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<(x-\mu)/(\sigma))`

Substitute the given into the formula

`\begin{gathered} P(Z<(21-32)/(5)) \\ =P(Z<-(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