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The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $440 and standard deviation $20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1

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Answer:

$465.6 should be budgeted.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the z-score of a measure X is given by:


Z = (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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean $440 and standard deviation $20.

This means that
\mu = 440, \sigma = 20

How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?

The 100 - 10 = 90th percentile should be budgeted, which is X when Z has a p-value of 0.9, so X when Z = 1.28. Then


Z = (X - \mu)/(\sigma)


1.28 = (X - 440)/(20)


X - 440 = 1.28*20


X = 465.6

$465.6 should be budgeted.

User Jon Smark
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