a student obtained the mean n standaard deviation of 100 observation as 40 n 5.1 respectively it was later found that one observation was wrongly copied as 50 the correct figure being 40. find the correct mean and S.D.

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Let the hundred observations be

x_{1}, x_{2}, x_{3}, …....x_{99}, x_{100}

Suppose that the correct value of x100 was 40 and that by mistake the student had copied it as 50. Then

`(x1 + x2 + x3 + x99 + 50)/100` = 40

Therefore (x_{1} + x_{2} + x_{3} + x_{99}) = 4000 – 50

= 3950

Now, `(x1 + x2 + x3 + x99 + 40) / 100 ` = `(3950 + 40 )/ 100`

= `3990 / 100`

= 39.9

Which is the true mean.

(SD)^{2} = `((x2) / (n - M^2))`

(5.1)^{2} =` ((x^2 1 + x^2 2 + x^2 3 +........+x^2 99 + x^2 100) /(100 - 40^2)) ` ^{ }^{ }

= `((x^2 1 + x^2 2 + x^2 3 +........+x^2 99 ) /100)` - `(502 / (100 - 40^2))`

Therefore x^{2}_{1} + x^{2}_{2} + x^{2}_{3} +........+x^{2}_{99} /100 = (5.1)^{2 } + 40^{2} - 25

Now (SD)^{2} =( x^{2}_{1} + x^{2}_{2} + x^{2}_{3} +........+x^{2}_{99} /100 + x^{2}_{100} / 100) - ( 39.9)^{2}

= `((( 5.1)^2 + 402 - 25 + 40^2) / (100 - (39.9)^2 )) ` ^{ }

= 26.01 + 1600 -25 + 16 - (39.9)^{2}

= 17.01 + 1600 - (39.9)^{2}

= 1617.01 - 1592.01

= 25

Therefore = 5, which is the true for standard deviation