Heights of women are normally distributed with a mean of 64 inches and a standard deviation of 3.5 inches. Find the minimum height needed to be in the top 10 percent of women. O 72.5 inches 59.5 inches 68.5 inches 47.5 inches
Heights of women are normally distributed with a mean of 64 inches and a standard deviation of 3.5 inches. Find the minimum height needed to be in the top 10 percent of women. O 72.5 inches 59.5 inches 68.5 inches 47.5 inches
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Problem Statement:**
Heights of women are normally distributed with a mean of 64 inches and a standard deviation of 3.5 inches. Find the minimum height needed to be in the top 10 percent of women.
**Options:**
- ○ 72.5 inches
- ○ 59.5 inches
- ○ 68.5 inches
- ○ 47.5 inches
**Solution Explanation (for Educational Website):**
To solve this problem, we need to determine the z-score that corresponds to the top 10 percent of a normal distribution. This is typically done by looking up the cumulative distribution function (CDF) for a standard normal distribution.
1. **Find the Z-Score for the Top 10%:**
- The top 10% corresponds to the 90th percentile.
- From standard normal distribution tables (or using a calculator), the z-score for 0.90 (90th percentile) is approximately 1.28.
2. **Use the Z-Score Formula:**
\[
z = \frac{X - \mu}{\sigma}
\]
Where:
- \( z \) = Z-score
- \( X \) = Value in the distribution
- \( \mu \) = Mean of the distribution (64 inches)
- \( \sigma \) = Standard deviation of the distribution (3.5 inches)
3. **Calculate the Height:**
\[
1.28 = \frac{X - 64}{3.5}
\]
\[
X - 64 = 1.28 \times 3.5
\]
\[
X = 1.28 \times 3.5 + 64
\]
\[
X = 4.48 + 64
\]
\[
X = 68.48
\]
Therefore, the minimum height needed to be in the top 10 percent of women is approximately 68.5 inches.
**Correct Answer:**
- ○ 68.5 inches](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20a0d4d5-788e-4de6-b89d-c8e91545f77f%2F0bf52724-fb3d-453a-afa8-49fc75a1b554%2Fwnals5s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Heights of women are normally distributed with a mean of 64 inches and a standard deviation of 3.5 inches. Find the minimum height needed to be in the top 10 percent of women.
**Options:**
- ○ 72.5 inches
- ○ 59.5 inches
- ○ 68.5 inches
- ○ 47.5 inches
**Solution Explanation (for Educational Website):**
To solve this problem, we need to determine the z-score that corresponds to the top 10 percent of a normal distribution. This is typically done by looking up the cumulative distribution function (CDF) for a standard normal distribution.
1. **Find the Z-Score for the Top 10%:**
- The top 10% corresponds to the 90th percentile.
- From standard normal distribution tables (or using a calculator), the z-score for 0.90 (90th percentile) is approximately 1.28.
2. **Use the Z-Score Formula:**
\[
z = \frac{X - \mu}{\sigma}
\]
Where:
- \( z \) = Z-score
- \( X \) = Value in the distribution
- \( \mu \) = Mean of the distribution (64 inches)
- \( \sigma \) = Standard deviation of the distribution (3.5 inches)
3. **Calculate the Height:**
\[
1.28 = \frac{X - 64}{3.5}
\]
\[
X - 64 = 1.28 \times 3.5
\]
\[
X = 1.28 \times 3.5 + 64
\]
\[
X = 4.48 + 64
\]
\[
X = 68.48
\]
Therefore, the minimum height needed to be in the top 10 percent of women is approximately 68.5 inches.
**Correct Answer:**
- ○ 68.5 inches
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