An unnoticed mechanical issue has led to a defective rate of

An unnoticed mechanical issue has led to a defective rate of 22% for a particular product referred to as Widget Z. An extensive inspection will be made of a random sample of 50 such widgets. Let X = the number of sampled widgets that are defective. You are asked to determine the probability that at most 5 of the sampled widgets will be defective.

Solution

1) THE BINOMAIL DISTRIBUTION IS NOT FAVOURABLE TO USE WHEN THE SAMPLE IS TOO LARGE AS THE CASE IN THIS CASE WHERE N = 50, WHEN THE NUMBER IS TOO LARGE THE BINOMIAL CAN BE USED AS THE NORMAL DISTRIBUTION INSTEAD

2) PROBABILITY OF GETTING DEFECTED = 0.22

PROBABILITY OF GETTING NON DEFECTED = 1 -0.22 = 0.78

NOW WE HAVE TO CALCULATE P (DEFECTIVE AT MOST 5 )= P(0)+P(1)+P(2)+P(3)+P(4)+P(5)

P(0)= 50C0(0.22)^0*(0.78)^50 = ALMOST 0

P(1) = 50C1*(O.22)^1*(0.78)^49 = 0.0005

P(2) = 50C2*(0.22)^2*(0.78)^48 = 0.003

P(3)= 50C3*9.33)^3*(0.78)^47 = 0.007

P(4) = 50C4*(0.22)^4*(0.78)^46 = 0.012

P(5)= 50C5*(0.22)^5*(.78)^45 = 0.018

HENCE P(X<=5) = TOTAL OF ALL = 0.0305