Question -
| 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 -
(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