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Course Documents

Chapter 1 - Intro

Chapter 2 - Methods for Describing Sets of Data

Chapter 3 - Probability

Chapter 4 - Discrete Random Variables

Chapter 5 - Normal Random Variables

Chapter 6 - Sampling Distributions

Chapter 7 - Confidence Intervals

Chapter 8 - Tests of Hypothesis: One Sample

Chapter 9 - Confidence Intervals and Hypothesis Tests: Two Samples

Sample Exam I: Chapters 1 & 2

Sample Exam II: Chapters 3 & 4

Sample Exam III: Chapters 5 & 6

Sample Exam IV: Chapters 7 & 8

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Recently asked questions for category: Discrete Probability Distributions

A Stats Professor student asks:

Dear Professor, I came across two more questions as I was studying.

For Now You Try It Section 4.2 Discrete Random Variables

For Number one the part that asks if it would be unusual for someone to roll 3 or 4 sevens out of four rolls how do you know to use within 2 standard deviations of the mean? And on the test will you speficy by saying capture 75%? Also how do we know whether we use a z-score or chevy's theorem or empirical to figure out if it is unusual?

Problem 1:
A pair of six-sided die is rolled four times. We record the number of sevens rolled
over the four rolls. Calculate the standard deviation for the number of sevens
rolled. Would it be unusual for someone to roll 3 or 4 sevens out of four rolls?

Solution:
The standard deviation is: σ = 0.746 . Based on the following interval it
would be unusual to roll more than two sevens out of four rolls:
( 2 , 2 ) (0.667 2*0.746,0.667 2*0.746) ( 0.825, 2 µ σ µ σ − + = − + = − .159) .

For Problem 2:
I am not sure what information to use to go about finding the standard deviation?
Converting the above probability distribution into a game, players can risk $4 to bet
that a seven will not be rolled in four throws of the dice. If the player wins, he will
be paid $7 (a three-dollar profit), but if the player loses he/she loses his/her $4 bet.
What is the expected value for this game? Is this a smart bet to make in the long
run?

Solution:
The standard deviation isσ = 3.50 , but this is not very useful since there are
only two outcomes for a trial. We know either a person will win 3 dollars
or will lose 4 dollars. The standard deviation is not needed to capture the
variation. It is easier to see the variation by simply viewing the win and
loss probabilities.

Thank you.

Last reply:  Monday, October 21, 2013 9:16 AM EST        Total Replies: 1

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