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Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.

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Question

Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.

Sum
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Solution

Given that `barx = 50, n = 100` and S.D. `(sigma) = 4`

`barx = (sumx_i)/N`

⇒ 50 = `(sumx_i)/100`

⇒ `sumx_i` = 5000

And variance `sigma^2 = (sumf_ix_i^2)/N - ((sumf_ix_i)/N)^2`

`(4)^2 = (sumf_ix_i^2)/100 - (50)^2`

⇒ 16 = `(sumf_ix_i^2)/100 - 2500`

∴ `sumf_ix_i^2 = (2500 + 16) xx 100`

⇒ `sumf_ix_i^2 = 2516 xx 100` = 251600

Hence, the required sum are 5000 and 251600.

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Chapter 15: Statistics - Exercise [Page 279]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 15 Statistics
Exercise | Q 13 | Page 279

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