The personal office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records they have found that the length of a first interview is normally distributed with a mean u = 34 minutes and a standard deviation o = 7 minutes. What is the probability that the average length of time for nine interviews will be 40 minutes or longer? %3D
The personal office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records they have found that the length of a first interview is normally distributed with a mean u = 34 minutes and a standard deviation o = 7 minutes. What is the probability that the average length of time for nine interviews will be 40 minutes or longer? %3D
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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#1 It must be in the format of the second picture. Hopefully it helps and is not incomplete for ya.
![**Statistical Analysis Example**
**Random Variable:**
\(\bar{X}\)
**Given:**
\[ P(\bar{X} \leq 6820) = 0.0030 \]
### Question
Determine the probability of \(\bar{X}\) less than or equal to 6820.
### Sample
- Sample Size: \( n = 50 \)
- Sample Mean: \(\bar{X} = 6820\)
### Sampling Distribution
- **Type:** Normal
- **Justification:** Central Limit Theorem (CLT), \( n \geq 30 \)
- **Test Statistic:** \( Z \)
#### Calculations:
- Population Mean: \(\mu = 7500\)
- Population Standard Deviation: \(\sigma = 1750\)
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1750}{\sqrt{50}} = 247.4874 \]
\[ Z = \frac{\bar{X} - \mu}{\sigma_{\bar{x}}} = \frac{6820 - 7500}{247.4874} = -2.75 \]
### Diagram
- Normal distribution curve illustrating the Z-score of -2.75 positioned on the left of the mean. The area under the curve to the left of -2.75 represents the probability \( P(Z \leq -2.75) = 0.0030 \). This shaded area reflects the probability of \( \bar{X} \leq 6820 \).
### Conclusion
The probability that the sample mean is less than or equal to 6820 is 0.0030, supported by the Z-score calculation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff59662fb-eaec-4b3d-a1b9-c848915bfa7b%2Fc5f1911d-8e50-4788-84a5-7bef6005d50e%2F9l82m7i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Statistical Analysis Example**
**Random Variable:**
\(\bar{X}\)
**Given:**
\[ P(\bar{X} \leq 6820) = 0.0030 \]
### Question
Determine the probability of \(\bar{X}\) less than or equal to 6820.
### Sample
- Sample Size: \( n = 50 \)
- Sample Mean: \(\bar{X} = 6820\)
### Sampling Distribution
- **Type:** Normal
- **Justification:** Central Limit Theorem (CLT), \( n \geq 30 \)
- **Test Statistic:** \( Z \)
#### Calculations:
- Population Mean: \(\mu = 7500\)
- Population Standard Deviation: \(\sigma = 1750\)
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1750}{\sqrt{50}} = 247.4874 \]
\[ Z = \frac{\bar{X} - \mu}{\sigma_{\bar{x}}} = \frac{6820 - 7500}{247.4874} = -2.75 \]
### Diagram
- Normal distribution curve illustrating the Z-score of -2.75 positioned on the left of the mean. The area under the curve to the left of -2.75 represents the probability \( P(Z \leq -2.75) = 0.0030 \). This shaded area reflects the probability of \( \bar{X} \leq 6820 \).
### Conclusion
The probability that the sample mean is less than or equal to 6820 is 0.0030, supported by the Z-score calculation.
![### Probability and Statistics: Interview Lengths Analysis
**Problem 1:**
The personal office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records, they have found that the length of a first interview is normally distributed with a mean (\( \mu \)) of 34 minutes and a standard deviation (\( \sigma \)) of 7 minutes. What is the probability that the average length of time for nine interviews will be 40 minutes or longer?
**Solution Approach:**
- Use the Central Limit Theorem to determine the sampling distribution of the sample mean.
- The sample size (\( n \)) is 9, and therefore, the standard deviation of the sample mean (\( \sigma_{\bar{x}} \)) is \( \sigma/\sqrt{n} = 7/\sqrt{9} \).
- Calculate the z-score for the average of 40 minutes.
- Use a z-table or normal distribution calculator to find the probability.
This solution involves calculating the probability for a sample mean, a common problem in statistics involving normal distribution.
**Answer:**
[The final answer would be calculated based on the steps given above.]
**Note:**
This problem illustrates the concept of sampling distributions and gives a practical application of normal distribution in job interviewing processes.
---
Feel free to add any additional diagrams or visual aids to further illustrate the problem-solving steps on the educational website.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff59662fb-eaec-4b3d-a1b9-c848915bfa7b%2Fc5f1911d-8e50-4788-84a5-7bef6005d50e%2Flg8tadp.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability and Statistics: Interview Lengths Analysis
**Problem 1:**
The personal office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records, they have found that the length of a first interview is normally distributed with a mean (\( \mu \)) of 34 minutes and a standard deviation (\( \sigma \)) of 7 minutes. What is the probability that the average length of time for nine interviews will be 40 minutes or longer?
**Solution Approach:**
- Use the Central Limit Theorem to determine the sampling distribution of the sample mean.
- The sample size (\( n \)) is 9, and therefore, the standard deviation of the sample mean (\( \sigma_{\bar{x}} \)) is \( \sigma/\sqrt{n} = 7/\sqrt{9} \).
- Calculate the z-score for the average of 40 minutes.
- Use a z-table or normal distribution calculator to find the probability.
This solution involves calculating the probability for a sample mean, a common problem in statistics involving normal distribution.
**Answer:**
[The final answer would be calculated based on the steps given above.]
**Note:**
This problem illustrates the concept of sampling distributions and gives a practical application of normal distribution in job interviewing processes.
---
Feel free to add any additional diagrams or visual aids to further illustrate the problem-solving steps on the educational website.
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