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 4.
Find the probability that a person will wait for more than 7 minutes. Round your answer to four decimal places.

Answer 1

Answer:

0.3075 = 30.75% probability that a person will wait for more than 7 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.

The standard deviation is the square root of the variance.

In this problem, we have that:

$\mu = 6, \sigma = \sqrt{4} = 2$

Find the probability that a person will wait for more than 7 minutes.

This is 1 subtracted by the pvalue of Z when X = 7. So

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

$Z = \frac{7 - 6}{2}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

1 - 0.6915 = 0.3075

0.3075 = 30.75% probability that a person will wait for more than 7 minutes.