Question 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:
Calculating the median:
Median is the middle term of the observation in ascending order, Xi,
Here, the middle term is 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 Therefore, Median = 25
Mean Deviation,
MD= 1/50 × 450
= 9
Therefore, mean deviation is 9.
Question 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 |
(ii)
| Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |
| Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |
Solution:
(i)
Mean = 17900/50
= 358
Also, the number of observations, N=50
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
= 1/50 × 7896
= 157.92
Therefore, mean deviation is 157.92.
(ii)
Mean = 13630/106
= 128.58
Also, the number of observations, N=106
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
= 1/106 × 1272.6
= 12.005
Question 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:
= 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
= 1/110 × 1644
= 14.94
Therefore, mean deviation is 14.94.
Question 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:
Number of observations, N = 96
So, N/2 = 96/2 = 48
The cumulative frequency just greater than 48 is 59, and therefore, the corresponding value of x is 38.25
Hence, 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 Number of observations, N = 96.
= 1/96 × 980.5
= 10.21
∴ The mean deviation is 10.21
Question 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:
Number of observations, 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.
Therefore, Median = 28
Now,
= 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 the median of observation = 478/50 = 9.56
And, Mean deviation from mean of observation = 472/50 = 9.44
∴ The mean deviation from the median is 9.56 and from the mean is 9.44.
Question 6. Calculate mean deviation about median age for the age distribution of 100 persons given below:
| Age | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 45-50 | 50-55 |
| Number of persons | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
Solution:
Converting the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
Age xi fi Cumulative Frequency |di| = |xi –38| fi |di| 15.5-20.5 18 5 5 20 100 20.5-25.5 23 6 11 15 90 25.5-30.5 28 12 23 10 120 30.5-35.5 33 14 37 5 70 35.5-40.5 38 26 63 0 0 40.5-45.5 43 12 75 5 60 45.5-50.5 48 16 91 10 160 50.5-55.5 53 9 100 15 135 N = 100 Total = 735 We have, N = 100
So, N/2 = 100/2 = 50
The cumulative frequency just greater than N/2 is 63 and the corresponding class is 35.5-40.5.
l=35.5, f=26, h= 5, F =37
Therefore, Median = l + (N/2 – F)/f * h = 35.5 + 50-37/26 * 5 =38
Question 7. Calculate the mean deviation from the median of the following data:
| Class interval | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |
| Frequency | 4 | 6 | 8 | 5 | 2 |
Solution:
fi xi fixi |xi-9.2| fi|xi-9.2| 0-4 4 2 8 7.2 28.8 4-8 6 6 36 3.2 19.2 8-12 8 10 80 0.8 6.4 12-16 5 14 70 4.8 24.0 16-20 2 18 36 8.8 17.6 N=25 Total=230 total = 96.0 Mean = 230/25 =9.2
Mean Deviation=96/25 = 3.84
Question 8. Calculate the mean deviation from the median of the following data:
| Class interval | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
| Frequency | 4 | 5 | 3 | 6 | 2 |
Solution:
fi xi fixi |xi-14.1| fi|xi-14.1| 0-6 4 3 12 11.1 44.4 6-12 5 9 45 5.1 25.5 12-18 3 15 45 0.9 2.7 18-24 6 21 126 6.9 41.4 24-30 2 27 54 12.9 25.8 N=20 Total=282 total = 139.8 Mean = 282/20 =14.1
Mean Deviation= 139.8/20 = 6.99