Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.2
Question 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:
Median is the middle term of the observation in ascending order,
So, Median = 61
Let us assume,
xi =Heights in inches
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
= 1/197 × 336
= 1.70
Therefore, the mean deviation is 1.70.
Question 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:
Median is the middle term of the observation in ascending order,
We know, Median is the even term, (3+5)/2 = 4
So, Median = 8
Let us assume
xi =Number of calls
fi = Frequency
N = 245
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 = 245 Total = 366
= 1/245 × 336
= 1.49
Therefore, mean deviation is 1.49.
Question 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:
Calculating the median,
We know, Number of observations, N = 50
Median = (50)/2 = 25
Therefore, 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
= 1/50 × 136
= 2.72
Therefore, the mean deviation is 2.72.
Question 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 |
(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 |
Solution:
(i) We know,
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 Now, Mean = 234/26
= 9
= 88/26
= 3.3
∴ The mean deviation is 3.3
(ii) We know,
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 Mean = 350/25
= 14
= 158/25
= 6.32
∴ The mean deviation is 6.32
(iii) We know,
= 4000/80
= 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
= 1280/80
= 16
∴ The mean deviation is 16
Question 5. Find the mean deviation 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 |
Solution:
(i) We know,
Number of observations, N = 21
Median = (21)/2 = 10.5
Therefore, 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
= 1/21 × 108
= 5.14
∴ The mean deviation is 5.14
(ii) We know,
Number of observations, N =50
Median = (50)/2 = 25
Therefore, 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
= 1/50 × 625
= 12.5
Therefore, the mean deviation is 12.5
(iii) We know,
Number of observations, 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
= 1/20 × 25
= 1.25
∴ The mean deviation is 1.25
Please Login to comment...