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प्रश्न
Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.
उत्तर
`x_i` | `x_i - a` | `(x_i - a)^2` |
`a` | 0 | 0 |
`a + d` | `d` | `d^2` |
`a + 2d` | `2d` | `4d^2` |
— | — | — |
— | — | — |
— | — | — |
`a + (n - 1)d` | `(n - 1)d` | `(n - 1)^2d^2` |
We know that `sumx_i = n/2 [2a + (n - 1)d]`
∴ Mean = `(sumx_i)/n`
= `1/n[n/2 {2a + (n - 1)d}]`
= `1/2[2a + (n - 1)d]`
= `a + (n - 1)/2 d`
∴ `sum(x_i - a) = d[1 + 2 + 3 + ... + (n - 1)]`
= `(d(n - 1)n)/2`
And `sum(x_i - a)^2 = d^2[1^2 + 2^2 + 3^2 + ... + (n - 1)^2]`
= `d^2 * (n(n - 1)(2n - 1))/6`
∴ `sigma = sqrt((sum(x_i - a)^2)/n - ((sum(x_i - a))/n)^2`
= `sqrt((d^2n(n - 1)(2n - 1))/(6n) - ((dn(n - 1))/(2n))^2`
= `sqrt((d^2(n - 1)(2n - 1))/6 - (d^2(n - 1)^2)/4`
= `dsqrt((n - 1)/2((2n - 1)/3 - (n - 1)/3))`
= `dsqrt((n - 1)/2 [(4n - 2 - 3n + 3)/6]`
= `dsqrt(((n - 1)/2)((n + 1)/6)`
= `dsqrt((n^2 - 1)/12)`
Hence, the required S.D. = `dsqrt((n^2 - 1)/12)`.
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