Stat 2000: Assignment 3 Tips (Classroom Lecture Sections)
Published: Wed, 03/06/13
My tips for Assignment 3 are coming below, but first a couple of announcements.
Please note that I am planning on splitting my final exam seminar for
Stat 2000 into two days since we will have to cover Lesson 6 in Volume 1 as well as all of Volume 2. The plan is to meet from 9:00 am to 6:00 pm each day. Each day will cost $40 or, if you attend Day One, you can attend Day Two for half-price (you will pay a total of $60, in other words).
I plan to teach Day One on Easter Sunday, March 31 and Day Two 2 weeks later on Sunday, April 14. I will send more details and start taking registrations once everything is finalized next week.
Make sure you do: Tips on How to Do Well in Stat 2000
Did you read my Tips on what kind of calculator you should get?
Did you miss my Tips for Assignment 2?
Tips for Assignment 2 (Classroom Lecture Sections A01, A02, A03, etc.)
Don't have my book? You can download a free sample containing Lesson 3 at my website here:
You need to study Lesson 6: Discrete Probability Distributions (this
also includes the Binomial and Poisson Distributions; if you are using a
considerably older edition of my book, you may have those two
distributions taught in a separate Lesson 7).
Extra Tips for Probability
I am surprised to
see a lot of probability questions on this assignment that are from Stat
1000, so I have included a handout from my Basic Stats 1 study
book with a more thorough discussion of making two-way tables and Venn
diagrams. Those of you who have my Basic Stats 1 book, should study
Lesson 5: Introduction to Probability.
Conditional Probability
Here is another handout explaining the approach to determine a conditional probability:
Essentially, in conditional probability, when it says "given A" is
telling you that we know for sure that event A has occurred, so we are
now only interested in outcomes that belong to A. That becomes the
"whole". P(B|A) wants the fraction of that "whole" that also belongs to
B.
We want P(B|H). I first look through my Venn diagram and find all
the bits that belong to H, since we know for sure the person is a hockey
fan. There are four bits in the H circle so I add those bits up: 33 +
31 + 8 + 5 = 77%. Now, I gather all the bits in that H circle that
represent people who are also basketball fans. There are two bits: 8 + 5
= 13%. Thus, the probability a person is a basketball fan if they are a
hockey fan is 13%/77% or .13/.77 = .1688.
Here is a couple of extra conditional probability questions I have added to question 4 in my probability handout above:
Here is a couple of extra conditional probability questions I have added to question 16 in my probability handout above:
Additional Tips to help with the Assignment
Question 1:
If
you are ever asked to decide if a particular situation is binomial or
not, remember, to be binomial, four conditions must be satisfied:
(i) There must be a fixed number of trials, n.(ii) Each trial must be independent.(iii) Each trial can have only two possible outcomes, success or failure, and the probability of success on each trial must have a constant value, p.(iv) X, the number of successes, is a discrete random variable whereX = 0, 1, 2, ... n.
Hints for question 1: If you are reading off
numbers from a randomly selected row in the random number table, note
that every row has 40 digits. That is like 40 trials looking for
whatever digit you may be looking for. What is the probability that, at
any moment on the table, the next digit is a 0, or a 1, or a 2, etc.?
If you are selecting objects, are you sampling with replacement (independent trials) or without replacement (dependent trials)?
Question 2 (h) and (j) are examples of conditional probability.
I also discuss conditional probability in the handout, above.
Do you notice what kind of method you need to use to solve question 2 (k) and (l)?
Question 3 is a two-way table problem similar to
my
question 13 in the handout above.
For question 3(d), go through each outcome in your sample space and
determine the value of X first. For example, one of the outcomes in the
sample space you listed in part (a) would be "13" meaning X1=1 and X2=3 (you might write it as 1:3 or 1;3 or something
so that it is not mistaken for thirteen). Then X=2 since |X1 - X2| = |1 - 3| = |-2| = 2.
Note that question 3(e) is a conditional probability.
Questions 3(f) and (g) are using the formulas for mean and variance I show you at the start of Lesson 6 (questions 1, 2 and 3).
Questions 7 and 8 are obviously Poisson Distribution questions similar to my examples in Lesson 6.