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.

Respuesta :

We were given the following details:

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

[tex]\begin{gathered} \mu=5min \\ \sigma=3min \end{gathered}[/tex]

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

[tex]\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ The\text{ probability that a person will wait more than 1 minute implies that: }X=1 \\ Z=\frac{1-5}{3} \\ Z=-\frac{4}{3} \\ At\text{ Z =}-\frac{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}[/tex]