Question

the time spent waiting in the line is approximately normally distributed. the mean waiting time is 6 minutes and the variance of the waiting time is 9. find the probability that a person will wait for more than 4 minutes. round your answer to four decimal places.

Explanation:

Step1: Calculate the standard deviation

Since variance $\sigma^{2}=9$, then standard deviation $\sigma = \sqrt{9}=3$.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 6$ (mean), $x = 4$ (value). So $z=\frac{4 - 6}{3}=\frac{-2}{3}\approx - 0.67$.

Step3: Find the probability

We want $P(X>4)$. Using the property $P(X>x)=1 - P(X\leq x)$. For $x = 4$, $z\approx - 0.67$. Looking up in the standard normal table (or using a calculator with normalcdf function), $P(Z\leq - 0.67)\approx0.2514$. Then $P(X>4)=1 - 0.2514 = 0.7486$.

Answer:

$0.7486$