= A population of values has a normal distribution with μ size n = 97 is drawn. 14.9 and o 39.2. A random sample of a. Find the probability that a single randomly selected value is greater than 7.3. Round your answer to four decimal places. P(X > 7.3) = b. Find the probability that a sample of size n = 97 is randomly selected with a mean greater than 7.3. Round your answer to four decimal places. P(M > 7.3) = 0.2119
= A population of values has a normal distribution with μ size n = 97 is drawn. 14.9 and o 39.2. A random sample of a. Find the probability that a single randomly selected value is greater than 7.3. Round your answer to four decimal places. P(X > 7.3) = b. Find the probability that a sample of size n = 97 is randomly selected with a mean greater than 7.3. Round your answer to four decimal places. P(M > 7.3) = 0.2119
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![A population of values has a normal distribution with \( \mu = 14.9 \) and \( \sigma = 39.2 \). A random sample of size \( n = 97 \) is drawn.
a. Find the probability that a single randomly selected value is greater than 7.3. *Round your answer to four decimal places.*
\[ P(X > 7.3) = \]
b. Find the probability that a sample of size \( n = 97 \) is randomly selected with a mean greater than 7.3. *Round your answer to four decimal places.*
\[ P(M > 7.3) = 0.2119 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbfe2c9c4-bb55-4af5-832e-87464c5c3020%2Fbb09c3d8-03ae-4ff5-b87c-b87df5512a2f%2Fa1zbowc_processed.png&w=3840&q=75)
Transcribed Image Text:A population of values has a normal distribution with \( \mu = 14.9 \) and \( \sigma = 39.2 \). A random sample of size \( n = 97 \) is drawn.
a. Find the probability that a single randomly selected value is greater than 7.3. *Round your answer to four decimal places.*
\[ P(X > 7.3) = \]
b. Find the probability that a sample of size \( n = 97 \) is randomly selected with a mean greater than 7.3. *Round your answer to four decimal places.*
\[ P(M > 7.3) = 0.2119 \]
![**Transcription for Educational Website:**
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The manager of a computer retail store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.4 years and a standard deviation of 0.4 years. He then randomly selects records on 36 laptops sold in the past and finds that the mean replacement time is 4.2 years.
a. Find the probability that 36 randomly selected laptops will have a mean replacement time of 4.2 years or less. Round your answer to four decimal places.
\[ P(M \leq 4.2 \text{ years}) = \_ \_ \_ \_ \]
b. Based on the result above, does it appear that the computer store has been given laptops of lower than average quality?
- ○ No. The probability from the sample shows it is unlikely that the owner is given laptops lower than average quality.
- ○ Yes. The probability from the sample shows that there's a possibility that the owner is given laptops lower than average quality.
---
**Explanation:**
The problem involves a normally distributed dataset where the mean (\(\mu\)) is 4.4 years, and the standard deviation (\(\sigma\)) is 0.4 years. The mean replacement time from a sample of 36 laptops is calculated. The task is to calculate the probability that the sample mean is 4.2 years or less.
Using the central limit theorem, the sampling distribution of the sample mean (M) will be normally distributed with:
- Mean (\(\mu_M\)) = 4.4 years
- Standard Error (SE) = \(\sigma / \sqrt{n} = 0.4 / \sqrt{36} = 0.0667\)
The computation involves finding the standardized score (z-score) and using the z-table or normal distribution calculator for the probability.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbfe2c9c4-bb55-4af5-832e-87464c5c3020%2Fbb09c3d8-03ae-4ff5-b87c-b87df5512a2f%2F3gh1y1h_processed.png&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website:**
---
The manager of a computer retail store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.4 years and a standard deviation of 0.4 years. He then randomly selects records on 36 laptops sold in the past and finds that the mean replacement time is 4.2 years.
a. Find the probability that 36 randomly selected laptops will have a mean replacement time of 4.2 years or less. Round your answer to four decimal places.
\[ P(M \leq 4.2 \text{ years}) = \_ \_ \_ \_ \]
b. Based on the result above, does it appear that the computer store has been given laptops of lower than average quality?
- ○ No. The probability from the sample shows it is unlikely that the owner is given laptops lower than average quality.
- ○ Yes. The probability from the sample shows that there's a possibility that the owner is given laptops lower than average quality.
---
**Explanation:**
The problem involves a normally distributed dataset where the mean (\(\mu\)) is 4.4 years, and the standard deviation (\(\sigma\)) is 0.4 years. The mean replacement time from a sample of 36 laptops is calculated. The task is to calculate the probability that the sample mean is 4.2 years or less.
Using the central limit theorem, the sampling distribution of the sample mean (M) will be normally distributed with:
- Mean (\(\mu_M\)) = 4.4 years
- Standard Error (SE) = \(\sigma / \sqrt{n} = 0.4 / \sqrt{36} = 0.0667\)
The computation involves finding the standardized score (z-score) and using the z-table or normal distribution calculator for the probability.
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