Based on the image, it appears that the task involves using the standard normal distribution (z-scores) to calculate the number of students reporting readings more than 88 g. The mean is 88 g, and the standard deviation is 1 g, which means the z-score for 88 g would be 0 (since $z = \frac{x - \mu}{\sigma}$, where $\mu$ is the mean and $\sigma$ is the standard deviation).

To estimate how many students report readings more than 88 g, we would first calculate the area under the standard normal curve for readings above a z-score of 0 (because 88 g corresponds to a z-score of 0). From the table, we know that for $z = 0$, the area to the left of the z-score is 0.5. Thus, the area to the right of the z-score, which represents the probability of getting a reading greater than 88 g, is also 0.5.

Given that there are 170 students, the expected number of students reporting readings more than 88 g would be: $170 \times 0.5 = 85 \text{ students}.$

Would you like any further explanation on the steps taken?

Here are some follow-up questions:

1. How do you calculate z-scores for values not equal to the mean?
2. What does the area under the curve represent in a normal distribution?
3. How can you use z-scores to find probabilities for values other than 88 g?
4. Why is the area to the right of the z-score important for this problem?
5. What are the implications of a standard deviation change on the z-score?

Tip: The empirical rule (68-95-99.7) is a useful tool for estimating probabilities in normal distributions without needing to consult tables.