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

Answer:

$Z = 3.33$

Step-by-step explanation:

Z-score:

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.

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

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

$Z = \frac{17 - 13}{1.2}$

$Z = 3.33$