RD Sharma Solutions for Class 11 Maths Chapter 32–Statistics

Class 11: Maths Chapter 32 solutions. Complete Class 11 Maths Chapter 32 Notes.

RD Sharma Solutions for Class 11 Maths Chapter 32–Statistics

RD Sharma 11th Maths Chapter 32, Class 11 Maths Chapter 32 solutions

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=1n∑ni=1|di|MD=1n∑i=1n|di|

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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:

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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – M|

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – M|

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – M|

The number of observations are Even then Median = (40+52.3)/2 = 46.15

Median = 46.15

Number of observations, ‘n’ = 10

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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 ¯¯¯¯¯X–M.DX¯–M.D and ¯¯¯¯¯X+M.DX¯+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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/10 × 90

= 9

Now¯¯¯¯¯X–M.DX¯–M.D = 45.5 – 9 = 36.5¯¯¯¯¯X+M.DX¯+M.D = 45.5 + 9 = 54.5

So, There are total 6 observation between ¯¯¯¯¯X–M.DX¯–M.D and ¯¯¯¯¯X+M.DX¯+M.D

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

We know that,MD=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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=1n∑ni=1|di|MD=1n∑i=1n|di|

Where, |d_i| = |x_i – 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

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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:

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 x_i =Heights in inches

And, f_i = Number of students

N=197MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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:

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 x_i =Number of calls

And, f_i = Frequency

N = 245MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/245 × 336

= 1.49

∴ The mean deviation is 1.49.

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

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

N = 50MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/50 × 136

= 2.72

∴ The mean deviation is 2.72.

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

(i)

Solution:

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

By using the formula,

= 234/26

= 9

= 88/26

= 3.3

∴ The mean deviation is 3.3

(ii)

Solution:

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

By using the formula,

= 350/25

= 14

= 158/25

= 6.32

∴ The mean deviation is 6.32

(iii)

Solution:

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

By using the formula,

= 4000/80

= 50

= 1280/80

= 16

∴ The mean deviation is 16

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

(i)

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

N = 21MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/21 × 108

= 5.14

∴ The mean deviation is 5.14

(ii)

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

N = 50MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/50 × 625

= 12.5

∴ The mean deviation is 12.5

(iii)

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

N = 20MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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:

Solution:

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

Median is the middle term of the X_i,

Here, the middle term is 25

So, Median = 25

MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/50 × 450

= 9

∴ The mean deviation is 9

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

(i)

Solution:

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

By using the formula,

= 17900/50

= 358

N = 50MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/50 × 7896

= 157.92

∴ The mean deviation is 157.92

(ii)

Solution:

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

By using the formula,

= 13630/106

= 128.58

N = 106MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/106 × 1272.6

= 12.005

∴ The mean deviation is 12.005

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

Solution:

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

By using the formula,

= 5390/110

= 49

N = 110MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 1/110 × 1644

= 14.94

∴ The mean deviation is 14.94

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

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

N = 96MD=1n∑ni=1|di|MD=1n∑i=1n|di|

= 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:

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

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:

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

Calculate the mean and variance.

Solution:

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

Solution:

(ii)

Solution:

4. Find the standard deviation for the following data:

(i)

Solution:

(ii)

Solution:

EXERCISE 32.6 PAGE NO: 32.41

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

Solution:

2. Calculate the standard deviation for the following data:

Solution:

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

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

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

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:

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:

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:

(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:

Which of the distributions is more variable?

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 32: Download PDF

RD Sharma Solutions for Class 11 Maths Chapter 32–Statistics

Download PDF: RD Sharma Solutions for Class 11 Maths Chapter 32–Statistics PDF

Chapterwise RD Sharma Solutions for Class 11 Maths :

• Chapter 1–Sets
• Chapter 2–Relations
• Chapter 3–Functions
• Chapter 4–Measurement of Angles
• Chapter 5–Trigonometric Functions
• Chapter 6–Graphs of Trigonometric Functions
• Chapter 7–Values of Trigonometric Functions at Sum or Difference of Angles
• Chapter 8–Transformation Formulae
• Chapter 9–Values of Trigonometric Functions at Multiples and Submultiples of an Angle
• Chapter 10–Sine and Cosine Formulae and their Applications
• Chapter 11–Trigonometric Equations
• Chapter 12–Mathematical Induction
• Chapter 13–Complex Numbers
• Chapter 14–Quadratic Equations
• Chapter 15–Linear Inequations
• Chapter 16–Permutations
• Chapter 17–Combinations
• Chapter 18–Binomial Theorem
• Chapter 19–Arithmetic Progressions
• Chapter 20–Geometric Progressions
• Chapter 21–Some Special Series
• Chapter 22–Brief review of Cartesian System of Rectangular Coordinates
• Chapter 23–The Straight Lines
• Chapter 24–The Circle
• Chapter 25–Parabola
• Chapter 26–Ellipse
• Chapter 27–Hyperbola
• Chapter 28–Introduction to Three Dimensional Coordinate Geometry
• Chapter 29–Limits
• Chapter 30–Derivatives
• Chapter 31–Mathematical Reasoning
• Chapter 32–Statistics
• Chapter 33–Probability

About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.

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