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 = $631 and standard deviation σ = $46.

(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 0.06? (Round your answer to the nearest dollar.) $

Answer 1

Solution :

Given that ,

mean = = 631

standard deviation = = 46

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

= 1 - P[(x -) / < (646-631) /46 ]

= 1 - P(z <0.33 )

Using z table

= 1 - 0.6293

= 0.3707

probability= 0.3707

b)

Solution:-

Given that,

mean = = 631

standard deviation = = 46

Using standard normal table,

P(Z > z) = 0.06

= 1 - P(Z < z) = 0.06

= P(Z < z) = 1 - 0.06

= P(Z < z ) = 0.94

= P(Z <1.56 ) = 0.94

z =1.56

Using z-score formula,

x = z * +

x = 1.56* 46+631

x = 702.76

x= $ 703