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.

Given: 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

To find: the correct mean and the variance.

As per given criteria,

Number of reading, n=10

Mean of the given readings before correction,

But we know,

Substituting the corresponding values, we get

⇒ ∑x_i=45× 10=450

It is said one reading 25 was wrongly taken as 52,

So ∑x_i=450-52+25=423

So the correct mean after correction is

Also given the variance of the 10 readings is 16 before correction,

i.e., σ²=16

But we know

Substituting the corresponding values, we get

It is said one reading 25 was wrongly taken as 52, so

So the correct variance after correction is

σ²=1833.1-(42.3)²=1833.1-1789.29

σ²=43.81

Hence the corrected mean and variance is 42.3 and 43.81 respectively.