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.

PLZZZZZZZ HELP

 Let the hundred observations be

  x1, x2, x3, …....x99, x100

  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  (x1 + x2 + x3 + x99) = 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 x21  + x22  + x23 +........+x299  /100 = (5.1) +  402 - 25

  Now  (SD)2 =( x21  + x22  + x23 +........+x299  /100  + x2100 / 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

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