Employees at a regional hospital spend an average of $6 per day on lunch, with a standard deviation $4. A simple random sample of 100 employees is studied.

(a)

What is the expected value of the sampling distribution of the sample mean?

$

What is the standard deviation of the sampling distribution of the sample mean? (Round your answer to the nearest cent.)

$

What is the shape of the sampling distribution of the sample mean?

skewed to the rightexponential    uniformskewed to the leftnormal

(b)

What is the probability that the sample mean will be at least $5.00? (Round your answer to four decimal places.)

(c)

What is the probability that the sample mean will be at least $6.50? (Round your answer to four decimal places.)

### 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**.

---

### 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**.

---

### 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**.

---

### 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

---

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.

Employees at a regional hospital spend an average of $6 per day on lunch, with a standard deviation of $4. A simple random sample of 100 employees is studied. (a) What is the expected value of the sampling distribution of the sample mean? What is the standard deviation of the sampling distribution of the sample mean? What is the shape of the sampling distribution of the sample mean? (b) What is the probability that the sample mean will be at least $5.00? (c) What is the probability that the sample mean will be at least $6.50?

This problem explores the expected value, standard deviation, and shape of the sampling distribution of lunch costs at a hospital. It uses the Central Limit Theorem and Z-scores to calculate the probabilities of sample means being at least $5.00 and $6.50. Suitable for advanced high school or college-level statistics.

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