Answer to Question #268155 in Statistics and Probability for shabs

Question

duration of a lunch break for one employee at a manufacturing company follows a normal distribution with mean μ minutes and standard deviation 5 minutes. The probability that a lunch break lasts for more than 52 minutes is 0.25. Find the value of μ.

A. 55.37

B. 46.25

C. 57.75

D. 48.63

Answer 1

$\begin{aligned} &P[x>52]=0.25 \\ &P\left\{\frac{x-\mu}{b}>\frac{(52-\mu)}{b}\right\}=0.25 \\ \end{aligned}$

Use the Normal distribution approximation

$\begin{aligned} z &=\frac{x-\mu}{b} \\ p\left\{z_{2}<\frac{52-\mu}{5}\right\} &=1-0.25 \\ p\left\{z<\frac{52-\mu}{5}\right\} &=0.75 \\ \frac{52-\mu}{5} &=0.68 \quad \rightarrow \text { from z-table } \\ 52-\mu &=5 \times 0.68 \\ 52-\mu &=3.4 \\ \mu &=52.3 .4 \\ \mu &=48.6 \end{aligned}$

Option (D) is correct