Question

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 = $597 and standard deviation σ = $39. (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.)

Answer 1

Answer: The probability that the actual costs will exceed the budgeted amount is 0.104.

Step-by-step explanation:

Since 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, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = amount of money spent.

µ = mean

σ = standard deviation

From the information given,

µ = $597

σ = $39

The probability that the actual costs will exceed the budgeted amount($646) is expressed as

P(x > 646) = 1 - P(x ≤ 646)

For x = 646,

z = (646 - 597)/39 = 1.26

Looking at the normal distribution table, the probability corresponding to the z score is 0.896

P(x > 646) = 1 - 0.896

P(x > 646) = 0.104

Answer 2

Answer:

Probability that the actual costs will exceed the budgeted amount is 0.1038.

Step-by-step explanation:

We are given that 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 = $597 and standard deviation σ = $39.

Let X = amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant

So, X ~ N($\mu=597,\sigma^{2}=39^{2}$)

Now, the z score probability distribution is given by;

         Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean

           $\sigma$ = standard deviation

So, probability that the actual costs will exceed the budgeted amount is given by = P(X > $646)

    P(X > 646) = P( $\frac{X-\mu}{\sigma}$ > $\frac{646-597}{39}$ ) = P(Z > 1.26) = 1 - P(Z $\leq$ 1.26)

                                                      = 1 - 0.89617 = 0.1038

Therefore, probability that the actual costs will exceed the budgeted amount is 0.1038.