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Let's solve this problem step by step.

The problem involves a normal distribution with:
- Mean (μ) = 174.5 cm
- Standard deviation (σ) = 6.9 cm
- A population of 1000 students.

We are asked to find how many students:
1. Have heights less than 160.0 cm.
2. Have heights between 171.5 cm and 182.0 cm.

We will use the standard normal distribution (Z-scores) to convert these height values into Z-scores and then use the areas provided.

### (a) Heights less than 160.0 cm:
The Z-score is calculated using the formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Where:
- \( X = 160.0 \) cm
- \( \mu = 174.5 \) cm
- \( \sigma = 6.9 \) cm

So,
\[
Z = \frac{160.0 - 174.5}{6.9} = \frac{-14.5}{6.9} \approx -2.10
\]
Using the standard normal table or calculator, we can find the probability for \( Z = -2.10 \). Since the problem doesn't provide this value directly, we estimate from standard normal tables that \( P(Z < -2.10) \approx 0.0179 \).

This means about 1.79% of the students have heights less than 160.0 cm.

For 1000 students:
\[
0.0179 \times 1000 = 17.9 \approx 18 \text{ students}
\]

### (b) Heights between 171.5 cm and 182.0 cm:
Now we calculate the Z-scores for 171.5 cm and 182.0 cm.

For 171.5 cm:
\[
Z = \frac{171.5 - 174.5}{6.9} = \frac{-3.0}{6.9} \approx -0.43
\]

For 182.0 cm:
\[
Z = \frac{182.0 - 174.5}{6.9} = \frac{7.5}{6.9} \approx 1.09
\]

From the problem statement, we are given:
- The area from \( Z = 0 \) to \( Z = 0.43 \) is 0.1664.
- The area from \( Z = 0 \) to \( Z = 1.09 \) is 0.3621.

So the probability between \( Z = -0.43 \) and \( Z = 1.09 \) is:
\[
P(-0.43 < Z < 1.09) = 0.3621 + 0.1664 = 0.5285
\]

This means about 52.85% of students have heights between 171.5 cm and 182.0 cm.

For 1000 students:
\[
0.5285 \times 1000 = 528.5 \approx 529 \text{ students}
\]

### Final Answers:
- (a) About **18 students** have heights less than 160.0 cm.
- (b) About **529 students** have heights between 171.5 cm and 182.0 cm.

Would you like further clarification or details on any part of the solution?

Here are some related questions:
1. How would the result change if the population were 5000 students?
2. What is the probability of having a height greater than 182.0 cm?
3. What percentage of students have heights greater than 174.5 cm?
4. How do we calculate the area under a normal curve between two Z-values?
5. How can we use a calculator or software to compute these probabilities directly?

**Tip:** Z-scores help us standardize any normal distribution, making it easy to look up probabilities or areas under the curve in standard tables.

The heights of 1000 students are normally distributed with a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. Assuming that the heights recorded are normally distributed, how many of these students would you expect to have heights (a) less than 160.0 centimeters, (b) between 171.5 and 182.0 centimeters?

Solve a normal distribution problem involving 1000 students with an average height of 174.5 cm and a standard deviation of 6.9 cm. Find how many students have heights below 160.0 cm or between 171.5 cm and 182.0 cm using Z-scores and probability calculations. Ideal for students in Grade 12 or college-level statistics.

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