Question

1. Approximately 20% of the lightbulbs produced by a company are defective (and the rest are non-defective). Suppose 3 lightbulbs are selected randomly. Let Y be the random variable showing number of defective lightbulbs.

a)Complete the following probability distribution given in the following table. (You can use binomial distribution formula)

y	p(y)
0	0.512
1
2
3	0.008

1. Find the mean and variance of the above probability distribution

Answer 1

Ans:

a)

P(y=k)=3Ck*0.2k*(1-0.2)3-k

P(y=1)=3C1*0.2*(1-0.2)^2=0.384

P(y=2)=3C2*0.2^2*(1-0.2)=0.096

y	P(y)
0	0.512
1	0.384
2	0.096
3	0.008

b)

mean=np=3*0.2=0.6

or mean=0*0.512+1*0.384+2*0.096+3*0.008=0.6

Variance=np(1-p)=3*0.2*(1-0.2)=0.48

Variance=(0-0.6)^2*0.512+(1-0.6)^2*0.384+(2-0.6)^2*0.096+(3-0.6)^2*0.008=0.48