**Title: Understanding the Sampling Distribution of Sample Means**The annual salary for a specific occupation is normally distributed, with a mean salary of approximately $135,000 and a standard deviation of about $15,000. Random samples of 38 individuals are taken from this population, and the mean of each sample is calculated. The task is to determine the mean and standard deviation of the sampling distribution of these sample means, and then sketch a graph of this distribution.**1. Calculate the Mean and Standard Deviation:**- **Mean ($\mu_x$)**: Since the sample mean is an unbiased estimator of the population mean, $\mu_x = \mu = 135,000$.- **Standard Deviation ($\sigma_x$)**: The standard deviation of the sampling distribution ($\sigma_x$) is given by the formula: \[ \sigma_x = \frac{\sigma}{\sqrt{n}} \] Where: - $\sigma = 15,000$ (population standard deviation) - $n = 38$ (sample size) Calculating $\sigma_x$, we get: \[ \sigma_x = \frac{15,000}{\sqrt{38}} \approx 2,433 \]**2. Graph of the Sampling Distribution:**The task includes sketching a graph of the sampling distribution. Below are the descriptions of the given options: - **Option A:** Shows a normal distribution curve centered around $135,000 with a wide spread.- **Option B:** Displays a normal distribution curve centered at $135,000, narrower in spread than Option A, featuring a range from $127,000 to $143,000.- **Option C:** Presents a flat curve with a much wider spread, centered far below $135,000 at around $13,000.- **Option D:** Depicts a narrow normal distribution curve, incorrectly centered at $206,000.The correct choice, considering the calculated mean ($135,000) and standard deviation ($2,433), is **Option B**. This option visually represents the normal distribution characteristics of the sampling distribution with the appropriately centered mean and narrower spread due to the sample size.

The mean of the sampling distribution of these sample means is obtained below: From the given…

The annual salary for one particular occupation is normally distributed, with a mean of about $135,000 and a standard deviation of about $15,000. Random samples of 38 are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of these sample means. Then, sketch a graph of the sampling distribution. The mean is u = and the standard deviation is o: = (Round to the nearest integer as needed. Do not include the $ symbol in your answers.) Sketch a graph of the sampling distribution. Choose the correct answer below. OA. OB. OC. OD. Question Viewer 127.000 135.000 143.000 Mean salary (in dolars) 13,900 21 900 29 900 Mean aary n dolars 20 700 21.900 23 100 90,000 135 000 180.000 Mean salary (in dolars)

The annual salary for one particular occupation is normally distributed, with a mean of about $135,000 and a standard deviation of about $15,000. Random samples of 38 are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of these sample means. Then, sketch a graph of the sampling distribution. The mean is u = and the standard deviation is o: = (Round to the nearest integer as needed. Do not include the $ symbol in your answers.) Sketch a graph of the sampling distribution. Choose the correct answer below. OA. OB. OC. OD. Question Viewer 127.000 135.000 143.000 Mean salary (in dolars) 13,900 21 900 29 900 Mean aary n dolars 20 700 21.900 23 100 90,000 135 000 180.000 Mean salary (in dolars)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

**Title: Understanding the Sampling Distribution of Sample Means**

The annual salary for a specific occupation is normally distributed, with a mean salary of approximately $135,000 and a standard deviation of about $15,000. Random samples of 38 individuals are taken from this population, and the mean of each sample is calculated. The task is to determine the mean and standard deviation of the sampling distribution of these sample means, and then sketch a graph of this distribution.

**1. Calculate the Mean and Standard Deviation:**

- **Mean ($\mu_x$)**: Since the sample mean is an unbiased estimator of the population mean, $\mu_x = \mu = 135,000$.

- **Standard Deviation ($\sigma_x$)**: The standard deviation of the sampling distribution ($\sigma_x$) is given by the formula:

\[
\sigma_x = \frac{\sigma}{\sqrt{n}}
\]

Where:
- $\sigma = 15,000$ (population standard deviation)
- $n = 38$ (sample size)

Calculating $\sigma_x$, we get:

\[
\sigma_x = \frac{15,000}{\sqrt{38}} \approx 2,433
\]

**2. Graph of the Sampling Distribution:**

The task includes sketching a graph of the sampling distribution. Below are the descriptions of the given options:

- **Option A:** Shows a normal distribution curve centered around $135,000 with a wide spread.
- **Option B:** Displays a normal distribution curve centered at $135,000, narrower in spread than Option A, featuring a range from $127,000 to $143,000.
- **Option C:** Presents a flat curve with a much wider spread, centered far below $135,000 at around $13,000.
- **Option D:** Depicts a narrow normal distribution curve, incorrectly centered at $206,000.

The correct choice, considering the calculated mean ($135,000) and standard deviation ($2,433), is **Option B**. This option visually represents the normal distribution characteristics of the sampling distribution with the appropriately centered mean and narrower spread due to the sample size.

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