Question

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Suppose that 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 ultimately replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty, would you question the \(20 \%\) figure? Explain.

Short answer

a. The probability of having between 75 and 100 mufflers replaced under warranty is approximately 0.7279 or 72.79%. b. The probability that at most 70 mufflers are replaced under warranty is approximately 0.1056 or 10.56%. c. Yes, we would question the 20% replacement figure. The probability of having fewer than 50 replacements is virtually zero, so the claim of 20% seems exaggerated.

Step-by-step solution

Defining Given Parameters

Before analyzing the problem, let's define all given parameters for our problem. The company replaces \(20 \%\) of its mufflers, implying that the success probability \(p = 0.20\). The sample size \(n = 400\). With these parameters, the mean number of replacements \(μ = np = 400 * 0.20 = 80\). The standard deviation can be obtained by \(σ= \sqrt{np(1-p)} = \sqrt{400*0.20*(1-0.20)} = 8\)

Analyzing for 75-100 Mufflers Replacement

For part a, the problem is asking about the approximate probability that 75 to 100 mufflers are replaced. This is a range on the normal curve, so we need to find the z-values for these. Now compute two z-scores using formula \(Z = (X - μ) / σ\), one for 75 and one for 100. That would be, for 75, Z1 = (75 - 80) / 8 = -0.625 and for 100, Z2 = (100 - 80) / 8 = 2.5

Calculating Probabilities for Ranges

Now turn these z-scores into probabilities using z-table or calculator. For \(Z1=-0.625\), the corresponding probability is \(0.2659\) (for less than or equal to 75) and for \(Z2=2.5\), the probability is \(0.9938\) (for less than or equal to 100). To get the probability for the range, subtract those: \(0.9938 - 0.2659 = 0.7279\). So, there's about a \(72.79\%\) chance of getting between 75 to 100 replacements.

Calculating Probability for at most 70 Mufflers

For part b, the z-score for 70 mufflers is \(Z = (70 - 80) / 8 = -1.25\). The corresponding probability for \(Z=-1.25\) is \(0.1056\). So, the probability of at most 70 mufflers are replaced under warranty is almost \(10.56\%\)

Suspicion around the 20% Replacement Figure

For part c, if fewer than 50 in 400 purchases were ever replaced under warranty the z-score for 50 replacements is \(Z = (50 - 80) / 8 = -3.75\). The corresponding probability is almost zero which is highly unlikely. So, indeed we should question the 20% replacement rate.