Question

. The time required for a technician to machine a specific component is normally distributed with a mean of 2 hours and a standard deviation of 17 minutes. (a.) What is the probability that the technician can machine one component in 1.5 hours or less? (b.) What is the probability that the technician will require at least 2.75 hours to complete one component? (be sure to work in hours)

Answer 1

Answer:

a) There is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

b) There is a 0.42% probability that the technician will require at least 2.75 hours to complete one component

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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.

In this problem, we have to be careful. The mean is in hours, while the standard deviation is in minutes. I am going to work with both in hours, as the problem states. 17 minutes is 0.283 hours, so:

$\mu = 2, \sigma = 0.283$

(a.) What is the probability that the technician can machine one component in 1.5 hours or less?

This probability is the pvalue of the Zscore when $X = 1.5$. So:

$Z = \frac{1.5 - 2}{0.283}$

$Z = \frac{-0.5}{0.283}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

(b.) What is the probability that the technician will require at least 2.75 hours to complete one component?

The pvalue of the score of $X = 2.75$ is the probability that the technican will require less than 2.75 hours to complete one component. The probability that he will require at least 2.75 hours to complete one component is 1 subtracted by this pvalue. So:

$Z = \frac{2.75 - 2}{0.283}$

$Z = \frac{0.75}{0.283}$

$Z = 2.65$

$Z = 2.65$ has a pvalue of 0.99598.

This means that the probability that the technican will require at least 2.75 hours to complete one component is 1 - 0.99598 = 0.0042 = 0.42%.