Question

A machine at Keats Corporation fills 64-ounce detergent jugs. The machine can be adjusted to pour, on average, any amount of detergent into these jugs. However, the machine does not pour exactly the same amount of detergent into each jug; it varies from jug to jug. It is known that the net amount of detergent poured into each jug has a normal distribution with a standard deviation of 0.38 ounce. The quality control inspector wants to adjust the machine such that at least 95% of the jugs have more than 64 ounces of detergent. What should the mean amount of detergent poured by this machine into these jugs be? Round your answer to 2 decimal places.

Answer 1

Answer:

The mean should be 64.63 ounces.

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:

$\sigma = 0.38$

The quality control inspector wants to adjust the machine such that at least 95% of the jugs have more than 64 ounces of detergent. What should the mean amount of detergent poured by this machine into these jugs be?

This is $\mu$, for which X = 64 will have a pvalue of 1-0.95 = 0.05. So when X = 64, Z = -1.645.

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

$-1.645 = \frac{64 - \mu}{0.38}$

$64 - \mu = -1.645*0.38$

$\mu = 64 + 1.645*0.38$

$\mu = 64.63$

The mean should be 64.63 ounces.