According to the Official Stage Magician Handbook, magic rabbit weight (Ibs.) is normally distributed with u = 6.2 and o = 1.5. %3D Consider obtaining a random sample of 1,000 magic rabbits. Would it be unusual for the sample mean to be at least 6.5? Why or why not? O No because the z-score corresponding to a sample mean of 6.5 would be z = 6.32. O Yes because the z-score corresponding to a sample mean of 6.5 would be z = 0.20. %3D O No because the z-score corresponding to a sample mean of 6.5 would be z = 0.20. O Yes because the z-score corresponding to a sample mean of 6.5 would be z = 6.32.
According to the Official Stage Magician Handbook, magic rabbit weight (Ibs.) is normally distributed with u = 6.2 and o = 1.5. %3D Consider obtaining a random sample of 1,000 magic rabbits. Would it be unusual for the sample mean to be at least 6.5? Why or why not? O No because the z-score corresponding to a sample mean of 6.5 would be z = 6.32. O Yes because the z-score corresponding to a sample mean of 6.5 would be z = 0.20. %3D O No because the z-score corresponding to a sample mean of 6.5 would be z = 0.20. O Yes because the z-score corresponding to a sample mean of 6.5 would be z = 6.32.
MATLAB: An Introduction with Applications
6th Edition
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Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Exploring the Probability of Unusual Sample Means in Normal Distributions**
According to the Official Stage Magician Handbook, the weight of magic rabbits (in pounds) is normally distributed with a mean (μ) of 6.2 and a standard deviation (σ) of 1.5.
Consider obtaining a random sample of 1,000 magic rabbits.
**Would it be unusual for the sample mean to be at least 6.5? Why or why not?**
**Options:**
1. No, because the z-score corresponding to a sample mean of 6.5 would be z = 6.32.
2. Yes, because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.
3. No, because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.
4. Yes, because the z-score corresponding to a sample mean of 6.5 would be z = 6.32.
**Explanation:**
To determine whether a sample mean is unusual, we need to calculate the z-score, which is a measure of how many standard deviations an element is from the mean. The formula for the z-score in the context of sample means is:
\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
Where:
- \(\bar{x}\) is the sample mean.
- \(\mu\) is the population mean.
- \(\sigma\) is the population standard deviation.
- \(n\) is the sample size.
Given:
- \(\bar{x} = 6.5\)
- \(\mu = 6.2\)
- \(\sigma = 1.5\)
- \(n = 1000\)
By substituting these values into the z-score formula, we can determine the z-score corresponding to a sample mean of 6.5 and assess whether it is unusual based on standard normal distribution criteria.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F899c9d7f-c405-4c72-8a00-0f711d96c18e%2F9b32b713-1ab9-4a47-b18d-e49ccc10ce10%2Ff0oc8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Exploring the Probability of Unusual Sample Means in Normal Distributions**
According to the Official Stage Magician Handbook, the weight of magic rabbits (in pounds) is normally distributed with a mean (μ) of 6.2 and a standard deviation (σ) of 1.5.
Consider obtaining a random sample of 1,000 magic rabbits.
**Would it be unusual for the sample mean to be at least 6.5? Why or why not?**
**Options:**
1. No, because the z-score corresponding to a sample mean of 6.5 would be z = 6.32.
2. Yes, because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.
3. No, because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.
4. Yes, because the z-score corresponding to a sample mean of 6.5 would be z = 6.32.
**Explanation:**
To determine whether a sample mean is unusual, we need to calculate the z-score, which is a measure of how many standard deviations an element is from the mean. The formula for the z-score in the context of sample means is:
\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
Where:
- \(\bar{x}\) is the sample mean.
- \(\mu\) is the population mean.
- \(\sigma\) is the population standard deviation.
- \(n\) is the sample size.
Given:
- \(\bar{x} = 6.5\)
- \(\mu = 6.2\)
- \(\sigma = 1.5\)
- \(n = 1000\)
By substituting these values into the z-score formula, we can determine the z-score corresponding to a sample mean of 6.5 and assess whether it is unusual based on standard normal distribution criteria.
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