# RD Sharma Solutions for Class 11 Maths Chapter 32 Statistics

## RD Sharma Solutions Class 11 Maths Chapter 32 – Avail Free PDF Updated for (2021-22)

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, we have learnt about methods of representing data graphically and in tubular form. Here, we shall study various methods of finding a representative value of the given data.

RD Sharma Solutions for Class 11 Maths Chapter 32 Statistics are provided here for students to practice and prepare for their board exams. 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 stepwise manner, as each step carries marks in the annual exam.

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 below mentioned links.

Chapter 32 – Statistics contains seven exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. 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.
• Variance of individual observations.
• Variance of a discrete frequency distribution.
• Variance of a grouped or continuous frequency distribution.
• Analysis of frequency distribution.

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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.

Median is the middle number of all the observation.

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

So, Median = 3020 and n = 7

By using the formula to calculate Mean Deviation,

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

= 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.

Median is the middle number of all the observation.

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

Here the Number of observations are Even then Median = (46+48)/2 = 47

Median = 47 and n = 10

By using the formula to calculate Mean Deviation,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$
 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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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.

Median is the middle number of all the observation.

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

Here the Number of observations are Even then Median = (42+44)/2 = 43

Median = 43 and n = 10

By using the formula to calculate Mean Deviation,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$
 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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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.

Median is the middle number of all the observation.

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

Here the Number of observations are Even then Median = (28+29)/2 = 28.5

Median = 28.5 and n = 10

By using the formula to calculate Mean Deviation,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$
 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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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.

Median is the middle number of all the observation.

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

Here the Number of observations are Even then Median = (47+48)/2 = 47.5

Median = 47.5 and n = 10

By using the formula to calculate Mean Deviation,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$
 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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

Where, |di| = |xi – M|

 xi |di| = |xi – 4400| 4000 400 4200 200 4400 0 4600 200 4800 400 Total 1200
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

Where, |di| = |xi – M|

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

= 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/10 × 167.4

= 16.74

∴ The Mean Deviation is 16.74.

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

We know that,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/10 × 166.4

= 16.64

∴ The Mean Deviation is 16.64

5. In question 1(iii), (iv), (v) find the number of observations lying between $$\overline{X} – M.D$$ and $$\overline{X} + M.D$$, 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,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/10 × 90

= 9

Now

$$\overline{X} – M.D$$ = 45.5 – 9 = 36.5

$$\overline{X} + M.D$$ = 45.5 + 9 = 54.5

So, There are total 6 observation between $$\overline{X} – M.D$$ and $$\overline{X} + M.D$$

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

We know that,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/10 × 48.8

= 4.88

Now

So, There are total 5 observation between
and

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

We know that,

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/10 × 86.8

= 8.68

Now

#### 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

 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

= 234/26

= 9

= 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,

 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

= 350/25

= 14

= 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,

 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

= 4000/80

= 50

= 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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.

Median is the middle term of the 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
$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

= 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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,

= 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/106 × 1272.6

= 12.005

∴ The mean deviation is 12.005

3. Compute mean deviation from 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,

= 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/110 × 1644

= 14.94

∴ The mean deviation is 14.94

4. The age distribution of 100 life-insurance policy holders 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

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 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 Mean,

= 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

Mean deviation from Median = 478/50 = 9.56

And, Mean deviation from Median = 472/50 = 9.44

∴ The Mean Deviation from the median is 9.56 and from 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

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

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

Solution:

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

Solution:

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:

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:

∴ The mean of new observation is 24 and 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:

#### 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:

= 100 [1.857 – 0.0987]

= 100 [1.7583]

Var (X) = 175.83

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:

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:

(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:

4. Find the standard deviation for the following data:

(i)

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

Solution:

(ii)

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

Solution:

#### 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:

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:

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:

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:

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:

#### 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 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, Standard deviation of the distribution A (σ – 9)

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

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

And, 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:

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

Solution:

= 22.86

∴ Means are 35 and 22.86

4. Calculate 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:

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:

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:

### 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

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