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 7.45 minutes with a standard deviation of 1.52 minutes.

Answer 1

Answer:

4.65% probability that a randomly selected customer takes more than 10 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 7.45, \sigma = 1.52$

Probability that a customer takes more than 10 minutes:

This is 1 subtracted by the pvalue of Z when X = 10. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10 - 7.45}{1.52}$

$Z = 1.68$

$Z = 1.68$ has a pvalue of 0.9535

1 - 0.9535 = 0.0465

4.65% probability that a randomly selected customer takes more than 10 minutes