Question

1. F & G industries manufacturers a motor that is marketed with a warranty that guarantees it will be replaces free of charge if it fails within the first 13,000 hours of operation. On average, F & G, motors operate for 15,000 hours with a standard deviation of 1,250 hours before failing. The number of operating hours before failure is approximately normally distributed. a. What is the probability that a motor will have to be replaced free of charge

Answer 1

The value is
`P(X < 13000) =0.054799`

Explanation:

From the question we are told that

The number of hours considered is
`x= 13000 \ hours`

The population mean is
`\mu = 15000\ hours`

The standard deviation is
`\sigma = 1250\ hours`

Generally the the probability that a motor will have to be replaced free of charge is mathematically represented as

`P(X < 13000) = P(( X - \mu )/(\sigma ) < (13000 - 15000)/(1250) )`

`(X -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \ X )`

`P(X < 13000) = P(Z<-1.6 )`

From the z table the probability of (Z<-1.6 ) is

`P(Z<-1.6 ) = 0.054799`

So

`P(X < 13000) =0.054799`