Question

A radio station claims that the amount of advertising per hour of broadcast time has an average of 13 minutes and a standard deviation equal to 1.2 minutes. You listen to the radio station for 1​ hour, at a randomly selected​ time, and carefully observe that the amount of advertising time is equal to 17 minutes. Calculate the​ z-score for this amount of advertising time.

Answer 1

`Z = 3.33`

Explanation:

Z-score:

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question:

`\mu = 13, \sigma = 1.2`

You listen to the radio station for 1​ hour, at a randomly selected​ time, and carefully observe that the amount of advertising time is equal to 17 minutes. Calculate the​ z-score for this amount of advertising time.

We have to find Z when X = 17. So

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

`Z = (17 - 13)/(1.2)`

`Z = 3.33`