Question

While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Answer 1

Given that `n = 10,  barx = 45` and `sigma^2 = 16`.

∴ `barx = (sumx_i)/n`

⇒ 45 = `(sumx_i)/10`

⇒ `sumx_i` = 450

Corrected `sumx_i` = 450 – 52 + 25

= 423

∴ Correct Mean `barx = 423/10` = 42.3

And `sigma^2 = (sumx_i^2)/n - ((sumx_i)/n)^2`

⇒ 16 = `(sumx_i^2)/10 - (45)^2`

⇒ 16 = `(sumx_i^2)/10 - 2025`

⇒ `(sumx_i^2)/10` = 2041

∴ `sumx_i^2` = 20410

∴ Correct `sumx_i^2` = 20410 – (52)² + (25)²

= 20410 – 2704 + 625

= 18331

And corrected variance `sigma^2 = 18331/10 - (42.3)^2`

= 1833.1 – 1789.3

= 43.8

Hence the required mean = 42.3 and variance = 43.8