Suppose the average client charge per hour for out-of-court work by lawyers in Saskatchewan is $125. Suppose further that a random telephone sample of 32 lawyers in Saskatchewan is taken and that the sample average charge per hour for out-of-court work is $110. If the population variance is $525, what is the probability of getting a sample mean of $110 or larger? (Round your answers to 4 decimal places, e.g. 0.7578.) P = What is the probability of getting a sample mean larger than $135 per hour? P = What is the probability of getting a sample mean of between $120 and $130 per hour? P =
Suppose the average client charge per hour for out-of-court work by lawyers in Saskatchewan is $125. Suppose further that a random telephone sample of 32 lawyers in Saskatchewan is taken and that the sample average charge per hour for out-of-court work is $110. If the population variance is $525, what is the probability of getting a sample mean of $110 or larger? (Round your answers to 4 decimal places, e.g. 0.7578.) P = What is the probability of getting a sample mean larger than $135 per hour? P = What is the probability of getting a sample mean of between $120 and $130 per hour? P =
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:**
Suppose the average client charge per hour for out-of-court work by lawyers in Saskatchewan is $125. Suppose further that a random telephone sample of 32 lawyers in Saskatchewan is taken and that the sample average charge per hour for out-of-court work is $110. If the population variance is $525, what is the probability of getting a sample mean of $110 or larger?
*(Round your answers to 4 decimal places, e.g. 0.7578.)*
\[ P = \_\_\_\_\_\_ \]
What is the probability of getting a sample mean larger than $135 per hour?
\[ P = \_\_\_\_\_\_ \]
What is the probability of getting a sample mean of between $120 and $130 per hour?
\[ P = \_\_\_\_\_\_ \]
**Explanation of Calculation Approach:**
To solve the questions, you would typically use the normal distribution to find the probabilities. The first step would be to calculate the standard deviation from the variance:
- Population Standard Deviation (\(\sigma\)) = \(\sqrt{\text{Population Variance}} = \sqrt{525}\)
Next, use the standard error of the sample mean:
- Standard Error = \(\frac{\sigma}{\sqrt{n}}\), where \(n\) is the sample size.
For each probability question, calculate the z-score using:
- \(Z = \frac{\bar{X} - \mu}{\text{Standard Error}}\), where \(\bar{X}\) is the sample mean and \(\mu\) is the population mean.
Find the probabilities using normal distribution tables or calculators for the respective z-scores. Round the answers to four decimal places.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ff8a9d1-27d7-4c01-984f-c0979e8297f8%2Fc2dc1df9-c9af-4657-ad13-4dd76457c2ca%2F2ko0oj_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Suppose the average client charge per hour for out-of-court work by lawyers in Saskatchewan is $125. Suppose further that a random telephone sample of 32 lawyers in Saskatchewan is taken and that the sample average charge per hour for out-of-court work is $110. If the population variance is $525, what is the probability of getting a sample mean of $110 or larger?
*(Round your answers to 4 decimal places, e.g. 0.7578.)*
\[ P = \_\_\_\_\_\_ \]
What is the probability of getting a sample mean larger than $135 per hour?
\[ P = \_\_\_\_\_\_ \]
What is the probability of getting a sample mean of between $120 and $130 per hour?
\[ P = \_\_\_\_\_\_ \]
**Explanation of Calculation Approach:**
To solve the questions, you would typically use the normal distribution to find the probabilities. The first step would be to calculate the standard deviation from the variance:
- Population Standard Deviation (\(\sigma\)) = \(\sqrt{\text{Population Variance}} = \sqrt{525}\)
Next, use the standard error of the sample mean:
- Standard Error = \(\frac{\sigma}{\sqrt{n}}\), where \(n\) is the sample size.
For each probability question, calculate the z-score using:
- \(Z = \frac{\bar{X} - \mu}{\text{Standard Error}}\), where \(\bar{X}\) is the sample mean and \(\mu\) is the population mean.
Find the probabilities using normal distribution tables or calculators for the respective z-scores. Round the answers to four decimal places.
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