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प्रश्न
Answer the following :
Given below is the frequency distribution of marks obtained by 100 students. Compute arithmetic mean and S.D.
| Marks | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| No. of students | 4 | 12 | 25 | 28 | 26 | 5 |
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उत्तर
Since data is not continuous, we have to make it continuous.
Let u = `(x - "A")/"h" = (x - 74.5)/10`
Calculation of variance of u:
| Marks C. I. |
Mid value (xi) |
No. of students (fi) |
ui = `(x_"i" - 74.5)/10` | fiui | fiui2 |
| 39.5 - 49.5 | 44.5 | 4 | −3 | −12 | 36 |
| 49.5 - 59.5 | 54.5 | 12 | −2 | −24 | 48 |
| 59.5 - 69.5 | 64.5 | 25 | −1 | −25 | 25 |
| 69.5 - 79.5 | 74.5 | 28 | 0 | 0 | 0 |
| 79.5 - 89.5 | 84.5 | 26 | 1 | 26 | 26 |
| 89.5 - 99.5 | 94.5 | 5 | 2 | 10 | 20 |
| N = 100 | `sumf_"i"u_"i"` = – 25 | `sumf_"i"u_"i"^2` = 155 |
`bar(u) = (sumf_"i"u_"i")/"N" = (-25)/100` = – 0.25
`bar(x) = bar(u) xx "h" + "A"`
= – 0.25 × 10 + 74.5
= 72
Var (u) = `sigma_u^2 = (sumf_"i"u_"i"^2)/"N" - (baru)^2`
= `155/100 - (-0.25)^2`
= 1.55 – 0.0625
= 1.4875
∴ Var (X) = h2 var (u) = (10)2 × 1.4875
= 100 × 1.4875
= 148.75
∴ S.D. = `sigma_x = sqrt("Var(X)")`
= `sqrt(148.75)`
= 12.2
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