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

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

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 = (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 = √(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 = (X - \mu)/(\sigma)`

`Z = (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.