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.

Explanation

Incorrect mean is `40`. There are `200` candidates.

`:. barX = 40`

So, `sum x_i = 40**200 = 8000`

This is incorrect `sum x_i` as `43` was misread as `34`.

`:.` Corect `sum x_i = 8000-34+43 = 8009`

`:.` Correct mean `= 8009/ 200 = 40.045`

Now, we will find the correct standard deviation.

We know,

...