We are estimating the spares requirement for a radar power s

We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (?) life of 6500 hours. The standard deviation (?) determined from testing is 750 hours. What is the likelihood that a power supply would fail between 4500 and 7670 hours?

Solution

? =6500hrs

? =750hrs

P ( 4500<X<7670 )=P ( 4500?6500< (X??)<7670?6500 )=P ( 4500?6500/750<(X??/?)<7670?6500/750)

Since Z=x??/? , 4500?6500.750=?2.67 and 7670?6500.750=1.56 we have:

P ( 4500<X<7670 )=P ( ?2.67<Z<1.56 )

Step 3: Use the standard normal table to conclude that:

P ( ?2.67<Z<1.56 )=0.9368

likelihood that a power supply would fail between 4500 and 7670 hours = 1-P ( 4500<X<7670 ) = 0.0632