An instructor in a large lecture class found, at the end of the semester, that the total point distribution in his class was approximately Normal, with a mean of 530 and a standard deviation of 80. About what percent scored between 490 and 590?

﻿﻿30.9%

﻿﻿77.3%

0 46.5%

To solve this, we need to use the properties of a **Normal distribution**. The question asks for the percentage of students scoring between 490 and 590 points, with a mean (\(\mu\)) of 530 and a standard deviation (\(\sigma\)) of 80.

### Steps:
1. **Standardize the values (z-scores)**:
   \[
   z = \frac{X - \mu}{\sigma}
   \]
   where \(X\) is the value we are converting to a z-score, \(\mu = 530\), and \(\sigma = 80\).

   - For \(X = 490\):
     \[
     z = \frac{490 - 530}{80} = \frac{-40}{80} = -0.5
     \]

   - For \(X = 590\):
     \[
     z = \frac{590 - 530}{80} = \frac{60}{80} = 0.75
     \]

2. **Find the area under the normal curve between these two z-scores**:
   - The area (or percentage) between a z-score of -0.5 and 0.75 can be found using z-tables or a standard normal distribution calculator.
   
   - The percentage corresponding to \(z = -0.5\) is approximately **30.85%**.
   - The percentage corresponding to \(z = 0.75\) is approximately **77.34%**.

3. **Calculate the difference**:
   - The percentage between the two z-scores is:
     \[
     77.34\% - 30.85\% = 46.49\%
     \]

Thus, about **46.5%** of students scored between 490 and 590 points.

### Final Answer:
- **46.5%** (corresponding to the third option).

---

Would you like a more detailed explanation or have any questions?

Here are 5 related questions to explore further:
1. How do you find the probability for values outside a specific range in a normal distribution?
2. What is the z-score for a value below the mean by 2 standard deviations?
3. How do you calculate the area for a given z-score using a z-table?
4. What is the probability that a student scores above 590 in this distribution?
5. How does changing the standard deviation affect the shape of the normal distribution?

**Tip**: Always standardize the values (convert to z-scores) when working with normal distributions to make calculations easier.

An instructor in a large lecture class found, at the end of the semester, that the total point distribution in his class was approximately Normal, with a mean of 530 and a standard deviation of 80. About what percent scored between 490 and 590?

Explore how to calculate the percentage of students scoring between 490 and 590 points using a normal distribution with a mean of 530 and a standard deviation of 80. This detailed solution involves z-scores and finding the area under the normal curve. Suitable for Grades 10-12.

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