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प्रश्न
Find the variance and S.D. of the following frequency distribution which gives the distribution of 200 plants according to their height:
| Height (in cm) | 14 – 18 | 19 – 23 | 24 – 28 | 29 – 33 | 34 – 38 | 39 – 43 | 44 – 48 |
| No. of plants | 5 | 18 | 44 | 70 | 36 | 22 | 5 |
उत्तर
Since data is not continuous, we have to make it continuous.
Let u = `(x - "A")/"h" = (x - 31)/5`
Calculation of variance of u:
| Height (in cm) C. I. |
Mid value (xi) | No. of plants (fi) | ui = `(x_"i" - 31)/5` | fiui | fiui2 |
| 13.5 – 18.5 | 16 | 5 | – 3 | – 15 | 45 |
| 18.5 – 23.5 | 21 | 18 | – 2 | – 36 | 72 |
| 23.5 – 28.5 | 26 | 44 | – 1 | – 44 | 44 |
| 28.5 – 33.5 | 31 | 70 | 0 | 0 | 0 |
| 33.5 – 38.5 | 36 | 36 | 1 | 36 | 36 |
| 38.5 – 43.5 | 41 | 22 | 2 | 44 | 88 |
| 43.5 – 48.5 | 46 | 5 | 3 | 15 | 45 |
| N = 200 | `sumf_"i"u_"i"` = 0 | `sumf_"i"u_"i"^2` = 330 |
`bar(u) = (sumf_"i"u_"i")/"n" = 0/200` = 0
Var (u) = `sigma_u^2 = (sumf_"i"u_"i"^2)/"n" - (baru)^2`
= `330/200 - 0^2`
= 1.65
∴ Var (X) = h2 Var (u) = (5)2 × 1.65 = 25 × 1.65 = 41.25
∴ S.D. = `sigma_x = sqrt("Var(X)")`
= `sqrt(41.25)`
= 6.42
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