We are asked to find the probability that the average number of defects exceeds 4 among 100 randomly selected cars, given that the mean number of defects is $\mu = 3.2$ and the standard deviation is $\sigma = 2.4$.

Step-by-Step Solution:

1. Determine the sampling distribution of the sample mean: The sample mean $\bar{X}$ of a random sample of size $n = 100$ from a population with mean $\mu = 3.2$ and standard deviation $\sigma = 2.4$ is normally distributed (by the Central Limit Theorem) with:

   • Mean: $\mu_{\bar{X}} = \mu = 3.2$
   • Standard deviation (standard error): $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2.4}{\sqrt{100}} = \frac{2.4}{10} = 0.24$

2. Set up the probability statement: We need to find the probability that the average number of defects exceeds 4: $P(\bar{X} > 4)$

3. Standardize the value: To find this probability, we will standardize the value using the Z-score formula: $Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{4 - 3.2}{0.24} = \frac{0.8}{0.24} \approx 3.33$

4. Find the probability: We now find $P(Z > 3.33)$. Using standard normal distribution tables or a calculator: $P(Z > 3.33) \approx 0.0004$

Thus, the probability that the average number of defects exceeds 4 is approximately 0.0004.

Final Answer:

The correct answer is 0.0004.

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Do you want further details or have any questions?

Here are some related questions:

1. What would the probability be if the sample size was 50 instead of 100?
2. How does the Central Limit Theorem apply to this problem?
3. What if the standard deviation of defects was higher, say 3.0, instead of 2.4?
4. How would the probability change if we were looking for $P(\bar{X} > 3.5)$?
5. What does a Z-score of 3.33 tell us about the likelihood of the event?

Tip: When dealing with the average of a large sample size, the Central Limit Theorem ensures that the distribution of the sample mean will approximate a normal distribution, regardless of the shape of the original population distribution.