Class 11

Math

JEE Main Questions

Statistics

For a group of 200 candidates the mean and S.D. were found to be 40 and 15 respectively. Later on it was found that the score 43 was misread as 34. Find the correct mean and correct S.D.

$∴X=40$

So, $∑x_{i}=40∗200=8000$

This is incorrect $∑x_{i}$ as $43$ was misread as $34$.

$∴$ Corect $∑x_{i}=8000−34+43=8009$

$∴$ Correct mean $=2008009 =40.045$

Now, we will find the correct standard deviation.

We know,

$σ=N1 ∑(x_{i})_{2}−(x)_{2} $

When incorrect observations was present, then standard deviation was $15$.

$∴15=2001 ∑(x_{i})_{2}−(40)_{2} $

$⇒225=2001 ∑(x_{i})_{2}−1600$

$⇒365000=∑(x_{i})_{2}$

This is the sum when observations were incorrect.

$∴$ Correct sum$=365000−(34)_{2}+(43)_{2}=365693$

$∴$ Correct $∑(x_{i})_{2}=365693$

$∴σ=2001 (365693)−(40.045)_{2} =14.995$

So, the correct standard deviation is $14.995$.