Question

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Only \(20 \%\) of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the probability that at most 70 mufflers are replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were replaced under warranty, would you question the \(20 \%\) figure? Explain.

Short answer

a. The approximate probability that between 75 and 100 mufflers are replaced under warranty is 0.6994. b. The approximate probability that at most 70 mufflers are replaced under warranty is 0.1314. c. Yes, the 20% figure would be questioned if fewer than 50 mufflers are replaced under warranty because the probability of observing this event is less than 0.05.

Step-by-step solution

- Calculation for Part a

Here we need to find the approximate probability that between 75 and 100 mufflers out of 400 purchases are replaced under warranty. Using the normal approximation to the binomial, we calculate the mean and the standard deviation of the distribution. The mean (\( μ \)) is given by \( np \), and the standard deviation (\( σ \)) is \( \sqrt{np(1-p)} \), where n is the number of trials and p is the probability of success on each trial. Here, \( n = 400 \) and \( p = 0.20 \). So, \( μ =np= 400 * 0.20 = 80 \) and \( σ = \sqrt{np(1-p)} = \sqrt{400*0.20*0.80} ≈ 8.94 \). To find the probability that between 75 and 100 mufflers are replaced under warranty, we need to find the z-scores corresponding to 75 and 100 mufflers. The z-score is given by \( z = (x - μ)/σ \), where x is the number of successes. So the z-scores for 75 and 100 mufflers are \( z_75 = (75 - 80)/8.94 ≈ -0.56 \) and \( z_100 = (100 - 80)/8.94 ≈ 2.24 \). Using a standard normal distribution table, we find the probabilities \( P(Z < -0.56) ≈ 0.2881 \) and \( P(Z < 2.24) ≈ 0.9875 \). The desired probability is \( P(-0.56 < Z < 2.24) = P(Z < 2.24) - P(Z < -0.56) ≈ 0.9875 - 0.2881 ≈ 0.6994 \)

- Calculation for Part b

Here we need to find the probability that at most 70 mufflers out of 400 purchases are replaced under warranty. Using the normal approximation to the binomial, the z-score for 70 mufflers is \( z_70 = (70 - 80)/8.94 ≈ -1.12 \). Using a standard normal distribution table, we find the probability \( P(Z < -1.12) ≈ 0.1314 \). So the approximate probability that at most 70 mufflers are replaced under warranty is \( P(Z < -1.12) ≈ 0.1314 \)

- Evaluation for Part c

Here we need to consider whether the 20% warranty replacement figure would be questioned if fewer than 50 out of 400 randomly selected purchases were replaced under warranty. Using the normal approximation to the binomial, the z-score for 50 mufflers is \( z_50 = (50 - 80)/8.94 ≈ -3.36 \). Using a standard normal distribution table, we find the probability \( P(Z < -3.36) ≈ 0.0004 \). This probability is very small, less than 0.05, which means that if indeed 20% of mufflers are replaced under warranty, then observing fewer than 50 warranty replacements in a sample of 400 would be considered a highly unlikely event. So, yes, we would question the 20% figure.