Let's work through the questions you provided:

Problem Breakdown:

1. Part (a): It asks for a description of the random variable $X$, which is the amount of change students carry. The student has written that $X$ represents the amount of change carried by daytime statistics students, which seems correct based on the problem context.

2. Part (b): The formula for the normal cumulative distribution function (normalCDF) appears here, but the lower and upper limits are both set to 0.88. It looks like a mistake, as we need different values. To compute the probability for a normal distribution, you'll want to set appropriate limits for the problem.

3. Part (c): The random variable $\bar{X}$ is defined as the average change carried by 25 students. Given the problem states that the mean $\mu = 0.88$ and the standard deviation $\sigma = 0.88$, this part involves calculating the standard error of the mean, which is $\frac{\sigma}{\sqrt{n}} = \frac{0.88}{\sqrt{25}} = 0.176$.

4. Part (d): In this part, they again apply normalCDF but with a standard error. The formula setup looks correct, though you’ll need to adjust limits when solving a specific range or interval.

5. Part (e): This part asks for the probability that an individual had between $0.80 and $1.00. To solve this: $P(0.80 < X < 1.00) = P\left(\frac{X - \mu}{\sigma} \right)$ The z-scores can be calculated as: $z_1 = \frac{0.80 - 0.88}{0.88}, \quad z_2 = \frac{1.00 - 0.88}{0.88}$ You'd then use normalCDF to find the probability between these z-scores.

6. Part (f): This asks for the probability that the average change of 25 students is between $0.80 and $1.00. This is similar to part (e), but we now use the standard error of the mean $\sigma_{\bar{X}} = 0.176$. The z-scores become: $z_1 = \frac{0.80 - 0.88}{0.176}, \quad z_2 = \frac{1.00 - 0.88}{0.176}$ Again, use normalCDF to calculate the area between these two z-scores, which represents the probability.

Do you want to go into the detailed calculations for these? Let me know if you need further clarification!

Follow-up Questions:

1. What is the general formula for calculating z-scores in normal distributions?
2. Why is the standard error used instead of the standard deviation when dealing with averages?
3. How do you calculate cumulative probabilities using normalCDF?
4. What are the key differences between solving probability for a single individual versus an average of a group?
5. How does the size of the sample (e.g., 25 students) affect the standard error?

Tip:

When solving normal distribution problems, always make sure to identify whether you're dealing with individual data points or sample means. This will help you decide whether to use the standard deviation or the standard error in your z-score calculations.