Industry standards suggest that 11 percent of new vehicles require warranty serv

Question

Industry standards suggest that 11 percent of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 10 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? Probability b. What is the probability exactly one of these vehicles requires warranty service? Probability c. Determine the probability that exactly two of these vehicles require warranty service. Probability d. Compute the mean and standard deviation of this probability distribution. Mean mu Standard deviation sigma

Explanation / Answer

Here p = 0.11 and n = 10

(a)
Probability that none of these vehicles requires warranty service = (1-p)^10 = (1 - 0.11)^10 = 0.3118

(b)
Probability that exactly one of these vehicles requires warranty service
P(X=1) = 10C1 * p^1 * (1-p)^9
= 10C1 * (0.11)^1 * ( 1-0.11)^9
= 0.3854

(c)
Probability that exactly two of these vehicles require warranty service
P(X=2) = 10C2 * p^2 * (1-p)^8
= 10C2 * (0.11)^2 * ( 1-0.11)^8
= 0.2143

(d)
mean = np = 10 * 0.11 = 1.1
std. dev. = sqrt(n*p*(1-p)) = sqrt (1.1*0.89) = 0.9894

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