Mr. Jackson is looking at test scores for his Algebra 2 class. The scores on the Unit 2 Test follow a normal distribution, with a mean of 89 and a stan- dard deviation of 3. What percent of learners scored 86 or better?
Mr. Jackson is looking at test scores for his Algebra 2 class. The scores on the Unit 2 Test follow a normal distribution, with a mean of 89 and a stan- dard deviation of 3. What percent of learners scored 86 or better?
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![**Mr. Jackson is looking at test scores for his Algebra 2 class. The scores on the Unit 2 Test follow a normal distribution, with a mean of 89 and a standard deviation of 3.**
**What percent of learners scored 86 or better?**
**Explain your reasoning.**
---
For educational analysis, we need to understand the properties of a normal distribution:
1. **Normal Distribution Overview:**
- A normal distribution is symmetric about its mean.
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% falls within two standard deviations.
- About 99.7% lies within three standard deviations.
2. **Given Data:**
- Mean (μ) = 89
- Standard Deviation (σ) = 3
3. **Z-Score Calculation:**
- To find the percentage of learners who scored 86 or better, we first convert the score of 86 to its corresponding Z-score using the formula:
\[ Z = \frac{(X - μ)}{σ} \]
Where \(X\) is the score of interest.
\[ Z = \frac{(86 - 89)}{3} = \frac{-3}{3} = -1 \]
4. **Using Z-Score to Find the Percentage:**
- A Z-score of -1 corresponds to the point one standard deviation below the mean.
- Referring to Z-score tables or standard normal distribution tables, a Z-score of -1 corresponds to approximately 15.87% of the data to the left of this Z-score.
- Therefore, the area to the right (i.e., learners scoring 86 or better) is:
\[ 100\% - 15.87\% = 84.13\% \]
Hence, approximately 84.13% of students scored 86 or better. This reasoning assumes a perfect normal distribution curve for the test scores.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81d865ca-ed2c-4f6d-a4cc-7f843242dec1%2F4e3f09c5-4109-45a7-bd3b-5b154730cf5c%2Fyfwxmh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mr. Jackson is looking at test scores for his Algebra 2 class. The scores on the Unit 2 Test follow a normal distribution, with a mean of 89 and a standard deviation of 3.**
**What percent of learners scored 86 or better?**
**Explain your reasoning.**
---
For educational analysis, we need to understand the properties of a normal distribution:
1. **Normal Distribution Overview:**
- A normal distribution is symmetric about its mean.
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% falls within two standard deviations.
- About 99.7% lies within three standard deviations.
2. **Given Data:**
- Mean (μ) = 89
- Standard Deviation (σ) = 3
3. **Z-Score Calculation:**
- To find the percentage of learners who scored 86 or better, we first convert the score of 86 to its corresponding Z-score using the formula:
\[ Z = \frac{(X - μ)}{σ} \]
Where \(X\) is the score of interest.
\[ Z = \frac{(86 - 89)}{3} = \frac{-3}{3} = -1 \]
4. **Using Z-Score to Find the Percentage:**
- A Z-score of -1 corresponds to the point one standard deviation below the mean.
- Referring to Z-score tables or standard normal distribution tables, a Z-score of -1 corresponds to approximately 15.87% of the data to the left of this Z-score.
- Therefore, the area to the right (i.e., learners scoring 86 or better) is:
\[ 100\% - 15.87\% = 84.13\% \]
Hence, approximately 84.13% of students scored 86 or better. This reasoning assumes a perfect normal distribution curve for the test scores.
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