Answer to Question #246197 in Statistics and Probability for Pearl

Question

A pizza parlour will deliver a pizza take-away order to the customer if s/he lives within a 5 kilometre radius from the pizza parlour. If the customer lives within this radius, it is found that the time taken to receive the pizza is normally distributed with a mean time of 45 minutes and a standard deviation of 8 minutes.

If the time to receiving the pizza exceeds a particular time period, the customer gets the pizza free of charge. What should the approximate time limit (to the nearest minute) until delivery for the pizza be if the pizza parlour wants no more than 10% of the customers to get a free pizza?

Answer 1

Let $X=$ the time taken to receive the pizza: $X\sim N(\mu, \sigma^2).$

Given $\mu=45\ min, \sigma=8\ min.$

$$P(X>x)=1-P(X\leq x)$$

$$=1-P(Z\leq \dfrac{x-\mu}{\sigma})=1-P(Z\leq \dfrac{x-45}{8})=0.1$$

$$P(Z\leq \dfrac{x-45}{8})=0.9$$

$$\dfrac{x-45}{8}\approx1.281552$$

$$x\approx45+8\cdot1.281551$$

$$x=55\ min$$

The approximate time limit should be $55\ min.$