A 5594 or 5594 B 0594 or 594 C 4406 or 4406 D 0154 or 154 We

A) .5594 or 55.94%

B) .0594 or 5.94%

C) .4406 or 44.06%

D) .0154 or 1.54%

We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (U) life of 6500 hours. The standard deviation ( Sigma ) determined from testing is 750 hours. What is the likelihood that a power supply would fail in less thann 4880 hours? A) .5594 or 55.94% B) .0594 or 5.94% C) .4406 or 44.06% D) .0154 or 1.54%

Solution

P(X<4880) = P((X-mean)/s <(4880-6500)/750)

=P(Z<-2.16) =0.0154 (from standard normal table)

Answer: D) .0154 or 1.54%