Question

While preparing for their comeback tour, The Flaming Rogers find that the average time it takes their sound tech to set up for a show is 56.1 minutes, with a standard deviation of 6.4 minutes. If the band manager decides to include only the fastest 23% of sound techs on the tour, what should the cutoff time be for concert setup? Assume the times are normally distributed.

Answer 1

Answer:

The cutoff time be for concert setup should be of 51.4 minutes.

Step-by-step 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:

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

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 p-value, we get the probability that the value of the measure is greater than X.

Average time it takes their sound tech to set up for a show is 56.1 minutes, with a standard deviation of 6.4 minutes.

This means that $\mu = 56.1, \sigma = 6.4$

If the band manager decides to include only the fastest 23% of sound techs on the tour, what should the cutoff time be for concert setup?

The cutoff time would be the 23rd percentile of times, which is X when Z has a p-value of 0.23, so X when Z = -0.74.

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

$-0.74 = \frac{X - 56.1}{6.4}$

$X - 56.1 = -0.74*6.4$

$X = 51.4$

The cutoff time be for concert setup should be of 51.4 minutes.