Question

A factory uses a special kind of lubricant to maintain its two milling machines. Weekly lubricant usage for each machine is an independent random variable (zero correlation). The first machine has a mean usage of 50.6 gallons and standard deviation of 12 gallons. The second machine has a mean usage of 64.4 gallons and standard deviation of 16 gallons.
Suppose that at the beginning of the week, the factory has a total of 135 gallons of lubricant in stock. The factory will not receive any replenishment of lubricant from its supplier until the end of the week. Assume that the total lubricant usage (of the two machines combined) follows a normal distribution. What is the probability that the factory will run out of lubricant before the next replenishment arrives?

Answer 1

Answer:

0.1587 = 15.87% probability that the factory will run out of lubricant before the next replenishment arrives

Step-by-step explanation:

When the distribution is normal, we use 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.

Addition of normal variables.

When we add two normal variables, the mean is the sum of the separate means, and the standard deviation is the square root of the sum of the variances. So

The first machine has a mean usage of 50.6 gallons and standard deviation of 12 gallons. The second machine has a mean usage of 64.4 gallons and standard deviation of 16 gallons.

Total consumption is of both machines, combined. So

$\mu = 50.6 + 64.4 = 115$

$\sigma = \sqrt{12^2 + 16^2} = 20$

What is the probability that the factory will run out of lubricant before the next replenishment arrives?

Probability of consuming more than 135 gallons, which is 1 subtracted by the pvalue of Z when X = 135. So

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

$Z = \frac{135 - 115}{20}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

1 - 0.8413 = 0.1587

0.1587 = 15.87% probability that the factory will run out of lubricant before the next replenishment arrives