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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.

The number of students reporting readings more than 88 g is ________. (Round to the nearest whole number as needed.)

Learn how to calculate the number of students with mass readings above 88g using the standard normal distribution. This problem involves calculating the z-score, applying the properties of the normal curve, and estimating the number of students out of 170. Ideal for students studying statistics in high school or early college.

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