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A college's data about the incoming freshmen indicates that the mean of their high school GPAs was 3.4 , with a standard deviation of 0.30 ; the distribution was roughly mound-shaped and only slightly skewed. The students are randomly assigned to freshman writing seminars in groups of 25 . What might the mean GPA of one of these seminar groups be? Describe the appropriate sampling distribution model, including shape, center, and spread, with attention to assumptions and conditions. Make a sketch using the 68-95-99.7 Rule. Describe the appropriate sampling distribution model, including shape, center, and spread. A. N( \( \square \), (Type integers or decimals.) B. The sampling distribution model is not Normal because the sample size is less than 30 .

Ask by Lindsey Boyd. in the United States
Mar 08,2025

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The sampling distribution of the mean GPA for a group of 25 students is approximately Normal with a mean of 3.4 and a standard error of 0.06.

Solución

1. **Identify Population Parameters** The incoming freshmen have a population mean \[ \mu = 3.4 \] and a population standard deviation \[ \sigma = 0.30. \] 2. **Determine the Sample Size and the Sampling Distribution of the Mean** For a random sample (or group) of \( n = 25 \) students, the sampling distribution of the sample mean, \(\bar{x}\), has: - **Center (Mean):** \[ \mu_{\bar{x}} = \mu = 3.4. \] - **Spread (Standard Error):** \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.30}{\sqrt{25}} = \frac{0.30}{5} = 0.06. \] 3. **Assess the Shape of the Sampling Distribution** The original population distribution is "roughly mound-shaped and only slightly skewed." Because the distribution is roughly symmetric, the sampling distribution of the mean will be nearly Normal even for \( n = 25 \). Therefore, the sampling distribution model is approximately: \[ \bar{x} \sim N(3.4, 0.06). \] 4. **Apply the 68-95-99.7 Rule** Under the normal distribution, the 68-95-99.7 Rule states that: - About 68% of the values fall within one standard error of the mean: \[ 3.4 \pm 0.06, \quad \text{i.e., between } 3.34 \text{ and } 3.46. \] - About 95% of the values lie within two standard errors: \[ 3.4 \pm 2(0.06), \quad \text{i.e., between } 3.28 \text{ and } 3.52. \] - About 99.7% of the values lie within three standard errors: \[ 3.4 \pm 3(0.06), \quad \text{i.e., between } 3.22 \text{ and } 3.58. \] 5. **Conclusion** The appropriate sampling distribution model is: **A. \( N(3.4, 0.06) \)** Option B is incorrect because the original distribution is roughly mound-shaped; with a slightly skewed distribution and \( n = 25 \), the sampling distribution is still approximately Normal.

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The mean GPA of one of the seminar groups can be expected to be around 3.4, as this is the mean GPA of the incoming freshmen. The sampling distribution of the sample mean GPA will be approximately normal due to the Central Limit Theorem, as the group size is 25, which is greater than 30% of the population, and the original distribution of GPAs is roughly mound-shaped. The sampling distribution will be \( N(3.4, \frac{0.30}{\sqrt{25}}) = N(3.4, 0.06) \). This means the standard error is 0.06. According to the 68-95-99.7 rule, about 68% of the seminar groups would have GPAs between 3.34 (3.4 - 0.06) and 3.46 (3.4 + 0.06), 95% would be between 3.28 and 3.52, and 99.7% would fall within 3.22 and 3.58. The model is valid because even though the sample size is less than 30, the shape of the original distribution is key, ensuring that the sample means will still be roughly normally distributed, meeting assumptions for using the normal approximation in sampling distributions.

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