https://file.aiquickdraw.com/mathbot/file/1728084385902hhm5qili.jpg

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 \(\lambda\).

You are interested in finding out the mean number of customers entering a 24-hour convenience store every 10 minutes. You suspect this can be modeled by the Poisson distribution with a mean of λ = 4.43 customers. You are to randomly pick n = 79 10-minute time frames, and observe the number of customers who enter the convenience store in each. After which, you are to average the 79 counts you have. That is, compute the value of X̄.

Solve a Poisson distribution problem to find the expected number of customers entering a store, standard deviation, and probability bounds for a sample of 79 time frames. Key statistical concepts such as the Z-score and normal approximation are covered.

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