Question

The length of time to perform an oil change at a certain shop is normally distributed with mean of 29.5 minutes and standard deviation 3 minutes. What is the probability that a mechanic can complete 16 oil changes in an eight-hour day

Answer 1

Answer:

56.75% probability that a mechanic can complete 16 oil changes in an eight-hour day

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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 problem, we have that:

$\mu = 29.5, \sigma = 3$

What is the probability that a mechanic can complete 16 oil changes in an eight-hour day

16 oil changes in an eight hour days is 16/8 = 2 oil changes per hour, that is, one each 60/2 = 30 minutes.

So this probability is the pvalue of Z when X = 30. So

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

$Z = \frac{30 - 29.5}{3}$

$Z = 0.17$

$Z = 0.17$ has a pvalue of 0.5675

56.75% probability that a mechanic can complete 16 oil changes in an eight-hour day