Part (a)

1. Expected Value of the Sampling Distribution of the Sample Mean:

The expected value (mean) of the sampling distribution of the sample mean, $\mu_{\bar{x}}$, is equal to the population mean $\mu$.

$\mu_{\bar{x}} = \mu = 6$

So, the expected value is $6.00.

2. Standard Deviation of the Sampling Distribution of the Sample Mean:

The standard deviation of the sampling distribution of the sample mean, $\sigma_{\bar{x}}$, is given by:

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

where:

• $\sigma = 4$ (standard deviation of the population),
• $n = 100$ (sample size).

Now, calculate the standard deviation:

$\sigma_{\bar{x}} = \frac{4}{\sqrt{100}} = \frac{4}{10} = 0.40$

So, the standard deviation is $0.40.

3. Shape of the Sampling Distribution of the Sample Mean:

According to the Central Limit Theorem, when the sample size is large (typically $n \geq 30$), the sampling distribution of the sample mean is approximately normal regardless of the shape of the population distribution.

So, the shape is normal.

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Part (b)

We need to find the probability that the sample mean will be at least $5.00, i.e., $P(\bar{x} \geq 5.00)$.

1. The mean of the sampling distribution is $\mu_{\bar{x}} = 6$.
2. The standard deviation of the sampling distribution is $\sigma_{\bar{x}} = 0.40$.

First, calculate the $z$-score:

$z = \frac{x - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{5.00 - 6.00}{0.40} = \frac{-1.00}{0.40} = -2.50$

Now, use a standard normal table (or calculator) to find the probability corresponding to $z = -2.50$:

$P(z \geq -2.50) = 1 - P(z < -2.50) = 1 - 0.0062 = 0.9938$

So, the probability that the sample mean will be at least $5.00 is 0.9938.

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Part (c)

We need to find the probability that the sample mean will be at least $6.50, i.e., $P(\bar{x} \geq 6.50)$.

1. The mean of the sampling distribution is $\mu_{\bar{x}} = 6$.
2. The standard deviation of the sampling distribution is $\sigma_{\bar{x}} = 0.40$.

First, calculate the $z$-score:

$z = \frac{x - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{6.50 - 6.00}{0.40} = \frac{0.50}{0.40} = 1.25$

Now, use a standard normal table (or calculator) to find the probability corresponding to $z = 1.25$:

$P(z \geq 1.25) = 1 - P(z < 1.25) = 1 - 0.8944 = 0.1056$

So, the probability that the sample mean will be at least $6.50 is 0.1056.

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Summary of Results:

• Expected value of sample mean: $6.00
• Standard deviation of sample mean: $0.40
• Shape of distribution: Normal
• Probability that sample mean is at least $5.00: 0.9938
• Probability that sample mean is at least $6.50: 0.1056

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Would you like more details or explanations on any part? Here are 5 related questions you might find useful:

1. How does the Central Limit Theorem apply to different sample sizes?
2. What if the population distribution was non-normal—how would it affect the results?
3. How can we calculate probabilities for a range of sample mean values (e.g., between $5 and $6)?
4. What is the significance of the $z$-score in these problems?
5. How would the results change if the sample size was different?

Tip: The Central Limit Theorem ensures that as the sample size increases, the distribution of the sample mean approaches normality even if the original population is not normally distributed.