Solution:

Given:

• Sample size $n = 49$
• Population mean $\mu = 77$
• Population standard deviation $\sigma = 7$
• The population is skewed right.

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Part (a): Describe the sampling distribution of $\bar{x}$

According to the Central Limit Theorem (CLT), when the sample size is large enough (generally $n \geq 30$), the sampling distribution of the sample mean $\bar{x}$ will be approximately normal—even if the population distribution is skewed.

Since $n = 49$, which is large, the sampling distribution of $\bar{x}$ will be approximately normal with:

• Mean: $\mu_{\bar{x}} = \mu = 77$
• Standard deviation (standard error):
  $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{7}{\sqrt{49}} = \frac{7}{7} = 1$

Correct answer for Part (a):

A. The distribution is approximately normal.

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Part (b): $P\left( \bar{x} > 78.35 \right)$

This asks for the probability that the sample mean $\bar{x}$ is greater than 78.35. Since the sampling distribution is approximately normal, we can use the z-score formula:

$z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}}$

$z = \frac{78.35 - 77}{1} = 1.35$

Now, we look up the z-score of 1.35 in a standard normal table, which gives:

$P(Z < 1.35) \approx 0.9115$

Thus:

$P(\bar{x} > 78.35) = 1 - 0.9115 = 0.0885$

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Part (c): $P\left( \bar{x} \leq 74.9 \right)$

Similarly, we calculate the z-score for 74.9:

$z = \frac{74.9 - 77}{1} = -2.1$

Looking up the z-score of -2.1:

$P(Z < -2.1) \approx 0.0179$

Thus:

$P(\bar{x} \leq 74.9) = 0.0179$

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Part (d): $P(75.65 < \bar{x} < 78.6)$

We need to find the probability that $\bar{x}$ is between 75.65 and 78.6. This involves calculating two z-scores.

1. For 75.65: $z = \frac{75.65 - 77}{1} = -1.35$ $P(Z < -1.35) \approx 0.0885$

2. For 78.6: $z = \frac{78.6 - 77}{1} = 1.6$ $P(Z < 1.6) \approx 0.9452$

Now, we subtract the two probabilities:

$P(75.65 < \bar{x} < 78.6) = 0.9452 - 0.0885 = 0.8567$

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Summary of Results:

• (a): The sampling distribution is approximately normal (Answer: A).
• (b): $P(\bar{x} > 78.35) = 0.0885$.
• (c): $P(\bar{x} \leq 74.9) = 0.0179$.
• (d): $P(75.65 < \bar{x} < 78.6) = 0.8567$.

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Would you like more details or have any questions?
Here are five related questions you could explore:

1. How does the sample size affect the standard error of the mean?
2. What happens to the sampling distribution if the population distribution is symmetric?
3. How does skewness affect the sampling distribution for small sample sizes?
4. What is the difference between a z-distribution and a t-distribution?
5. Why is the Central Limit Theorem important in statistics?

Tip: Always verify the sample size to ensure that the normal approximation is valid for the sampling distribution.