Question

Poisson Distribution - Probability and Statistics, Schuams 4th ed

I have the answers to the following questions, but am not totally sure how to solev using Poissons theorom and would apprciate an explanation.

4.91) Given that 3% of the electric bulbs manufactured by a company are defective, in a sample of 100 bulbs find the probability that A) More than 5 will be defective B) between 1 and 3 will be defective C) Less than or equal to 2 bulbs will be defective

Answers:

A) 0.0838

B) 0.5976

C) 0.4232

Answer 1

Solution:

Given: n = 100 bulbs,
3% of the electric bulbs manufactured by a company are defective, in a sample of 100 bulbs
p=0.03
Mean=m=np= 0.03×100=3
The pmf for the random variable X is given as,
P (X=x)={[exp(-m)]×(m^x)}/x!
={[exp(-3)]×(3^x)}/x!
P(x>5) =1-P(x<=5)
P(x<=5) =P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
={[exp(-3)]×(3^0)}/0! +{[exp(-3)]×(3^1)}/1! +{[exp(-3)]×(3^2)}/2! +{[exp(-3)]×(3^3)}/3!  
+{[exp(-3)]×(3^4)}/4! +{[exp(-3)]×(3^5)}/5!  
=0.04979+0.14936+0.22404+0.22404+0.16803+0.10082
=0.9161
P(x>5) =1-0.9161=0.0839
the probability that the more than 5 will be defective is 0.0839

B)P(1<=X<=3)=P(x<=3)-P(X=0)

P(x<=3)={[exp(-3)]×(3^0)}/0!+{[exp(-3)]×(3^1)}/1!

+{[exp(-3)]×(3^2)}/2! +{[exp(-3)]×(3^3)}/3!
=0.049790.14936+0.22404+0.2240
=0.64723

P(X=0)={[exp(-3)]×(3^0)}/0!
=0.04979

P(1<=X<=3)=P(x<=3)-P(X=0)

=0.64723 -0.04979
=0.59744
The probability that the between 1 and 3 will be defective is 0.59744

C)P(X<=2)=P(X=0)+P(X=1)+P(X=2)
={[exp(-3)]×(3^0)}/0! +{[exp(-3)]×(3^1)}/1!

+{[exp(-3)]×(3^2)}/2!  
=0.04979+0.14936+0.22404
=0.4232
The probability that the Less than or equal to 2 bulbs will be defective 0.4232