RD Chapter 32 Statistics Ex 32.2 Solutions
Question - 1 : - Calculate the meandeviation 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 |
Answer - 1 : -
To find the mean deviation from the median, firstly let uscalculate 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=
= 1/197 × 336
= 1.70
∴ Themean deviation is 1.70.
Question - 2 : - The number oftelephone calls received at an exchange in 245 successive on2-minute intervalsis 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 meandeviation about the median.
Answer - 2 : -
To find the mean deviation from the median, firstly let uscalculate 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=
= 1/245 × 336
= 1.49
∴ Themean deviation is 1.49.
Question - 3 : - Calculate the meandeviation about the median of the following frequency distribution:
| xi | 5 | 7 | 9 | 11 | 13 | 15 | 17 |
| fi | 2 | 4 | 6 | 8 | 10 | 12 | 8 |
Answer - 3 : -
To find the mean deviation from the median, firstly let uscalculate 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=
= 1/50 × 136
= 2.72
∴ Themean deviation is 2.72.
Question - 4 : - Find the meandeviation from the mean for the following data:
(i)
| xi | 5 | 7 | 9 | 10 | 12 | 15 |
| fi | 8 | 6 | 2 | 2 | 2 | 6 |
(ii)
| xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
(iii)
| xi | 10 | 30 | 50 | 70 | 90 |
| fi | 4 | 24 | 28 | 16 | 8 |
Answer - 4 : -
(i)
To find the mean deviation from the mean, firstly let uscalculate 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
∴ Themean deviation is 3.3
(ii)
To find the mean deviation from the mean, firstly let uscalculate 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
∴ Themean deviation is 6.32
(iii)
To find the mean deviation from the mean, firstly let uscalculate 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
∴ Themean deviation is 16
Question - 5 : - Find the meandeviation from the median for the following data :
(i)
| xi | 15 | 21 | 27 | 30 |
| fi | 3 | 5 | 6 | 7 |
(ii)
| xi | 74 | 89 | 42 | 54 | 91 | 94 | 35 |
| fi | 20 | 12 | 2 | 4 | 5 | 3 | 4 |
(iii)
| Marks obtained | 10 | 11 | 12 | 14 | 15 |
| No. of students | 2 | 3 | 8 | 3 | 4 |
Answer - 5 : -
(i)
To find the mean deviation from the median, firstly let uscalculate 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=
= 1/21 × 108
= 5.14
∴ Themean deviation is 5.14
(ii)
To find the mean deviation from the median, firstly let uscalculate 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=
= 1/50 × 625
= 12.5
∴ Themean deviation is 12.5
(iii)
To find the mean deviation from the median, firstly let uscalculate 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=
= 1/20 × 25
= 1.25
∴ Themean deviation is 1.25