Question

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 32 minutes and a standard deviation of 5 minutes. What is the probability that a randomly selected customer will have to wait less than 21 minutes, to the nearest thousandth?

Answer 1

Given:

$\begin{gathered} mean(\mu)=32 \\ Standard-deviation(\sigma)=5 \end{gathered}$

To Determine: The probability that a randomly selected customer will have to wait less than 21 minutes, to the nearest thousandth

Solution

Using normal distribution formula below

$P(Z<\frac{x-\mu}{\sigma})$

Substitute the given into the formula

$\begin{gathered} P(Z<\frac{21-32}{5}) \\ =P(Z<-\frac{11}{5}) \\ =P(Z<-2.2) \end{gathered}$$\begin{gathered} P(X<-2.2)=1-P(X>-2.2) \\ =1-0.986097 \\ =0.01390345 \\ \approx0.014(Nearest\text{ thousandth\rparen} \end{gathered}$

Hence, the probability that a randomly selected customer will have to wait less than 21 minutes, to the nearest thousandth is 0.014