The problem shown in the image involves a Poisson distribution with the following key parameters:

• Poisson distribution parameter ($\lambda$): 4.43 (the average number of customers every 10 minutes)
• Sample size ($n$): 79 (79 ten-minute time frames)

Part (a) - Expected Value of $\bar{X}$

Since the mean of the Poisson distribution is $\lambda$, the expected value of $\bar{X}$ (the sample mean) is simply the population mean:

$E(\bar{X}) = \lambda = 4.43$

Part (b) - Standard Deviation of $\bar{X}$

For a Poisson distribution, the standard deviation of the sample mean is calculated as:

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$ where $\sigma$ is the standard deviation of the population, which is the square root of $\lambda$:

$\sigma = \sqrt{\lambda} = \sqrt{4.43}$ Thus, the standard deviation of $\bar{X}$ is:

$\sigma_{\bar{X}} = \frac{\sqrt{4.43}}{\sqrt{79}} \approx \frac{2.105}{8.888} \approx 0.2368$

Part (c) - Probability $P(3.8 \leq \bar{X} \leq 4.7)$

For this, we can standardize the sample mean using the normal approximation. The Z-score for the bounds (3.8 and 4.7) is given by:

$Z = \frac{\bar{X} - \lambda}{\sigma_{\bar{X}}}$

• For $3.8$: $Z_1 = \frac{3.8 - 4.43}{0.2368} \approx -2.69$

• For $4.7$: $Z_2 = \frac{4.7 - 4.43}{0.2368} \approx 1.14$

Using a Z-table, the probabilities for these Z-scores are:

$P(Z_1 \leq Z \leq Z_2) = P(-2.69 \leq Z \leq 1.14)$

From Z-tables:

• $P(Z_1 = -2.69) \approx 0.0036$
• $P(Z_2 = 1.14) \approx 0.8729$

Thus, the probability is:

$P(3.8 \leq \bar{X} \leq 4.7) = 0.8729 - 0.0036 = 0.8693$

Part (d) - 91% of the Time, the Value of $\bar{X}$ Will Exceed What Value?

We need to find the Z-score corresponding to a cumulative probability of $1 - 0.91 = 0.09$ (since we are looking for the lower tail), which from Z-tables gives $Z = -1.34$. Now, use the Z-score formula to find the corresponding $\bar{X}$:

$\bar{X} = \lambda + Z \cdot \sigma_{\bar{X}} = 4.43 + (-1.34) \cdot 0.2368$

$\bar{X} \approx 4.43 - 0.3173 = 4.1127$

So, 91% of the time, the observed value of $\bar{X}$ will exceed approximately 4.1127.

Would you like more details or explanations for any part? Here are five related questions you can explore further:

1. How does the Poisson distribution approximate to a normal distribution for large $n$?
2. How do Z-scores translate probabilities for non-standard normal distributions?
3. Why is $\sigma_{\bar{X}}$ calculated using the square root of $n$?
4. How would the results change if $\lambda$ were much larger or smaller?
5. What is the impact of changing the confidence level in Part (d)?

Tip: The Poisson distribution becomes more symmetric and approximates the normal distribution well for large values of $\lambda$.