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

  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)²  +  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