Answer to Question #249177 in Statistics and Probability for Bongs

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. Paul has relocated and now lives within the 5 kilometre radius. He orders a pizza from the pizza parlour. The probability that it takes between 43 and 49 minutes for him to receive his pizza is

Answer 1

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

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

$$P(43<X<49)=P(X<49)-P(X\leq 43)$$

$$=P(Z< \dfrac{49-\mu}{\sigma})-P(Z\leq \dfrac{43-\mu}{\sigma})$$

$$=P(Z< \dfrac{49-45}{8})-P(Z\leq \dfrac{43-45}{8})$$

$$=P(Z< 0.5)-P(Z\leq-0.25)$$

$$\approx0.69146246-0.40129367$$

$$\approx0.290169$$

The probability that it takes between 43 and 49 minutes for him to receive his pizza is $0.290169.$