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