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

Answer:

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

Step-by-step 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

   $P(X < 13000) = P(\frac{ X - \mu }{\sigma } < \frac{13000 - 15000}{1250} )$

$\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )$

     $P(X < 13000) = P(Z<-1.6 )$

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

     $P(Z<-1.6 ) = 0.054799$

So

    $P(X < 13000) =0.054799$