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The problem provides a normally distributed set of data, where the mean height is 175 cm and the standard deviation is 5 cm. The question asks for the percentage of students between 170 cm and 180 cm in height.

We can solve this using the properties of a normal distribution.

1. **Calculate z-scores** for 170 cm and 180 cm:
   - For 170 cm: 
     \[
     z = \frac{170 - 175}{5} = \frac{-5}{5} = -1
     \]
   - For 180 cm: 
     \[
     z = \frac{180 - 175}{5} = \frac{5}{5} = 1
     \]

2. **Use the z-table**: The z-scores correspond to percentages in a standard normal distribution.
   - The percentage of values below a z-score of 1 is approximately 84.13%.
   - The percentage of values below a z-score of -1 is approximately 15.87%.

3. **Find the percentage between z = -1 and z = 1**:
   \[
   84.13\% - 15.87\% = 68.26\%
   \]

Thus, approximately **68%** of the students are between 170 cm and 180 cm in height.

The correct answer is **C. 68**.

Would you like further details or have any questions about this?

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Here are 5 related questions:
1. How would the percentage change if the range were 165 cm to 185 cm?
2. What is the probability of a student being shorter than 170 cm?
3. How would you compute the z-score for a height of 160 cm?
4. How does the standard deviation affect the distribution of student heights?
5. What percentage of students are taller than 180 cm?

**Tip:** Understanding how z-scores relate to percentages in a normal distribution is key to solving problems involving the normal curve.

The measurement of the height of 600 students of a college is normally distributed with a mean of 175 centimeters and a standard deviation of 5 centimeters. What percent of students are between 170 cm and 180 cm in height?

Find out what percentage of students fall between 170 cm and 180 cm in a normally distributed set of heights. The problem involves using the mean, standard deviation, and z-scores to calculate the solution. Suitable for Grades 10-12.

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