MAK 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 μ = 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?
MAK 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 μ = 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?
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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![### Example Problem: Probability in Normal Distribution
#### Problem:
The personnel 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:
- Mean (μ) = 34 minutes
- Standard deviation (σ) = 7 minutes.
**Question**: What is the probability that the average length of time for nine interviews will be 40 minutes or longer?
#### Solution Steps:
1. **Identify the given variables**:
- Mean (μ) = 34 minutes
- Standard deviation (σ) = 7 minutes
- Sample size (n) = 9
2. **Calculate the standard error of the mean (SEM)** using the formula:
\[ SEM = \frac{\sigma}{\sqrt{n}} \]
3. **Plug in the values**:
\[ SEM = \frac{7}{\sqrt{9}} = \frac{7}{3} \approx 2.33 \]
4. **Calculate the Z-score** to find the probability using the Z-formula:
\[ Z = \frac{X - \mu}{SEM} \]
where X is the desired average time (40 minutes in this case).
5. **Plug in the values**:
\[ Z = \frac{40 - 34}{2.33} \approx \frac{6}{2.33} \approx 2.58 \]
6. **Find the probability** from the Z-table corresponding to the calculated Z-score (2.58).
7. From the Z-table, a Z-score of 2.58 has a cumulative probability of about 0.9951.
8. **Calculate the probability of the average interview time being 40 minutes or longer**:
Since we need 40 minutes or longer,
\[ P(X \ge 40) = 1 - P(X < 40) = 1 - 0.9951 = 0.0049 \]
#### Answer:
The probability that the average length of time for nine interviews will be 40 minutes or longer is approximately 0.0049, or 0.49%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92f7f593-d0d3-42fe-aced-ab767687b37f%2F1b7cd18e-775e-4bc2-8972-3c0dc34cc0a4%2Flsyuwjc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Problem: Probability in Normal Distribution
#### Problem:
The personnel 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:
- Mean (μ) = 34 minutes
- Standard deviation (σ) = 7 minutes.
**Question**: What is the probability that the average length of time for nine interviews will be 40 minutes or longer?
#### Solution Steps:
1. **Identify the given variables**:
- Mean (μ) = 34 minutes
- Standard deviation (σ) = 7 minutes
- Sample size (n) = 9
2. **Calculate the standard error of the mean (SEM)** using the formula:
\[ SEM = \frac{\sigma}{\sqrt{n}} \]
3. **Plug in the values**:
\[ SEM = \frac{7}{\sqrt{9}} = \frac{7}{3} \approx 2.33 \]
4. **Calculate the Z-score** to find the probability using the Z-formula:
\[ Z = \frac{X - \mu}{SEM} \]
where X is the desired average time (40 minutes in this case).
5. **Plug in the values**:
\[ Z = \frac{40 - 34}{2.33} \approx \frac{6}{2.33} \approx 2.58 \]
6. **Find the probability** from the Z-table corresponding to the calculated Z-score (2.58).
7. From the Z-table, a Z-score of 2.58 has a cumulative probability of about 0.9951.
8. **Calculate the probability of the average interview time being 40 minutes or longer**:
Since we need 40 minutes or longer,
\[ P(X \ge 40) = 1 - P(X < 40) = 1 - 0.9951 = 0.0049 \]
#### Answer:
The probability that the average length of time for nine interviews will be 40 minutes or longer is approximately 0.0049, or 0.49%.
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