Question

The score 28 and 40 bound the middle 95% of the distribution. What percent is left at the tails of the distribution

Answer 1

$$100\%-95\%=5\%$$

5% is left at the tails of the distribution.

$$P(Z<z_1)=0.025$$

$$z_1=-1.96$$

$$P(Z>z_2)=0.025$$

$$z_1=1.96$$

$$\dfrac{28-\mu}{\sigma}=-1.96$$

$$\dfrac{40-\mu}{\sigma}=1.96$$

Then

$$\mu-28=40-\mu$$

$$\mu=34$$

$$\dfrac{40-34}{\sigma}=1.96$$

$$\sigma=\dfrac{6}{1.96}=3.0612$$