Question -
Answer -
Let the assumed mean,A = 92.5 and h = 5
Let us make the tableof the given data and append other columns after calculations.
| Height (class) | Number of children Frequency fi | Midpoint Xi | Yi = (xi – A)/h | Yi2 | fiyi | fiyi2 |
| 70 – 75 | 2 | 72.5 | -4 | 16 | -12 | 48 |
| 75 – 80 | 1 | 77.5 | -3 | 9 | -12 | 36 |
| 80 – 85 | 12 | 82.5 | -2 | 4 | -14 | 28 |
| 85 – 90 | 29 | 87.5 | -1 | 1 | -7 | 7 |
| 90 – 95 | 25 | 92.5 | 0 | 0 | 0 | 0 |
| 95 – 100 | 12 | 97.5 | 1 | 1 | 9 | 9 |
| 100 – 105 | 10 | 102.5 | 2 | 4 | 12 | 24 |
| 105 – 110 | 4 | 107.5 | 3 | 9 | 18 | 54 |
| 110 – 115 | 5 | 112.5 | 4 | 16 | 12 | 48 |
| N = 60 | | | | 6 | 254 |
Mean,
Where, A = 92.5, h = 5
So, x̅ = 92.5 +((6/60) × 5)
= 92.5 + ½
= 92.5 + 0.5
= 93
Then, Variance,
σ2 =(52/602) [60(254) – 62]
= (1/144) [15240 – 36]
= 15204/144
= 1267/12
= 105.583
Hence, standarddeviation = σ = √105.583
= 10.275
∴ Mean = 93,variance = 105.583 and Standard Deviation = 10.275