Assuming that the heights of college women are normally distributed with mean 67 inches and standard deviation 2 inches, what percentage of women are taller than 61 inches? 97.7% 99.9% 15.9% 50.0% 0.1%
Assuming that the heights of college women are normally distributed with mean 67 inches and standard deviation 2 inches, what percentage of women are taller than 61 inches? 97.7% 99.9% 15.9% 50.0% 0.1%
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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![### Understanding Normal Distribution and Heights of College Women
**Problem Statement:**
Assuming that the heights of college women are normally distributed with a mean of 67 inches and a standard deviation of 2 inches, what percentage of women are taller than 61 inches?
**Possible Answers:**
- 97.7%
- 99.9%
- 15.9%
- 50.0%
- 0.1%
**Explanation:**
To solve this problem, we need to use the properties of the normal distribution. According to the empirical rule (also known as the 68-95-99.7 rule), for a normal distribution:
1. About 68% of values fall within one standard deviation of the mean.
2. About 95% of values fall within two standard deviations of the mean.
3. About 99.7% of values fall within three standard deviations of the mean.
The z-score formula for a given value \(X\) in a normal distribution is:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( \mu \) is the mean
- \( \sigma \) is the standard deviation
- \( X \) is the value for which we are calculating the z-score
For \(X = 61\):
\[ Z = \frac{61 - 67}{2} = \frac{-6}{2} = -3 \]
A z-score of -3 means that 61 inches is 3 standard deviations below the mean. Looking at standard normal distribution tables, a z-score of -3 corresponds to a percentile of about 0.1%, meaning 0.1% of the women are shorter than 61 inches. Therefore, the percentage of women taller than 61 inches is approximately 100% - 0.1% = 99.9%.
Therefore, the correct answer is:
- **99.9%**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6805edd-9acc-4ccd-9c7a-22df1558ea0b%2F6e941491-a822-4569-bcf0-ea4776b2a1a7%2Fovt0yt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Normal Distribution and Heights of College Women
**Problem Statement:**
Assuming that the heights of college women are normally distributed with a mean of 67 inches and a standard deviation of 2 inches, what percentage of women are taller than 61 inches?
**Possible Answers:**
- 97.7%
- 99.9%
- 15.9%
- 50.0%
- 0.1%
**Explanation:**
To solve this problem, we need to use the properties of the normal distribution. According to the empirical rule (also known as the 68-95-99.7 rule), for a normal distribution:
1. About 68% of values fall within one standard deviation of the mean.
2. About 95% of values fall within two standard deviations of the mean.
3. About 99.7% of values fall within three standard deviations of the mean.
The z-score formula for a given value \(X\) in a normal distribution is:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( \mu \) is the mean
- \( \sigma \) is the standard deviation
- \( X \) is the value for which we are calculating the z-score
For \(X = 61\):
\[ Z = \frac{61 - 67}{2} = \frac{-6}{2} = -3 \]
A z-score of -3 means that 61 inches is 3 standard deviations below the mean. Looking at standard normal distribution tables, a z-score of -3 corresponds to a percentile of about 0.1%, meaning 0.1% of the women are shorter than 61 inches. Therefore, the percentage of women taller than 61 inches is approximately 100% - 0.1% = 99.9%.
Therefore, the correct answer is:
- **99.9%**
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