Question

A sample of final exam scores is normally distributed with a mean equal to 25 and a variance equal to 16.

what percentage of scores are between 21 and 29?

Answer 1

The answer is "0.6827".

Explanation:

`\bold{X \to N(\mu = 25, \sigma^2 = 16)}`

`\to P(21 < X < 29) = P(\frac {21-25}{4} < (X- \mu)/(\sigma)< (29-25)/(4))`

`= P(-1 < Z< 1)\\\\= P(Z> -1)-P(Z>1)\\\\= 1-P(Z< -1)-P(Z>1)\\\\= 1-P(Z> 1)-P(Z>1)\\\\= 1- 2 * P(Z>1)\\\\=1-2 * 0.15866 \\\\ = 1- 0.31732\\\\=0.6827`