Question

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

Answer 1

Answer:

The answer is "0.6827".

Step-by-step explanation:

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

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

                              $= P(-1 < Z< 1)\\\\= P(Z> -1)-P(Z>1)\\\\= 1-P(Z< -1)-P(Z>1)\\\\= 1-P(Z> 1)-P(Z>1)\\\\= 1- 2 \times P(Z>1)\\\\=1-2 \times 0.15866 \\\\ = 1- 0.31732\\\\=0.6827$