The heights of 18 year-old men are approximately normally distributed, with a mean 68 inches and a standard deviation of 3 inches. If a random sample of fifty 18-year old men selected, what is the probability that the mean height is between 67 and 69 inches? The mean is and the standard deviation (round to two decimals) is The probability is
The heights of 18 year-old men are approximately normally distributed, with a mean 68 inches and a standard deviation of 3 inches. If a random sample of fifty 18-year old men selected, what is the probability that the mean height is between 67 and 69 inches? The mean is and the standard deviation (round to two decimals) is The probability is
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![**Understanding Normal Distribution in Heights of 18-Year-Old Men**
The heights of 18-year-old men are approximately normally distributed, with a mean of 68 inches and a standard deviation of 3 inches.
Let's calculate the probability for a specific scenario:
### Problem
If a random sample of fifty 18-year-old men is selected, what is the probability that the mean height is between 67 and 69 inches?
To solve this, we will follow these steps:
1. **Determine the Sample Mean and Standard Deviation:**
- **The mean (μ)** is given as 68 inches.
- **The standard deviation of the sample mean (σ/√n)** can be calculated using the standard deviation (σ) and the sample size (n).
2. **Calculate the Standard Deviation of the Sample Mean:**
Given:
- Population standard deviation (σ) = 3 inches
- Sample size (n) = 50
\[
\text{Standard deviation (σ)} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{50}} \approx 0.42 \text{ inches}
\]
3. **Find the Z-Scores for 67 and 69 inches:**
\[
Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}
\]
For 67 inches:
\[
Z_{67} = \frac{67 - 68}{0.42} \approx -2.38
\]
For 69 inches:
\[
Z_{69} = \frac{69 - 68}{0.42} \approx 2.38
\]
4. **Determine the Probability Using Z-Scores:**
Referring to the standard normal distribution table, find the probabilities corresponding to \(Z = \pm 2.38\).
\[
P(X < 69) - P(X < 67) = P(Z < 2.38) - P(Z < -2.38)
\]
\[
P(Z < 2.38) \approx 0.9913 \quad \text{and} \quad P(Z < -2.38) \approx 0.0087
\]
\[
P(67 <](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa41b47d-a27d-4623-80de-b3b064f82b52%2F5b0b3bf8-2862-4456-b0cb-2b65d125fd15%2Fa066okn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding Normal Distribution in Heights of 18-Year-Old Men**
The heights of 18-year-old men are approximately normally distributed, with a mean of 68 inches and a standard deviation of 3 inches.
Let's calculate the probability for a specific scenario:
### Problem
If a random sample of fifty 18-year-old men is selected, what is the probability that the mean height is between 67 and 69 inches?
To solve this, we will follow these steps:
1. **Determine the Sample Mean and Standard Deviation:**
- **The mean (μ)** is given as 68 inches.
- **The standard deviation of the sample mean (σ/√n)** can be calculated using the standard deviation (σ) and the sample size (n).
2. **Calculate the Standard Deviation of the Sample Mean:**
Given:
- Population standard deviation (σ) = 3 inches
- Sample size (n) = 50
\[
\text{Standard deviation (σ)} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{50}} \approx 0.42 \text{ inches}
\]
3. **Find the Z-Scores for 67 and 69 inches:**
\[
Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}
\]
For 67 inches:
\[
Z_{67} = \frac{67 - 68}{0.42} \approx -2.38
\]
For 69 inches:
\[
Z_{69} = \frac{69 - 68}{0.42} \approx 2.38
\]
4. **Determine the Probability Using Z-Scores:**
Referring to the standard normal distribution table, find the probabilities corresponding to \(Z = \pm 2.38\).
\[
P(X < 69) - P(X < 67) = P(Z < 2.38) - P(Z < -2.38)
\]
\[
P(Z < 2.38) \approx 0.9913 \quad \text{and} \quad P(Z < -2.38) \approx 0.0087
\]
\[
P(67 <
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