Answer:
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
![\mu = 30*4 = 120](https://tex.z-dn.net/?f=%5Cmu%20%3D%2030%2A4%20%3D%20120)
![\sigma = 11.5\sqrt{4} = 11.5*2 = 23](https://tex.z-dn.net/?f=%5Csigma%20%3D%2011.5%5Csqrt%7B4%7D%20%3D%2011.5%2A2%20%3D%2023)
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.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{150 - 120}{23}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B150%20-%20120%7D%7B23%7D)
![Z = 1.3](https://tex.z-dn.net/?f=Z%20%3D%201.3)
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