Question

The fast food restaurant two blocks away serves customers in an average of 62 seconds with a standard deviation of 24.5 seconds. If the manager wants to advertize that 95% of the time, they serve customers within X seconds, what is the value of X

Answer 1

Answer:

X = 101.48

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 of 62 seconds with a standard deviation of 24.5 seconds.

This means that $\mu = 62, \sigma = 24.5$

If the manager wants to advertize that 95% of the time, they serve customers within X seconds, what is the value of X?

This is the 95th percentile of times, which is X when Z has a p-value of 0.95, so X when Z = 1.645.

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

$1.645 = \frac{X - 62}{24.5}$

$X - 62 = 1.645*24$

$X = 101.48$