Question

suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points. the instructor wanted to \pass\ anyone who scored above 69. what proportion of exams will have passing scores? a. 0.1587 b. 0.75 c. 0.2266 d. 0.7734 e. -0.75

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 69$, $\mu=75$, and $\sigma = 8$. So $z=\frac{69 - 75}{8}=\frac{- 6}{8}=-0.75$.

Step2: Find the proportion

We want $P(X>69)$, which is equivalent to $P(Z>-0.75)$ in the standard normal distribution. Since $P(Z > z)=1 - P(Z\leq z)$, and looking up $P(Z\leq - 0.75)$ in the standard - normal table gives $0.2266$. Then $P(Z>-0.75)=1 - 0.2266 = 0.7734$.

Answer:

D. 0.7734