The amount of money spent weekly on cleaning maintenance and

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 ? = $607 and standard deviation ? = $46.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 0.12?

Solution

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

X is aproximately normally distributed with mean = $ 607 and Standard Deviation = $ 46

Let b denotes the amount which will 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 0.12.

So P( X > b) = 0.12

=> P( Z > (b - 607)/46) = 0.12 ... ( Z follows standard normal distribution)

=> 1 - P( Z <= (b - 607)/46) = 0.12

=> P( Z <= (b - 607)/46) = 1-0.12 = 0.88

=> P( Z <= (b - 607)/46) = 0.88

Using standard normal probability table we get,

(b - 607)/46 = 1.17

b = 46*1.17 + 607

= 660.82 that is approximately $ 661

So approximately $661 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 0.12.