Here’s how to solve the problem step by step without rounding until the end.

(a) What can you expect the value of $\bar{X}$ to be?

The expected value of the sample mean $\bar{X}$ is the same as the population mean $\lambda$, since $\bar{X}$ is an unbiased estimator of $\lambda$.
Thus:

$E(\bar{X}) = \lambda = 4.24$

(b) Find the value of the standard deviation of $\bar{X}$.

For a Poisson distribution, the standard deviation of the sample mean $\bar{X}$ is given by:

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

Where:

• $\sigma = \sqrt{\lambda}$, since for Poisson, variance = mean = $\lambda$,
• $n = 52$.

Substitute the values: $\sigma = \sqrt{4.24} \quad \text{and} \quad \sigma_{\bar{X}} = \frac{\sqrt{4.24}}{\sqrt{52}}$

Calculating: $\sigma = 2.058052, \quad \sigma_{\bar{X}} = \frac{2.058052}{7.211102} \approx 0.285435$

Thus, $\sigma_{\bar{X}} = 0.285435$.

(c) Find the probability that the mean of the sample of 52 is between 1.2 and 4.8.

We calculate the z-scores for $X = 1.2$ and $X = 4.8$ using:

$z = \frac{X - \mu}{\sigma_{\bar{X}}}$

Where:

• $\mu = 4.24$,
• $\sigma_{\bar{X}} = 0.285435$.

Step 1: Calculate $z$-scores

For $X = 1.2$: $z_1 = \frac{1.2 - 4.24}{0.285435} = \frac{-3.04}{0.285435} \approx -10.650$

For $X = 4.8$: $z_2 = \frac{4.8 - 4.24}{0.285435} = \frac{0.56}{0.285435} \approx 1.962$

Step 2: Find probabilities

Using the standard normal distribution table:

• $P(z < -10.650)$ is practically 0,
• $P(z < 1.962) \approx 0.9751$.

Thus: $P(1.2 \leq \bar{X} \leq 4.8) = P(z_2) - P(z_1) \approx 0.9751 - 0 \approx 0.9751$

(d) 91% of the time, the observed value of $\bar{X}$ will exceed what value?

For this, find the $z$-score corresponding to the 91st percentile (i.e., $1 - 0.91 = 0.09$ to the left):

From the z-table:

• $P(z < -1.340) = 0.0900$.

Now, use the z-score formula to find $X$:

$z = \frac{X - \mu}{\sigma_{\bar{X}}} \implies X = \mu + z \cdot \sigma_{\bar{X}}$

Substitute: $X = 4.24 + (-1.340)(0.285435) \approx 4.24 - 0.382477 \approx 3.8575$

Summary of Answers

(a) $\bar{X} = 4.24$
(b) $\sigma_{\bar{X}} = 0.285435$
(c) $P(1.2 \leq \bar{X} \leq 4.8) = 0.9751$
(d) Observed $\bar{X} > 3.8575$ for 91% of the time.

Would you like further explanations or computations?
Here are 5 related questions:

1. How is the Poisson distribution applied in real-world scenarios?
2. What is the relationship between sample size and standard error?
3. How do you interpret z-scores in probability?
4. Can you explain why the expected value equals the population mean?
5. How does increasing $n$ affect $\sigma_{\bar{X}}$?

Tip: Always verify assumptions (e.g., Poisson distribution validity) when solving statistical problems!