To answer this question, we need to break it into two parts:

Part 1: Percentage of Samples Between 98 and 103 for $n = 36$

First, let's define the parameters:

• Population mean $\mu = 100$
• Population standard deviation $\sigma = 20$
• Sample size $n = 36$

For a sample mean, the standard deviation is calculated as: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{36}} = \frac{20}{6} = 3.33$

Now, we need to find the z-scores for 98 and 103 using the formula for z-scores: $z = \frac{x - \mu}{\sigma_{\bar{x}}}$ where $x$ is the value we are interested in, $\mu$ is the population mean, and $\sigma_{\bar{x}}$ is the standard error.

For $x = 98$:

$z = \frac{98 - 100}{3.33} = \frac{-2}{3.33} = -0.60$

For $x = 103$:

$z = \frac{103 - 100}{3.33} = \frac{3}{3.33} = 0.90$

Next, we need to find the probability between these z-scores. Using a standard normal distribution table or a calculator:

• The cumulative probability for $z = -0.60$ is approximately 0.2743.
• The cumulative probability for $z = 0.90$ is approximately 0.8159.

To find the percentage of samples between 98 and 103, we subtract these cumulative probabilities: $P(98 \leq \bar{x} \leq 103) = P(z = 0.90) - P(z = -0.60) = 0.8159 - 0.2743 = 0.5416$

Thus, the percentage is: $0.5416 \times 100 = 54.16\%$

Part 2: Mean Representing the 99th Percentile

For the 99th percentile, we need to find the z-score corresponding to a cumulative probability of 0.99. From the standard normal distribution table, the z-score for the 99th percentile is approximately $z = 2.33$.

Now, we can use the z-score formula to find the sample mean: $z = \frac{x - \mu}{\sigma_{\bar{x}}}$ Rearranging to solve for $x$: $x = \mu + z \cdot \sigma_{\bar{x}} = 100 + 2.33 \cdot 3.33 = 100 + 7.77 = 107.77$

So, the mean representing the 99th percentile is approximately 107.77.

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Summary of Results:

1. The percentage of samples between 98 and 103 is 54.16%.
2. The mean representing the 99th percentile is 107.77.

Would you like a detailed explanation of any step or have further questions?

Related Questions:

1. How do you calculate the z-score for other percentiles, such as the 95th percentile?
2. What would the percentage of samples be if the sample size was increased to $n = 64$?
3. How would the standard error change if we had a sample size of $n = 16$?
4. What happens to the distribution of sample means as the sample size increases?
5. How do you interpret the z-scores in the context of hypothesis testing?

Tip: The z-score formula is a critical tool in understanding how values relate to the overall distribution, especially when dealing with sample means or standard deviations!