The scores on a standardized test are normally distributed with a mean of 95 and standard deviation of 20. What test score is 0.6 standard deviations below the mean?

Algebra and Trigonometry (6th Edition)
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ISBN:9780134463216
Author:Robert F. Blitzer
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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Understanding Standardized Test Scores

Given Problem:
The scores on a standardized test are normally distributed with a mean of 95 and a standard deviation of 20. What test score is 0.6 standard deviations below the mean?

To solve this problem, we need to understand the concept of z-scores and how they represent standard deviations from the mean in a normal distribution. 

#### Steps to Solve:

1. **Identify the mean (μ) and standard deviation (σ):**
   - Mean (μ) = 95
   - Standard Deviation (σ) = 20

2. **Calculate the z-score:**
   - A z-score of 0.6 below the mean is represented as -0.6.

3. **Use the formula:**
   \[
   X = μ + (z \times σ)
   \]
   where \( X \) is the test score, \( μ \) is the mean, \( z \) is the z-score, and \( σ \) is the standard deviation.

   \[
   X = 95 + (-0.6 \times 20)
   \]

4. **Perform the calculation:**
   \[
   X = 95 + (-12) = 95 - 12 = 83
   \]

Therefore, a test score that is 0.6 standard deviations below the mean is **83**.
Transcribed Image Text:### Understanding Standardized Test Scores Given Problem: The scores on a standardized test are normally distributed with a mean of 95 and a standard deviation of 20. What test score is 0.6 standard deviations below the mean? To solve this problem, we need to understand the concept of z-scores and how they represent standard deviations from the mean in a normal distribution. #### Steps to Solve: 1. **Identify the mean (μ) and standard deviation (σ):** - Mean (μ) = 95 - Standard Deviation (σ) = 20 2. **Calculate the z-score:** - A z-score of 0.6 below the mean is represented as -0.6. 3. **Use the formula:** \[ X = μ + (z \times σ) \] where \( X \) is the test score, \( μ \) is the mean, \( z \) is the z-score, and \( σ \) is the standard deviation. \[ X = 95 + (-0.6 \times 20) \] 4. **Perform the calculation:** \[ X = 95 + (-12) = 95 - 12 = 83 \] Therefore, a test score that is 0.6 standard deviations below the mean is **83**.
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