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.
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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