Nirmal,
I expect tat you are to assume that the distribution of defective pins X is a binomial distribution with n = 100
and p = 0.05 where p is the probability that a pin is defective. Thus X has approximately a normal distribution with
mean = n p = 5 and
variance = n p (np) = 100 0.05 (0.95) = 4.75
Now use your knowledge of the normal distribution to find P(X ≤ 10).
Penny
