The heights of adult women are approximately normally distributed with a mean of and a standard deviation of . Find the height such that of women are shorter than .
Table of contents
- 1. Intro to Stats and Collecting Data1h 14m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically2h 5m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables3h 6m
- 6. Normal Distribution and Continuous Random Variables2h 11m
- 7. Sampling Distributions & Confidence Intervals: Mean3h 23m
- Sampling Distribution of the Sample Mean and Central Limit Theorem19m
- Distribution of Sample Mean - Excel23m
- Introduction to Confidence Intervals15m
- Confidence Intervals for Population Mean1h 18m
- Determining the Minimum Sample Size Required12m
- Finding Probabilities and T Critical Values - Excel28m
- Confidence Intervals for Population Means - Excel25m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 25m
- 9. Hypothesis Testing for One Sample3h 29m
- 10. Hypothesis Testing for Two Samples4h 50m
- Two Proportions1h 13m
- Two Proportions Hypothesis Test - Excel28m
- Two Means - Unknown, Unequal Variance1h 3m
- Two Means - Unknown Variances Hypothesis Test - Excel12m
- Two Means - Unknown, Equal Variance15m
- Two Means - Unknown, Equal Variances Hypothesis Test - Excel9m
- Two Means - Known Variance12m
- Two Means - Sigma Known Hypothesis Test - Excel21m
- Two Means - Matched Pairs (Dependent Samples)42m
- Matched Pairs Hypothesis Test - Excel12m
- 11. Correlation1h 24m
- 12. Regression1h 50m
- 13. Chi-Square Tests & Goodness of Fit2h 21m
- 14. ANOVA1h 57m
6. Normal Distribution and Continuous Random Variables
Non-Standard Normal Distribution
Problem 7.2.57
Textbook Question
The ACT and SAT are two college entrance exams. The composite score on the ACT is approximately normally distributed with mean 21.1 and standard deviation 5.1. The composite score on the SAT is approximately normally distributed with mean 1026 and standard deviation 210. Suppose you scored 26 on the ACT and 1240 on the SAT. Which exam did you score better on? Justify your reasoning using the normal model.
Verified step by step guidance1
Identify the distributions for both exams: ACT scores are normally distributed with mean \( \mu_{ACT} = 21.1 \) and standard deviation \( \sigma_{ACT} = 5.1 \), and SAT scores are normally distributed with mean \( \mu_{SAT} = 1026 \) and standard deviation \( \sigma_{SAT} = 210 \).
Calculate the z-score for the ACT score of 26 using the formula:
\[ z_{ACT} = \frac{X_{ACT} - \mu_{ACT}}{\sigma_{ACT}} = \frac{26 - 21.1}{5.1} \]
Calculate the z-score for the SAT score of 1240 using the formula:
\[ z_{SAT} = \frac{X_{SAT} - \mu_{SAT}}{\sigma_{SAT}} = \frac{1240 - 1026}{210} \]
Interpret the z-scores: A higher z-score indicates a score further above the mean relative to the distribution's spread, meaning a better performance compared to other test takers.
Compare the two z-scores to determine on which exam you scored better. The exam with the higher z-score is the one where your score is relatively better.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Normal Distribution
The normal distribution is a continuous probability distribution characterized by its symmetric, bell-shaped curve. It is defined by its mean and standard deviation, which determine the center and spread of the data. Many natural phenomena, including test scores, approximate this distribution, allowing us to use it for probability and comparison.
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Z-Score
A z-score measures how many standard deviations a data point is from the mean. It standardizes different data points, enabling comparison across different scales or distributions. Calculated as (score - mean) / standard deviation, it helps determine relative performance on different exams.
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Comparing Scores Using the Normal Model
To compare scores from different tests with different scales, convert each score to its z-score. The higher z-score indicates a better relative performance compared to the test-taking population. This method uses the normal model to justify which score is better standardized.
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