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 9 magic rabbits. Would it be unusual for the sample mean to be at least 6.5? Why or why not? 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 = 0.60. O No because the z-score corresponding to a sample mean of 6.5 would be z = 0.60. Yes because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.

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**Understanding Z-Scores and Sample Means**

According to the Official Stage Magician Handbook, magic rabbit weight (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 9 magic rabbits.

**Question:**
Would it be unusual for the sample mean to be at least 6.5? Why or why not?

**Choices:**
1. No, because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.
2. Yes, because the z-score corresponding to a sample mean of 6.5 would be z = 0.60.
3. No, because the z-score corresponding to a sample mean of 6.5 would be z = 0.60.
4. Yes, because the z-score corresponding to a sample mean of 6.5 would be z = 0.20.

**Explanation:**

- **Mean (μ):** 6.2
- **Standard Deviation (σ):** 1.5
- **Sample Size (n):** 9
- **Sample Mean (X̄):** 6.5

To determine if it is unusual for the sample mean to be at least 6.5, we need to calculate the z-score. The formula for the z-score for a sample mean is:

\[ z = \frac{X̄ - μ}{\frac{σ}{\sqrt{n}}} \]

Given the values:
\[ z = \frac{6.5 - 6.2}{\frac{1.5}{\sqrt{9}}} \]

First, calculate the standard error (SE):
\[ SE = \frac{σ}{\sqrt{n}} = \frac{1.5}{\sqrt{9}} = \frac{1.5}{3} = 0.5 \]

Now, calculate the z-score:
\[ z = \frac{6.5 - 6.2}{0.5} = \frac{0.3}{0.5} = 0.6 \]

Therefore, the z-score corresponding to a sample mean of 6.5 is 0.6. Since a z-score of 0.6 is not significantly high or low (it is within the range of -2 to 2
Transcribed Image Text:**Understanding Z-Scores and Sample Means** According to the Official Stage Magician Handbook, magic rabbit weight (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 9 magic rabbits. **Question:** Would it be unusual for the sample mean to be at least 6.5? Why or why not? **Choices:** 1. No, because the z-score corresponding to a sample mean of 6.5 would be z = 0.20. 2. Yes, because the z-score corresponding to a sample mean of 6.5 would be z = 0.60. 3. No, because the z-score corresponding to a sample mean of 6.5 would be z = 0.60. 4. Yes, because the z-score corresponding to a sample mean of 6.5 would be z = 0.20. **Explanation:** - **Mean (μ):** 6.2 - **Standard Deviation (σ):** 1.5 - **Sample Size (n):** 9 - **Sample Mean (X̄):** 6.5 To determine if it is unusual for the sample mean to be at least 6.5, we need to calculate the z-score. The formula for the z-score for a sample mean is: \[ z = \frac{X̄ - μ}{\frac{σ}{\sqrt{n}}} \] Given the values: \[ z = \frac{6.5 - 6.2}{\frac{1.5}{\sqrt{9}}} \] First, calculate the standard error (SE): \[ SE = \frac{σ}{\sqrt{n}} = \frac{1.5}{\sqrt{9}} = \frac{1.5}{3} = 0.5 \] Now, calculate the z-score: \[ z = \frac{6.5 - 6.2}{0.5} = \frac{0.3}{0.5} = 0.6 \] Therefore, the z-score corresponding to a sample mean of 6.5 is 0.6. Since a z-score of 0.6 is not significantly high or low (it is within the range of -2 to 2
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