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,

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