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 $10. 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 0.05

$456.45 should be budgeted.

Step-by-step explanation:

Normal Probability Distribution:

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

In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean $440 and standard deviation $10.

This means that

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 0.05?

This is x for which

P(X > x) = 0.05

So this is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95, so X when Z = 1.645.

$456.45 should be budgeted.

step-by-step explanation:

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