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
Access answers to RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics
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,
x_{i} | |d_{i}| = |x_{i} – 3020| |
3011 | 9 |
2780 | 240 |
3020 | 0 |
2354 | 666 |
3541 | 521 |
4150 | 1130 |
5000 | 1980 |
Total | 4546 |
= 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,
x_{i} | |d_{i}| = |x_{i} – 47| |
38 | 9 |
70 | 23 |
48 | 1 |
34 | 13 |
42 | 5 |
55 | 8 |
63 | 16 |
46 | 1 |
54 | 7 |
44 | 3 |
Total | 86 |
= 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,
x_{i} | |d_{i}| = |x_{i} – 43| |
30 | 13 |
34 | 9 |
38 | 5 |
40 | 3 |
42 | 1 |
44 | 1 |
50 | 7 |
51 | 8 |
60 | 17 |
66 | 23 |
Total | 87 |
= 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,
x_{i} | |d_{i}| = |x_{i} – 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 |
= 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,
x_{i} | |d_{i}| = |x_{i} – 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 |
= 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,
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
x_{i} | |d_{i}| = |x_{i} – 10| |
4 | 6 |
7 | 3 |
8 | 2 |
9 | 1 |
10 | 0 |
12 | 2 |
13 | 3 |
17 | 7 |
Total | 24 |
= 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,
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
x_{i} | |d_{i}| = |x_{i} – 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 |
= 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,
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
x_{i} | |d_{i}| = |x_{i} – 50| |
38 | 12 |
70 | 20 |
48 | 2 |
40 | 10 |
42 | 8 |
55 | 5 |
63 | 13 |
46 | 4 |
54 | 4 |
44 | 6 |
Total | 84 |
= 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,
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
x_{i} | |d_{i}| = |x_{i} – 50| |
36 | 14 |
72 | 22 |
46 | 4 |
42 | 8 |
60 | 10 |
45 | 5 |
53 | 3 |
46 | 4 |
51 | 1 |
49 | 1 |
Total | 72 |
= 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,
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
x_{i} | |d_{i}| = |x_{i} – 55| |
57 | 2 |
64 | 9 |
43 | 12 |
67 | 12 |
49 | 6 |
59 | 4 |
44 | 11 |
47 | 8 |
61 | 6 |
59 | 4 |
Total | 74 |
= 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,
Where, |d_{i}| = |x_{i} – M|
x_{i} | |d_{i}| = |x_{i} – 4400| |
4000 | 400 |
4200 | 200 |
4400 | 0 |
4600 | 200 |
4800 | 400 |
Total | 1200 |
= 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,
Where, |d_{i}| = |x_{i} – M|
x_{i} | |d_{i}| = |x_{i} – 4400| |
3800 | 600 |
4000 | 400 |
4200 | 200 |
4400 | 0 |
4600 | 200 |
4800 | 400 |
5800 | 1400 |
Total | 3200 |
= 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,
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
x_{i} | |d_{i}| = |x_{i} – 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 |
= 1/10 × 167.4
= 16.74
∴ The Mean Deviation is 16.74.
(ii) Find the mean deviation from the mean also.
We know that,
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
x_{i} | |d_{i}| = |x_{i} – 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 |
= 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
Solution:
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
We know that,
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
x_{i} | |d_{i}| = |x_{i} – 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 |
= 1/10 × 90
= 9
Now
So, there are a total of 6 observations between
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
We know that,
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
x_{i} | |d_{i}| = |x_{i} – 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 |
= 1/10 × 48.8
= 4.88
Now
So, there are a total of 5 observations between
and
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
We know that,
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
x_{i} | |d_{i}| = |x_{i} – 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 |
= 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 x_{i} =Heights in inches
And, f_{i} = Number of students
x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M|
= |x_{i} – 61| |
f_{i} |d_{i}| |
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
= 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 x_{i} =Number of calls
And, f_{i} = Frequency
x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M|
= |x_{i} – 61| |
f_{i} |d_{i}| |
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
= 1/245 × 336
= 1.49
∴ The mean deviation is 1.49.
3. Calculate the mean deviation about the median of the following frequency distribution:
x_{i} | 5 | 7 | 9 | 11 | 13 | 15 | 17 |
f_{i} | 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.
x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M|
= |x_{i} – 61| |
f_{i} |d_{i}| |
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
= 1/50 × 136
= 2.72
∴ The mean deviation is 2.72.
4. Find the mean deviation from the mean for the following data:
(i)
x_{i} | 5 | 7 | 9 | 10 | 12 | 15 |
f_{i} | 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,
x_{i} | f_{i} | Cumulative Frequency (x_{i}f_{i}) | |d_{i}| = |x_{i} – Mean| | f_{i} |d_{i}| |
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)
x_{i} | 5 | 10 | 15 | 20 | 25 |
f_{i} | 7 | 4 | 6 | 3 | 5 |
Solution:
To find the mean deviation from the mean, firstly, let us calculate the mean.
By using the formula,
x_{i} | f_{i} | Cumulative Frequency (x_{i}f_{i}) | |d_{i}| = |x_{i} – Mean| | f_{i} |d_{i}| |
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)
x_{i} | 10 | 30 | 50 | 70 | 90 |
f_{i} | 4 | 24 | 28 | 16 | 8 |
Solution:
To find the mean deviation from the mean, firstly, let us calculate the mean.
By using the formula,
x_{i} | f_{i} | Cumulative Frequency (x_{i}f_{i}) | |d_{i}| = |x_{i} – Mean| | f_{i} |d_{i}| |
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)
x_{i} | 15 | 21 | 27 | 30 |
f_{i} | 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.
x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M| | f_{i} |d_{i}| |
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
= 1/21 × 108
= 5.14
∴ The mean deviation is 5.14
(ii)
x_{i} | 74 | 89 | 42 | 54 | 91 | 94 | 35 |
f_{i} | 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.
x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M| | f_{i} |d_{i}| |
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
= 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.
x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M| | f_{i} |d_{i}| |
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
= 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 X_{i.}
Here, the middle term is 25.
So, Median = 25
Class Interval | x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M| | f_{i} |d_{i}| |
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 |
= 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 | x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M| | f_{i} |d_{i}| |
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
= 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 | x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M| | f_{i} |d_{i}| |
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
= 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,
= 5390/110
= 49
Class Interval | x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M| | f_{i} |d_{i}| |
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
= 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 | x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – M| | f_{i} |d_{i}| |
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
= 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,
= 1350/50
= 27
Class Interval | x_{i} | f_{i} | Cumulative Frequency | |d_{i}| = |x_{i} – Median| | f_{i} |d_{i}| | F_{i}X_{i} | |X_{i} – Mean| | F_{i }|X_{i} – 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
(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 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:
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 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 (σ^{2} =18)
So, the standard deviation of the distribution A (σ – 9)
Similarly, the variation of the distribution of wages in plant B (σ^{2 }=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:
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:
= 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:
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:
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