Answered: You are told 20% of plain M&M’s are…

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

You are told 20% of plain M&M’s are green. In a random sample of 500 plain M&M’s you find that 123 are green. What is the probability of getting the result you found?

- Are the conditions required to use the Normal sampling distribution for

^p (sample proportion) met?

- what does the sampling distribution model look like?

- Is this an unusual result?

Expert Solution

Step 1

Given Information:

20% of plain M&M’s are green. i.e., $p = 0.20$

Sample size $n = 500$ plain M&M's

Out of which X = 123 are green.

Sample proportion is the number of successes divided by the sample size. It is given by the formula:

$\begin{array}{rcl} \hat{p} & = & \frac{X}{n} \\ = & \frac{123}{500} \\ = & 0.246 \end{array}$

Check if the conditions are met:

$n p = 500 (0.20) = 100 > 10$

$n (1 - p) = 500 (1 - 0.20) = 400 > 10$

Conditions required to use the Normal sampling distribution for $\hat{p}$ (sample proportion) met.

The sampling distribution model will be normal, with mean $μ_{\hat{p}} = p = 0.20$ and standard deviation : $\begin{array}{rcl} σ_{\hat{p}} & = & \sqrt{\frac{p q}{n}} \\ = & \sqrt{\frac{0.20 (1 - 0.20)}{500}} \\ = & \sqrt{\frac{0.16}{500}} \\ = & 0.01788854382 \\ \approx & 0.0179 \end{array}$

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