Business Statistics - Probability Distribution

Question

Industry standards suggest that 7% of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 9 Nissans yesterday. (Round your 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?

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

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

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

Answer 1

a. The probability that none of the 9 vehicles requires warranty service is:

P(X = 0) = (0.93)^9 ≈ 0.4031

b. The probability that exactly one of the 9 vehicles requires warranty service is:

P(X = 1) = (9C1)(0.07)(0.93)^8 ≈ 0.3906

c. The probability that exactly two of the 9 vehicles require warranty service is:

P(X = 2) = (9C2)(0.07)^2(0.93)^7 ≈ 0.1546

d. The mean of the probability distribution is:

μ = np = 9(0.07) = 0.63

The standard deviation of the probability distribution is:

σ = sqrt(np(1-p)) = sqrt(9(0.07)(0.93)) ≈ 0.7755