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. Required:
a. If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?
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 0.16? (Round your answer to the nearest dollar.)

a) 0.4121

b) $588

Step-by-step explanation:

Mean μ = $633

Standard deviation σ = $45.

Required:

a. If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?

We solve using 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 = 0.22222

Probability value from Z-Table:

P(x<646) = 0.58793

P(x>646) = 1 - P(x<646) = 0.41207

≈ 0.4121

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 0.16? (Round your answer to the nearest dollar.)

Converting 0.16 to percentage = 0.16 × 100% = 16%

The z score of 16%

= -0.994

We are to find x

Using z score formula

z = (x-μ)/σ

-0.994 = x - 633/45

Cross Multiply

-0.994 × 45 = x - 633

-44.73 = x - 633

x = -44.73 + 633

x = $588.27

Approximately to the nearest dollar, the amount 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 0.16

is $588

the answer is b; false. 2 dimensional things are a lot more different from 3 dimensional things.

i hope this ! : d

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