Question

The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $430 and standard deviation $10. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05? (Round your answer to the nearest cent.)

Answer 1

Budgeted amount is $446.44

Explanation:

Given data:

mean
`\mu  =  430`

standard deviation is given as
`\sigma  = 10`

`P(X\geq x) = 0.05`

`1 - P(X\leq x) = 0.05`

`P(X\leq x) = 0.95`

`P((X - \mu)/(\sigma) \leq (x - \mu)/(\sigma)) = 0.95`

`P(Z\leq z) = 0.95`

`(X - \mu)/(\sigma) = 1.644` { by using normal table or online calculator}

`(x - 430)/(10) = 1.644`

solving for x we have

x = 446.44

Budgeted amount is $446.44