6

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 85

(ii)

x: 2 3 4 5 6 7
f: 4 9 16 14 11 6

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 86

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 87

EXERCISE 32.6 PAGE NO: 32.41

1. Calculate the mean and S.D. for the following data:

Expenditure (in ₹): 0-10 10-20 20-30 30-40 40-50
Frequency: 14 13 27 21 15

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 88

2. Calculate the standard deviation for the following data:

Class: 0-30 30-60 60-90 90-120 120-150 150-180 180-210
Frequency: 9 17 43 82 81 44 24

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 89

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 90

3. Calculate the A.M. and S.D. for the following distribution:

Class: 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency: 18 16 15 12 10 5 2 1

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 91

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 92

4. A student obtained the mean and standard deviation of 100 observations as 40 and 5.1, respectively. It was later found that one observation was wrongly copied as 50, the correct figure is 40. Find the correct mean and S.D.

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 93

5. Calculate the mean, median and standard deviation of the following distribution

Class-interval 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70
Frequency: 2 3 8 12 16 5 2 3

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 94

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 95

EXERCISE 32.7 PAGE NO: 32.47

1. Two plants, A and B of a factory, show the following results about the number of workers and the wages paid to them.

Plant A Plant B
No. of workers 5000 6000
Average monthly wages ₹2500 ₹2500
The variance of the distribution of wages 81 100

In which plant – A or B – is there greater variability in individual wages?

Solution:

Variation of the distribution of wages in plant A (σ2 =18)

So, the standard deviation of the distribution A (σ – 9)

Similarly, the variation of the distribution of wages in plant B (σ2 =100)

So, the standard deviation of the distribution B (σ – 10)

And, the average monthly wages in both the plants is 2500,

Since the plant with a greater value of SD will have more variability in salary.

∴ Plant B has more variability in individual wages than plant A.

2. The means and standard deviations of heights and weights of 50 students in a class are as follows:

Weights Heights
Mean 63.2 kg 63.2 inch
Standard deviation 5.6 kg 11.5 inch

Which shows more variability, heights or weights?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 96

3. The coefficient of variation of the two distributions are 60% and 70%, and their standard deviations are 21 and 16, respectively. What is their arithmetic means?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 97

= 22.86

∴ Means are 35 and 22.86

4. Calculate the coefficient of variation from the following data:

Income (in ₹): 1000-1700 1700-2400 2400-3100 3100-3800 3800-4500 4500-5200
No. of families: 12 18 20 25 35 10

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 98

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 99

5. An analysis of the weekly wages paid to workers in two firms, A and B, belonging to the same industry, gives the following results:

Firm A Firm B
No. of wage earners 586 648
Average weekly wages ₹52.5 ₹47.5
The variance of the distribution of wages 100 121


(i) Which firm – A or B – pays out the larger amount as weekly wages?
(ii) Which firm – A or B – has greater variability in individual wages?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 100

6. The following are some particulars of the distribution of weights of boys and girls in a class:

Boys Girls
Number 100 50
Mean weight 60 kg 45 kg
Variance 9 4


Which of the distributions is more variable?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics - image 101

Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics

Exercise 32.1 Solutions

Exercise 32.2 Solutions

Exercise 32.3 Solutions

Exercise 32.4 Solutions

Exercise 32.5 Solutions

Exercise 32.6 Solutions

Exercise 32.7 Solutions

Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 32

Q1

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The RD Sharma Solutions for Class 11 Maths Chapter 32 are developed by considering the exercise-wise marks distribution as per the CBSE syllabus. Complex questions are provided with step-by-step explanations making it easier for students to understand the concepts at ease. Elaborated solutions enable students to understand the concepts efficiently. Simplified and logical language is used in order to boost confidence among students.
Q2

Is BYJU’S website providing Solutions for RD Sharma Class 11 Maths Chapter 32?

Yes, students can get free PDFs of RD Sharma Solutions for Class 11 Maths Chapter 32. The solutions are designed by an expert faculty team at BYJU’S in a unique way. Students who aim to score high in board exams are advised to solve the RD Sharma Textbook. Studying these books on a regular basis will improve time management and problem-solving abilities among students.
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Yes, students can download the RD Sharma Solutions for Class 11 Maths Chapter 32 in PDF format both online and offline at BYJU’S website. The solutions for the textbook questions are curated by highly professional teachers as per the current CBSE syllabus. Using the solutions PDF, students can also cross-check their answers to obtain an idea of solving complex problems effortlessly.

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