Question

Industry standards suggest that 18% 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 =

Answer 1

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

n=10 and p=0.18

P (k=0) = 10C0 * (0.18)^0 * (1-0.18)^(10-0)

=1 *(0.18)^0 * (1-0.18)^(10-0) = 0.1374

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

P (k=1) = 10C1 * (0.18)^1 * (1-0.18)^(10-1)

=10* (0.18)^1 * (1-0.18)^(10-1) = 0.3017

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

P (k=2) = 10C2 * (0.18)^2 * (1-0.18)^(10-2)

=45* (0.18)^2 * (1-0.18)^(10-2) = 0.2980

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

P (k<3) = P(k=0) +P(k=1) +P(k=2) = 0.1374+0.3017+0.2980 = 0.7371

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

mean = 10*0.18 = 1.80

standard deviation = square root(n * p * (1-p)) = square root(10 * 0.18 * (1-0.18)) = 1.2149