Question

Scores of an achievement test show that if follows a normal distribution. Its mean is 78 with a standard deviation of 8. Find the interval wherein the middle 80% of the scores lie?

Answer 1

$$\dfrac{1-0.8}{2}=0.1$$

$$P(X<x_1)=P(Z<\dfrac{x_1-\mu}{\sigma})$$

$$=P(Z<\dfrac{x_1-78}{8})=0.1$$

$$\dfrac{x_1-78}{8}\approx-1.281552$$

$$x_1=78-8(1.281552)$$

$$x_1\approx67.7476$$

$$P(X>x_2)=1-P(Z<\dfrac{x_2-\mu}{\sigma})$$

$$=1-P(Z<\dfrac{x_2-78}{8})=0.1$$

$$P(Z<\dfrac{x_2-78}{8})=0.9$$

$$\dfrac{x_2-78}{8}\approx1.281552$$

$$x_2=78+8(1.281552)$$

$$x_2\approx88.2524$$

Interval $(67.7476, 88.2524).$