A computer has 100 electronic components The probablility th

A computer has 100 electronic components, The probablility that a component will fail during one year of operation is 0.0022. Assuming the components function indepentantly, determine the probability that during one year of operation between 1 and 3 components inclusive, fail.

Solution

Normal Approximation to Binomial Distribution
Mean ( np ) =100 * 0.0022 = 0.22
Standard Deviation ( npq )= 100*0.0022*0.9978 = 0.4685
Normal Distribution = Z= X- u / sd                   
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1) = (1-0.22)/0.4685
= 0.78/0.4685 = 1.6649
= P ( Z <1.6649) From Standard Normal Table
= 0.95203
P(X < 3) = (3-0.22)/0.4685
= 2.78/0.4685 = 5.9338
= P ( Z <5.9338) From Standard Normal Table
= 1
P(1 < X < 3) = 1-0.95203 = 0.048