A manufacturer of luxury SUVs is concerned because warranty claims and customer dissatisfaction seems to be on the rise. The manufacturer knows that, based on experience with last year’s model, there is a probability of 4% that a customer will bring back their SUV within the first 6 months of ownership for a warranty repair. A sample of 5 SUVs are selected.
a) What is the appropriate probability distribution to use to describe this situation, and why is it appropriate? Explain your thinking fully.
b) What is the probability that none will need a warranty repair?
c) What is the probability that 2 or less will need a warranty repair?
a) The appropriate probability distribution to use in this situation is the binomial distribution. The binomial distribution is suitable when there are two possible outcomes for each trial (in this case, either the customer will bring back their SUV for a warranty repair or they won't), and when the trials are independent (the warranty claims for each SUV are not affected by the others).
b) To calculate the probability that none of the 5 SUVs will need a warranty repair, we can use the formula for the binomial distribution:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- n is the number of trials (in this case, 5 SUVs)
- k is the number of successes (in this case, 0 SUVs needing warranty repair)
- p is the probability of success (in this case, 4% or 0.04)
Plugging in the values:
P(X = 0) = (5 choose 0) * 0.04^0 * (1-0.04)^(5-0)
Calculating the values:
P(X = 0) = 1 * 1 * 0.96^5
= 0.884736
So, the probability that none of the SUVs will need a warranty repair is approximately 0.8847.
c) To calculate the probability that 2 or less SUVs will need a warranty repair, we need to calculate the probabilities of 0, 1, and 2 successes and sum them up.
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
We have already calculated the probability of 0 successes (P(X = 0)) in part b. Now, let's calculate the probabilities of 1 and 2 successes:
P(X = 1) = (5 choose 1) * 0.04^1 * (1-0.04)^(5-1)
P(X = 2) = (5 choose 2) * 0.04^2 * (1-0.04)^(5-2)
Calculating the values:
P(X = 1) = 5 * 0.04 * 0.96^4
= 0.3456
P(X = 2) = 10 * 0.04^2 * 0.96^3
= 0.3456
Now, summing up the probabilities:
P(X ≤ 2) = 0.884736 + 0.3456 + 0.3456
= 1.575936
So, the probability that 2 or less SUVs will need a warranty repair is approximately 1.576.
a) The appropriate probability distribution to use in this situation is the binomial distribution. The binomial distribution is appropriate because it deals with events that have only two possible outcomes, namely success (a customer brings back their SUV for a warranty repair) or failure (a customer does not bring their SUV back for a warranty repair). In this case, the probability of success (p) is given as 4%, and the probability of failure (q) can be calculated as 1 - p = 1 - 0.04 = 0.96. Additionally, the random variable of interest is the number of customers out of the sample of 5 who will need a warranty repair, which also fits the definition of a binomial distribution.
b) To calculate the probability that none of the 5 SUVs will need a warranty repair, we need to calculate the probability of 0 successes (p = 0) using the binomial distribution formula:
P(X = 0) = (n C x) * (p^x) * (q^(n-x))
where n is the number of trials (5), x is the number of successes (0), p is the probability of success (0.04), q is the probability of failure (0.96), and (n C x) represents the number of combinations of n items taken x at a time.
Plugging in the values, we have:
P(X = 0) = (5 C 0) * (0.04^0) * (0.96^(5-0))
Using the formula for combinations, (5 C 0) = 1:
P(X = 0) = 1 * 1 * 0.96^5 ≈ 0.815
Therefore, the probability that none of the 5 SUVs will need a warranty repair is approximately 0.815, or 81.5%.
c) To calculate the probability that 2 or fewer SUVs will need a warranty repair, we need to calculate the sum of the probabilities of 0, 1, and 2 successes using the binomial distribution formula.
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
We already know the probability of 0 successes (P(X = 0) ≈ 0.815) from part (b). Now, let's calculate the probabilities for 1 and 2 successes:
P(X = 1) = (5 C 1) * (0.04^1) * (0.96^(5-1))
= 5 * 0.04 * 0.96^4 ≈ 0.346
P(X = 2) = (5 C 2) * (0.04^2) * (0.96^(5-2))
= 10 * 0.04^2 * 0.96^3 ≈ 0.078
Summing up these probabilities:
P(X ≤ 2) ≈ 0.815 + 0.346 + 0.078 ≈ 1.239
Therefore, the probability that 2 or fewer SUVs will need a warranty repair is approximately 1.239, or 123.9%. However, probabilities cannot exceed 1, so the correct probability is 1.0 (100%).