Question

Electronic chips are produced with a probability 0.9 of not being defective. What is the approximate probability that more than 9075 of the next 10,000 are good?

Answer 1

Solution:
Given in the question
P(Good Electronic Chips) = 0.9
Number of sample n= 10000
Number of favourable cases x = 9075
We need to calculate probability that more than 9075 of the next 10,000 are good = ?
First we will check requirements for a normal approximation for the binomial distribution
np = 0.9*10000 = 9000>=5
n(1-p) = (1-0.9)*10000 = 1000>=5
so requirements are satisfied. Z-score can be calcuclated as
Z = (x-np)/sqrt(np(1-p)) = (9075 - 10000*0.9)/sqrt(10000*0.9*(1-0.9)) = (9075-9000)/sqrt(900) = 75/30 = 2.5
From Z table we found p-value
P(X>9075) = 1 - P(X<=9075) = 1 - 0.9938 = 0.0062
So there is 0.62% probability that more than 9075 of the next 10,000 are good.