In: Statistics and Probability
3. Historically, a textile company samples three sheets of a certain material as part of incoming inspection and the historical 10.5 blemishes are found on average. If a sample of one sheet of this material is tested, based on the Poisson distribution:
a) How many blemishes should be found on average in this sample?
b) What is the probability that 4 or more blemishes will be found?
X : Number of blemishes will be found in a sample of one sheet
Given,
Historical 10.5 blemishes are found on average in samples of three sheets of a certain material as part of incoming inspection
Therefore,
a) Number blemishes should be found on average in this sample of one sheet = 10.5/3 = 3.5
X follows Poisson distribution with mean : = 3.5
Probability mass function of a Poisson distribution
i.e
Probability that 'x' blemishes will be found is given by
b) Probability that 4 or more blemishes will be found = P(X4) = 1 - [P(X=0)+P(X=1)+P(X=2)+P(X=3)]
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.03020+0.10569+0.18496+0.21579= 0.53664
P(X4) = 1 - [P(X=0)+P(X=1)+P(X=2)+P(X=3)] = 1-0.53664 = 0.46336
Probability that 4 or more blemishes will be found = 0.46336