Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally distributed, with a mean of 115 pounds and a standard deviation of 37.2 pounds. Random samples of size 15 are drawn from this population and the mean of each sample is determined. G: = (Round to three decimal places as needed) Sketch a graph of the sampling distribution. Choose the correct graph below. OA. OB. OC. OD. 86.2 115 1438 958 115 1342 -105.4 9.6 124.6 -335.4 9.6 354.6
Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally distributed, with a mean of 115 pounds and a standard deviation of 37.2 pounds. Random samples of size 15 are drawn from this population and the mean of each sample is determined. G: = (Round to three decimal places as needed) Sketch a graph of the sampling distribution. Choose the correct graph below. OA. OB. OC. OD. 86.2 115 1438 958 115 1342 -105.4 9.6 124.6 -335.4 9.6 354.6
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:**Educational Text: Understanding the Sampling Distribution Using the Central Limit Theorem**
To explore the central limit theorem, we aim to identify the mean and standard error of the mean for a specific sampling distribution. Subsequently, we will sketch the appropriate graph for this distribution.
**Scenario:**
The per capita consumption of red meat by individuals in a certain country follows a normal distribution, with an established mean of 115 pounds and a standard deviation of 37.2 pounds. Random samples, each consisting of 15 individuals, are drawn from this population, and the mean for each sample is calculated.
**Calculations:**
1. **Mean of the Sampling Distribution (\( \mu_{\bar{x}} \)):**
- Since the sample mean is an unbiased estimator of the population mean, \( \mu_{\bar{x}} = \mu = 115 \) pounds.
2. **Standard Error of the Mean (\( \sigma_{\bar{x}} \)):**
- Calculated using the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
- \( \sigma_{\bar{x}} = \frac{37.2}{\sqrt{15}} \approx 9.606 \) (rounded to three decimal places).
**Graphing the Sampling Distribution:**
We are provided with four graph options (A, B, C, D) to sketch the correct sampling distribution:
- **Graph A:** The x-axis ranges from -105.4 to 124.6 with a centered mean at 9.6.
- **Graph B:** The x-axis spans from -335.4 to 354.6 with a centering point at 9.6.
- **Graph C:** The x-axis covers 86.2 to 143.8 with a centered mean at 115.
- **Graph D:** The x-axis runs from 95.8 to 134.2 with a centered mean at 115.
**Correct Graph:**
The appropriate graph should reflect a normal distribution centered at the mean of 115 with a standard error of approximately 9.606. Based on these criteria, **Graph D** represents the correct sampling distribution. It has a centered mean of 115 pounds and reflects the variability (standard error) expected in the sampling distribution.
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