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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
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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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.
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.
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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