A factory uses a special kind of lubricant to maintain its four milling machines. The weekly lubricant usage for each machine can be approximated with a normal distribution. This distribution has a mean of 30 gallons and a standard deviation of 11.5 gallons, and it can be considered as an independent random variable (zero covariance, zero correlation) from machine to machine. Suppose at the beginning of the week, the factory has a total of 150 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 (four machines combined) also follows a normal distribution. Required:
What is the probability that the factory will run out of lubricant before the next replenishment arrives?

0.0968 = 9.68% 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 and standard deviation , the zscore of a measure X is given by:

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.

This distribution has a mean of 30 gallons and a standard deviation of 11.5 gallons

Since there are four machines, we have that:

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

More than 150 gallons, so this is 1 subtracted by the pvalue of Z when X = 50.

has a pvalue of 0.9032

1 - 0.9032 = 0.0968

0.0968 = 9.68% probability that the factory will run out of lubricant before the next replenishment arrives

the answer is (-4,-4)

hope this ! : d