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? .

.4406 or 44.06% .

.4962 or 49.62% .

.9368 or 93.68%

.0556 or 5.56%

Solution

Normal Distribution
Mean ( u ) =6500
Standard Deviation ( sd )=750
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 4500) = (4500-6500)/750
= -2000/750 = -2.6667
= P ( Z <-2.6667) From Standard Normal Table
= 0.00383
P(X < 7670) = (7670-6500)/750
= 1170/750 = 1.56
= P ( Z <1.56) From Standard Normal Table
= 0.94062
P(4500 < X < 7670) = 0.94062-0.00383 = 0.9368