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This question is not about the average of a sample, but a single observation. Therefore, the Central Limit Theorem doesn't apply here because it concerns the behavior of the sample mean. Thus, we cannot answer this question with the information given.

Although the Central Limit Theorem can start to be useful with small sample sizes, 5 is too small for the theorem to apply, especially given that our population distribution is not normal (it's right-skewed), so we cannot answer this question using the Central Limit Theorem.

This question can be answered using the Central Limit Theorem since 50 is a large sample size. According to the theorem, regardless of the characteristics of the original distribution, the distribution of the sample mean is approximately normal with mean equal to the population mean, and standard deviation equal to the population standard deviation divided by the square root of the sample size. Here, we need to find the chance that the mean shower time of 50 showers exceeds 9 minutes. To do that, we first calculate the standard deviation of the sample which is \( \frac{2}{\sqrt{50}} \approx 0.28 \). The z-score for the sample mean of 9 minutes is: \( z = \frac{9 - 8.2}{0.28} \approx 2.857 \). To find the probability that our sample mean is greater than 9, we need to find the right tail from the z-score above. Using the standard normal distribution tables, this can be approximately calculated as \(1-P(Z<2.857)\). Using the z-table gives us that \(P(Z<2.857) \approx 0.9979\). So, \( 1- 0.9979 = 0.0021 \). Therefore, the probability that 50 showers will have a mean length greater than 9 minutes is approximately 0.0021 or 0.21%.