2) A normal distribution has a mean of µ = 10 with o = 3.5. What proportion of the scores in this distribution are greater than X = 5.5?
2) A normal distribution has a mean of µ = 10 with o = 3.5. What proportion of the scores in this distribution are greater than X = 5.5?
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:**
A normal distribution has a mean (\(\mu\)) of 10 and a standard deviation (\(\sigma\)) of 3.5. What proportion of the scores in this distribution are greater than \(X = 5.5\)?
**Solution:**
To find the proportion of scores greater than \(X = 5.5\), we can use the Z-score formula:
\[ Z = \frac{X - \mu}{\sigma} \]
**Steps:**
1. **Calculate the Z-score for \(X = 5.5\):**
\[
Z = \frac{5.5 - 10}{3.5} = \frac{-4.5}{3.5} \approx -1.29
\]
2. **Find the Probability:**
We look up the Z-score of -1.29 in the standard normal distribution table, or use a calculator, to find the probability that \(X \leq 5.5\).
- Probability (\(X \leq 5.5\)) ≈ 0.0985
3. **Calculate the Proportion Greater Than \(X = 5.5\):**
\[
\text{Proportion} (X > 5.5) = 1 - \text{Probability} (X \leq 5.5) = 1 - 0.0985 \approx 0.9015
\]
**Conclusion:**
Approximately 90.15% of the scores in this distribution are greater than \(X = 5.5\).
**Graphical Representation:**
This solution can be visualized using a bell curve where the area under the curve to the right of \(X = 5.5\) represents the proportion of the distribution greater than this value. The mean \(\mu = 10\) is the center of the distribution, and the standard deviation \(\sigma = 3.5\) defines the spread.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F26a21d52-7816-4917-a16a-44de3883c3a3%2F85bf783b-ae40-4c76-bf99-fa34477180ce%2F5bhfunp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A normal distribution has a mean (\(\mu\)) of 10 and a standard deviation (\(\sigma\)) of 3.5. What proportion of the scores in this distribution are greater than \(X = 5.5\)?
**Solution:**
To find the proportion of scores greater than \(X = 5.5\), we can use the Z-score formula:
\[ Z = \frac{X - \mu}{\sigma} \]
**Steps:**
1. **Calculate the Z-score for \(X = 5.5\):**
\[
Z = \frac{5.5 - 10}{3.5} = \frac{-4.5}{3.5} \approx -1.29
\]
2. **Find the Probability:**
We look up the Z-score of -1.29 in the standard normal distribution table, or use a calculator, to find the probability that \(X \leq 5.5\).
- Probability (\(X \leq 5.5\)) ≈ 0.0985
3. **Calculate the Proportion Greater Than \(X = 5.5\):**
\[
\text{Proportion} (X > 5.5) = 1 - \text{Probability} (X \leq 5.5) = 1 - 0.0985 \approx 0.9015
\]
**Conclusion:**
Approximately 90.15% of the scores in this distribution are greater than \(X = 5.5\).
**Graphical Representation:**
This solution can be visualized using a bell curve where the area under the curve to the right of \(X = 5.5\) represents the proportion of the distribution greater than this value. The mean \(\mu = 10\) is the center of the distribution, and the standard deviation \(\sigma = 3.5\) defines the spread.
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