Question

The time spent waiting in the line is approximately normally distributed. The mean waiting time is 55 minutes and the variance of the waiting time is 11. Find the probability that a person will wait for more than 33 minutes. Round your answer to four decimal places.

Answer 1

1 = 100% probability that a person will wait for more than 33 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 55 minutes and the variance of the waiting time is 11.

This means that
`\mu = 55, \sigma = √(11)`

Find the probability that a person will wait for more than 33 minutes.

This is 1 subtracted by the p-value of Z when X = 33. So

`Z = (X - \mu)/(\sigma)`

`Z = (33 - 55)/(√(11))`

`Z = -6.63`

`Z = -6.63` has a p-value of 0.

1 - 0 = 1

1 = 100% probability that a person will wait for more than 33 minutes.