Question

The time spent waiting in the line is approximately normally distributed. The mean waiting time is 5 minutes and the standard deviation of the waiting time is 3 minutes. Find the probability that a person will wait for more than 1 minute. Round your answer to four decimal places.

Answer 1

We were given the following details:

This is a normal distribution. Normal distributions are solved using the z-score

`\begin{gathered} \mu=5min \\ \sigma=3min \end{gathered}`

The z-score for a value, X is calculated using the formula:

`\begin{gathered} Z=(X-\mu)/(\sigma) \\ The\text{ probability that a person will wait more than 1 minute implies that: }X=1 \\ Z=(1-5)/(3) \\ Z=-(4)/(3) \\ At\text{ Z =}-(4)/(3)\text{, pvalue =}0.091759 \\ The\text{ probability that a person waits more than 1 minute is given by:} \\ P=1-0.091759 \\ P=0.908241\approx0.9082 \\ P=0.9082\text{ or }90.82\text{\%} \end{gathered}`