According to the Official Stage Magician Handbook, magic rabbit weight (Ibs.) is normally distributed with u = 6.2 and o = 1.5. Consider obtaining a random sample of 3 magic rabbits. Would it be unusual for the sample mean to be at least 6.5? Why or why not? Yes because the z-score corresponding to a sample mean of 6.5 would be z = 0.34. No because the z-score corresponding to a sample mean of 6.5 would be z = 0.34. O Yes because the z-score corresponding to a sample mean of 6.5 would be z =0.20. O No because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.
According to the Official Stage Magician Handbook, magic rabbit weight (Ibs.) is normally distributed with u = 6.2 and o = 1.5. Consider obtaining a random sample of 3 magic rabbits. Would it be unusual for the sample mean to be at least 6.5? Why or why not? Yes because the z-score corresponding to a sample mean of 6.5 would be z = 0.34. No because the z-score corresponding to a sample mean of 6.5 would be z = 0.34. O Yes because the z-score corresponding to a sample mean of 6.5 would be z =0.20. O No because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.
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Question
![### Assessing the Sample Mean of Magic Rabbits' Weight
**Concept: Normal Distribution**
- According to the Official Stage Magician Handbook, magic rabbit weight (in pounds) follows a normal distribution.
- The distribution is characterized by:
- Mean (μ) = 6.2 lbs.
- Standard Deviation (σ) = 1.5 lbs.
**Scenario: Random Sampling**
- Consider obtaining a random sample of 3 magic rabbits.
- The goal is to determine if it would be unusual for the sample mean to be at least 6.5 lbs.
**Question: Unusual Sample Mean?**
- Would it be unusual for the sample mean to be at least 6.5? Why or why not?
**Answer Choices:**
1. **Yes**, because the z-score corresponding to a sample mean of 6.5 would be \( z = 0.34 \).
2. **No**, because the z-score corresponding to a sample mean of 6.5 would be \( z = 0.34 \).
3. **Yes**, because the z-score corresponding to a sample mean of 6.5 would be \( z = 0.20 \).
4. **No**, because the z-score corresponding to a sample mean of 6.5 would be \( z = 0.20 \).
**Explanation:**
- To answer this, we calculate the z-score for the sample mean of 6.5. The formula for the z-score of the sample mean is:
\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
Where:
- \(\bar{x}\) = Sample mean
- \(\mu\) = Population mean
- \(\sigma\) = Population standard deviation
- \(n\) = Sample size
Substituting the values:
- \(\bar{x} = 6.5\)
- \(\mu = 6.2\)
- \(\sigma = 1.5\)
- \(n = 3\)
\[ z = \frac{6.5 - 6.2}{\frac{1.5}{\sqrt{3}}} \]
\[ z = \frac{0.3}{0.866} \]
\[ z \approx 0.35 \]
So, the z-score should be approximately 0.35.
To](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F899c9d7f-c405-4c72-8a00-0f711d96c18e%2Fece06393-8dfa-4b1c-8f7f-1f312ab7e598%2F3ns6bjl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Assessing the Sample Mean of Magic Rabbits' Weight
**Concept: Normal Distribution**
- According to the Official Stage Magician Handbook, magic rabbit weight (in pounds) follows a normal distribution.
- The distribution is characterized by:
- Mean (μ) = 6.2 lbs.
- Standard Deviation (σ) = 1.5 lbs.
**Scenario: Random Sampling**
- Consider obtaining a random sample of 3 magic rabbits.
- The goal is to determine if it would be unusual for the sample mean to be at least 6.5 lbs.
**Question: Unusual Sample Mean?**
- Would it be unusual for the sample mean to be at least 6.5? Why or why not?
**Answer Choices:**
1. **Yes**, because the z-score corresponding to a sample mean of 6.5 would be \( z = 0.34 \).
2. **No**, because the z-score corresponding to a sample mean of 6.5 would be \( z = 0.34 \).
3. **Yes**, because the z-score corresponding to a sample mean of 6.5 would be \( z = 0.20 \).
4. **No**, because the z-score corresponding to a sample mean of 6.5 would be \( z = 0.20 \).
**Explanation:**
- To answer this, we calculate the z-score for the sample mean of 6.5. The formula for the z-score of the sample mean is:
\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
Where:
- \(\bar{x}\) = Sample mean
- \(\mu\) = Population mean
- \(\sigma\) = Population standard deviation
- \(n\) = Sample size
Substituting the values:
- \(\bar{x} = 6.5\)
- \(\mu = 6.2\)
- \(\sigma = 1.5\)
- \(n = 3\)
\[ z = \frac{6.5 - 6.2}{\frac{1.5}{\sqrt{3}}} \]
\[ z = \frac{0.3}{0.866} \]
\[ z \approx 0.35 \]
So, the z-score should be approximately 0.35.
To
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