Question

A grocery stores studies how long it takes customers to get through the speed check lane. They assume that if it takes more than 10 minutes, the customer will be upset. Find the probability that a randomly selected customer takes more than 10 minutes if the average is 8.56 minutes with a standard deviation of 1.04 minutes.

Answer 1

The probability that a randomly selected customer takes more than 10 minutes will be 8.38%

Explanation

The average or mean$(\mu)$ is 8.56 minutes and standard deviation$(\sigma)$ is 1.04 minutes.

Formula for finding the z-score is:   $z= \frac{X-\mu}{\sigma}$

So, the z-score for $X=$ 10 minutes will be.....

$z(X=10)=\frac{10-8.56}{1.04}=\frac{1.44}{1.04}=1.3846... \approx 1.38$

Now, according to the standard normal distribution table,  $P(z<1.38)=0.9162$ . So.....

$P(X>10)=P(z>1.38)= 1-P(z<1.38)=1-0.9162= 0.0838= 8.38\%$

So, the probability that a randomly selected customer takes more than 10 minutes will be 8.38%