The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean μ = $633 and standard deviation σ = $45.
(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (Round your answer to four decimal places.)
(b) How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.09? (Round your answer to the nearest dollar.)
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(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (Round your answer to four decimal places.)

We solve this question using the z score formula

z = (x-μ)/σ

where:

x is the raw score

μ is the population mean

σ is the population standard deviation.

For x > $646

z = $646 - $633/$45

z = 0.28889

Probability value from Z-Table:

P(x<646) = 0.61367

P(x>646) = 1 - P(x<646) = 0.38633

Approximately = 0.3863

The probability that the actual costs will exceed the budgeted amount is 0.3863

(b) How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.09? (Round your answer to the nearest dollar.

Let the budgeted amount = x

Converting 0.09 to percentage = 0.09× 100% = 9%

100 - 9% = 91%

We find the z score of 91%

= 1.341

We are to find x

Using z score formula

z = (x-μ)/σ

1.341 = x - 633/45

Cross Multiply

1.341 × 45 = x - 633

60.345 = x - 633

x = 60.345 + 633

x = $693.345

Approximately to the nearest dollar is $693