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.

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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.
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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