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 = $37.

(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (Use 4 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 0.16? (Round to the nearest dollar.)
$

Answer 1

Solution :

Given ,

mean = = 597

standard deviation = = 37

(a)

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

= 1 - P[(x -) / < (646-597) / 37]

= 1 - P(z < 1.32)

Using z table

= 1 - 0.9066

= 0.0934

probability= 0.0934

(b)

Using standard normal table,

so,

P(Z > z) = 0.16

= 1 - P(Z < z) = 0.16

= P(Z < z) = 1 - 0.16

= P(Z < z ) = 0.84

= P(Z < 0.99 ) = 0.84

z =0.99

Using z-score formula,

x = z * +

x = 0.99* 37+597

x =633.63

x= $634