RD Chapter 32 Statistics Ex 32.3 Solutions
Question - 1 : - Compute the meandeviation from the median of the following distribution:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 5 | 10 | 20 | 5 | 10 |
Answer - 1 : -
To find the mean deviation from the median, firstly let uscalculate the median.
Median is the middle term of the Xi,
Here, the middle term is 25
So, Median = 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 |
MD=
= 1/50 × 450
= 9
∴ Themean deviation is 9
Question - 2 : - Find the meandeviation 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 |
Answer - 2 : -
(i)
To find the mean deviation from the mean, firstly let uscalculate the mean.
By using the formula,
= 17900/50
= 358
| 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 |
N = 50
MD=
n∑= 1/50 × 7896
= 157.92
∴ Themean deviation is 157.92
(ii)
To find the mean deviation from the mean, firstly let uscalculate the mean.
By using the formula,
= 13630/106
= 128.58
| 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 |
N = 106
MD=
= 1/106 × 1272.6
= 12.005
∴ Themean deviation is 12.005
Question - 3 : - Compute meandeviation 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 |
Answer - 3 : -
To find the mean deviation from the mean, firstly let uscalculate the mean.
By using the formula,
= 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 |
N = 110
MD=
= 1/110 × 1644
= 14.94
∴ Themean deviation is 14.94
Question - 4 : - The agedistribution of 100 life-insurance policy holders 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 meandeviation from the median age.
Answer - 4 : -
To find the mean deviation from the median, firstly let uscalculate the median.
N = 96
So, N/2 = 96/2 = 48
The cumulative frequency just greater than 48 is 59, and thecorresponding value of x is 38.25
So, 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 |
N = 96
MD=
= 1/96 × 980.5
= 10.21
∴ Themean deviation is 10.21
Question - 5 : - Find the meandeviation 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 |
Answer - 5 : -
To find the mean deviation from the median, firstly let uscalculate the median.
N = 50
So, N/2 = 50/2 = 25
The cumulative frequency just greater than 25 is 58, and thecorresponding value of x is 28
So, Median = 28
By using the formula to calculate Mean,
= 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 Median = 478/50 = 9.56
And, Mean deviation from Median = 472/50 = 9.44
∴ TheMean Deviation from the median is 9.56 and from mean is 9.44.