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The mean and standard deviation of a set of n1 observations are x¯1 and s1, respectively while the mean and standard deviation of another set of n2 observations are x¯2 and s2 - Mathematics

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प्रश्न

The mean and standard deviation of a set of n1 observations are `barx_1` and s1, respectively while the mean and standard deviation of another set of n2 observations are `barx_2` and  s2, respectively. Show that the standard deviation of the combined set of (n1 + n2) observations is given by

S.D. = `sqrt((n_1(s_1)^2 + n_2(s_2)^2)/(n_1 + n_2) + (n_1n_2 (barx_1 - barx_2)^2)/(n_1 + n_2)^2)`

बेरीज
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उत्तर

Let xi' = 1, 2, 3, 4, ..., n1

And yj' = 1, 2, 3, 4, ..., n2

∴ `barx_1 = 1/n_1 sum_(i = 1)^n x_i`

 And `barx_2 = 1/n_2 sum_(j = 1)^n y_j`

⇒ `sigma_1^2 = 1/n_1 sum_(i = 1)^(n_1) (x_i - barx_1)^2`

And `sigma_2^2 = 1/n_2 sum_(j = 1)^(n_2) (y_i - barx_2)^2`

Now mean of the combined series is given by

`barx = 1/(n_1 + n_2) [sum_(i = 1)^(n_1) + sum_(j = 1)^(n_2) y_j]`

= `(n_1 barx_1 + n_2 x_2)/(n_1 + n_2)`

Therefore, `sigma^2` of the combined series is

`sigma^2 = 1/(n_1 + n_2) [sum_(i = 1)^(n_1) (x_i - barx)^2 + sum_(j = 1)^(n_2) (y_j - barx)^2]`

Now, `sum_(i = 1)^(n_1) (x_i - barx)^2 = sum_(i = 1)^(n_1) (x_i - barx_j + bar_j - barx)^2`

= `sum_(i = 1)^(n_1) (x_i - x_j)^2 + n_1 (barx_j - barx)^2 + 2(barx_j - barx) sum_(i = 1)^(n_1) (x_i - barx_j)^2`

But `sum_(i = 1)^n (x_i - barx_i)` = 0

∵ The algebraic sum of the deviation of values of first series from their mean is zero

Also `sum_(i = 1)^(n_1) (x_i - barx)^2 = n_1s_1^2 + n_1(barx_1 - barx)^2`

= `n_1s_1^2 + n_1d_1^2`

Where `d_1 = (barx_1 - barx)`

Similarly, we have

`sum_(j = 1)^(n_2) (y_j - barx)^2 = sum_(j = 1)^(n_2) (y_j - barx_i + barx_i - barx)^2`

= `n_2s_2^2 + n_2d_2^2`

Where `d_2 = (barx_2 - barx)`

Now combined Standard Deviation (S.D.)

`sigma = sqrt((n_1(s_1^2 + d_1^2) + n_2(s_2^2 + d_2^2))/(n_1 + n_2))`

Where `d_1 = barx_1 - barx`

= `barx_1 - ((n_1barx_1 + n_2 barx_2)/(n_1 + n_2))`

= `(n_2(barx_1 - barx_2))/(n_1 + n_2)`

And `d_2 = barx_2 - barx`

= `barx_2 - ((n_1barx_1 + b_2barx_2)/(n_1 + n_2))`

= `(n_1(barx_2 - barx_1))/(n_1 + n_2)`

∴ `sigma^2 = 1/(n_1 + n_2)[n_1s_1^2 + n_2s_2^2 + (n_1n_2^2(barx_1 - barx_2)^2)/(n_1 + n_2)^2 + (n_2n_1^2(barx_2 - barx_1)^2)/(n_1 + n_2)^2]`

So, `sigma = sqrt((n_1s_1^2 + n_2s_2^2)/(n_1 + n_2) + (n_1n_2(barx_1 - barx_2)^2)/(n_1 + n_2)^2`

Hence proved.

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पाठ 15: Statistics - Exercise [पृष्ठ २७८]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 15 Statistics
Exercise | Q 7 | पृष्ठ २७८

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