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Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d. - Mathematics

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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.

योग

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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अध्याय 15: Statistics - Exercise [पृष्ठ २८०]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 15 Statistics
Exercise | Q 20 | पृष्ठ २८०

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