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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.

User Jedison
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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}

User Thomas Vanhelden
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