Question -
Answer -
Let the assumed mean A= 64. Here h = 1
We obtain thefollowing table from the given data.
| Xi | Frequency fi | Yi = (xi – A)/h | Yi2 | fiyi | fiyi2 |
| 60 | 2 | -4 | 16 | -8 | 32 |
| 61 | 1 | -3 | 9 | -3 | 9 |
| 62 | 12 | -2 | 4 | -24 | 48 |
| 63 | 29 | -1 | 1 | -29 | 29 |
| 64 | 25 | 0 | 0 | 0 | 0 |
| 65 | 12 | 1 | 1 | 12 | 12 |
| 66 | 10 | 2 | 4 | 20 | 40 |
| 67 | 4 | 3 | 9 | 12 | 36 |
| 68 | 5 | 4 | 16 | 20 | 80 |
| | | | 0 | 286 |
Mean,
Where A = 64, h = 1
So, x̅ = 64 + ((0/100)× 1)
= 64 + 0
= 64
Then, variance,
σ2 =(12/1002) [100(286) – 02]
= (1/10000) [28600 –0]
= 28600/10000
= 2.86
Hence, standarddeviation = σ = √2.886
= 1.691
∴ Mean = 64 andStandard Deviation = 1.691