Answered: The per capita consumption of red meat…

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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Question

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

109

pounds and a standard deviation of

38.3

pounds. Random samples of size

are drawn from this population and the mean of each sample is determined.

μx=nothing

σx=nothing

(Round to three decimal places as needed.)

Sketch a graph of the sampling distribution. Choose the correct graph below.

109 89.2128.8

A normal curve is over a horizontal x overbar axis labeled from 89.2 to 128.8 and is centered on 109.

9.9 -99.1118.9

A normal curve is over a horizontal x overbar axis labeled from negative 99.1 to 118.9 and is centered on 9.9.

9.9 -317.1336.9

A normal curve is over a horizontal x overbar axis labeled from negative 317.1 to 336.9 and is centered on 9.9.

109 79.3138.7

A normal curve is over a horizontal x overbar axis labeled from 79.3 to 138.7 and is centered on 109.

Click to select your answer(s).

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

The mean for the sampling distribution is,

$\begin{array}{rcl} μ_{\bar{x}} & = & μ \\ = & 109 \end{array}$

Thus, the mean for the sampling distribution is 109.

The standard deviation for the sampling distribution is,

$\begin{array}{rcl} σ_{\bar{x}} & = & \frac{σ}{\sqrt{n}} \\ = & \frac{38.3}{\sqrt{15}} \\ = & 9.889 \end{array}$

Thus, the standard deviation for the sampling distribution is 9.889.

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