Question

# Industry standards suggest that 12% of new vehicles require warranty service within the first year. Jones...

Industry standards suggest that 12% of new vehicles require warranty service within the first year. Jones Nissan, sold 10 Nissans yesterday. (Round the Mean answer to 2 decimal places and the other answers to 4 decimal places.)

a. What is the probability that none of these vehicles requires warranty service?

Probability

b. What is the probability that exactly one of these vehicles requires warranty service?

Probability

c. Determine the probability that exactly two of these vehicles require warranty service.

Probability

d. What is the probability that less than three of these vehicles require warranty service?

Probability

e. Compute the mean and standard deviation of this probability distribution.

 Mean µ = Standard deviation σ =

12% of the new vehicles require warranty service within the first year. John sold 10 Nissans yesterday.

Let, X be the random variable denoting the number of vehicles, out of these 10, requiring servicing in first year.

So, X~Binomial(10,0.12)

(a)

P(None of the vehicles require warranty servicing)

=P(X=0)

=(10C0)(0.12)0(0.88)10

=0.28

(b)

P(Exactly one requires warranty servicing)

=P(X=1)

=(10C1)(0.12)1(0.88)9

=10*0.12*0.32

=0.384

(c)

P(Exactly two vehicles requires warranty service)

=P(X=2)

=(10C2)(0.12)2*(0.88)8

=45*0.0144*0.36

=0.23

(d)

P(Less than three vahicles require warranty service)

=P(X=0)+P(X=1)+P(X=2)

=0.23+0.38+0.28

=0.89

(e)

X~Binomial(10,0.12)

Now, we know that if X~binomial(n,p), its expectation is np, and its standard deviation is sqrt(npq).

So,

E(X)

=10*0.12

=1.2

Standard Deviation

=sqrt(10*0.12*0.88)

=1.03