Question

nts, under any circumstances . You can only upload ONE combined file. . You must ALWAYS show all your work! For example, if I was going to answer the following question, here is how I would show my work: Question: Find the mean of the following data set (2,3.4,5,6 Answer Step 1: 2+3+4+5+6 20 Step 2: 20/5-4 Step 3: Mean 4 LabAssignment#2 Z-scores, Sampling Distributions, and the Central Limit Theorem Note: Be sure to label your answers with the correct number/letter, and show all your work. 1. Given the following contents of a bag of M&Ms: 5 red 2 blue 6 green 4 yellow 5 brown 6 orange a. What is the probability that the first M&M we choose will be either yellow or orange? b. What is the probability that the first M&M we choose will be green and the second M&M we choose (after first replacing the initial M&M) will be blue? 2. I am tutoring 5 students in statistics and these are their scores on a recent e (86, 88, 89, 90, 92) What is the average of these scores? What is the standard deviation of this population of scores? a. b.

Answer 1

Given the bag contains M&M as given below.

$5 \textup{ red},2\textup{ blue},6 \textup{ green}.4 \textup{ yellow},5 \textup{ yellow},6 \textup{ orange}$

a)Consider sampling with replacement ,

The probability that the first drawn card is either yellow/orange is found as follows)

$P\left ( \textup{yellow} \cup \textup{orange}\right )=P\left ( \textup{yellow} \right )+P\left ( \textup{orange}\right )\\ P\left ( \textup{yellow} \cup \textup{orange}\right )=\frac{4}{28}+\frac{6}{28}$

The required probability is $\frac{4}{28}+\frac{6}{28}={\color{Blue} 0.35714286}$

b)Consider sampling with replacement ,

The probability that the first drawn card is green is

$P\left ( \textup{yellow}\right )=\frac{4}{28}\\ P\left ( \textup{green} \right )=\frac{6}{28}$

$P\left ( \textup{green} \cap \textup{blue}\right )=\frac{6}{28}\left ( \frac{2}{28} \right )\\ {\color{Blue} P\left ( \textup{green} \cap \textup{yellow}\right )=0.015306122}$

c)The probability that the first drawn card is either yellow/orange is found as follows)

$P\left ( \textup{green} \right )=\frac{6}{28}$ and $P\left ( \textup{blue} \right )=\frac{2}{28}$

The required probability is $\frac{6}{28}\left (\frac{2}{28} \right )={\color{Blue} 0.0204}$ .

We are required to solve only one question. Please post the remaining questions as another post.