Question

A radio station claims that the amount of advertising per hour of broadcast time has an average of 10 minutes and a standard deviation equal to 5 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 8.2 minutes. Calculate the z-score for this amount of advertising time. Round your answer to 2 decimal places.

Answer 1

Answer: -0.36

Explanation:

Given: Mean :
`\mu=10\text{ minutes}`

Standard deviation :
`\sigma=5\text{ minutes}`

We know that the formula to calculate the z-score is given by :-

`z=(x-\mu)/(\sigma)`

For x=8.2 minutes, we have

`z=(8.2-10)/(5)=0.36`

Hence, the z-score for this amount of advertising time = -0.36