Question

The mean and standard deviation of 50 observations are 15 and 2 respectively. If it was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70 . If the correct mean is 16 , then the correct variance is equal to ?

Answer 1

Let the correct observation be x and incorrect observation be x'.

∴ \(x+ x' = 70 \)    .....(1)  (Given)

Incorrect sum of obs. = Mean x 50 = 15 x 50 = 750

∴ Correct sum = Incorrect sum - x' + x

50 x Correct mean = 750 - x' + x

⇒ \(750 - x' + x = 50 \times 16\)

⇒ \(x - x' = 800 - 750 = 50\)   .....(2)

Adding (1) & (2), we get

\(2x = 120\)

⇒ \(x =60\) & \(x' = 10\)

Incorrect standard deviation = 2 ⇒ incorrect variance = 4

⇒ \(\frac{(\sum{x_i}^2)_{incorect}}{n} - (Incorrect\, mean)^2 = 4\)

⇒ \((\sum {x_1}^2)_{incorrect }= 50 (4 + 15^2) = 50 \times 229 = 11450\)

\(\therefore(\sum {x_i}^2)_{correct} = (\sum {x_i}^2)_{incorrect} - x'^2 + x^2\)

\(= 11450 - 100 + 3600\)

\(= 14950\)

∴ Correct variance = \(\frac{(\sum {x_i}^2)_{correct}}n - (correct\, mean)^2\)

\(=\frac{ 14950 }{50} - 16^2\)

\(= 299 - 256\)

\(= 43\)