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 9 minutes.

Answer 1

Answer:

0.1587 = 15.87% probability that a person will wait for more than 9 minutes.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The mean waiting time is 6 minutes and the variance of the waiting time is 9.

This means that $\mu = 6, \sigma = \sqrt{9} = 3$

Find the probability that a person will wait for more than 9 minutes.

This is 1 subtracted by the p-value of Z when X = 9. So

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

$Z = \frac{9 - 6}{3}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

1 - 0.8413 = 0.1587

0.1587 = 15.87% probability that a person will wait for more than 9 minutes.