Stat 2000: Tips for Assignment 3
Published: Fri, 10/21/11
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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). I am also surprised by the rather complex probability questions on this assignment, so I have attached a handout from my Basic Stats 1 study book with a more thorough discussion of making two-way tables and Venn diagrams.
Question 1 is very similar to my question 18 in the handout I have given you.
Note that I also discuss conditional probability in the handout, but don't really show any examples of it. This is denoted P(B|A), pronounced "probability of B given A". The formula for this is
P(B|A) = P(A and B) divided by P(A)
What this really means is this:
Step 1: Look through your sample space or Venn diagram and find all the parts that belong to A and add those probabilities up. That is the denominator of your answer, P(A).
Step 2: Now go through the bits you just added up in Step 1 and collect the bits that also belong to B. Add that subset of bits up. That is the numerator of your answer, P(A and B) because that is the bits that are B and also A.
Step 3: Divide your answer in Step 2 by your answer in Step 1 to get the conditional probability.
Essentially, 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.
For example, if you look at my question 18 in the handout, I could add a part (d) that asks, "What
is the probability someone is a basketball fan if they are a hockey
fan?" Any probability question that asks, what is the probability of B if event A has occurred, you are doing conditional probability.
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.
Question 1(a) requires you to write the Sample Space which is kind of annoying because that is not helpful in solving the problem. You could use two-way tables, where the columns of your first table would be M and MC, the complement of M and the rows of your first table would be A and AC, etc. You can also read off the sample space from the Venn diagram you construct. Just look at each bit in your Venn diagram and describe what it represents. For example, in the hockey, football, basketball Venn diagram I construct in question 18 of the attached handout, the part of the diagram where I have 33% placed would be people who like hockey, but don't like football or basketball, so that would be HFCBC, whereas the part where I have placed 31% would be HFBC since those people like hockey and football but don't follow basketball. There are a total of 8 numbers in my Venn diagram, so that means there are 8 outcomes in the sample space.
Question 2 is similar to question 13 in my handout but also uses the formulas for mean and variance I teach in Lesson 6 of your book. By the way, in part d, what they mean is look at each outcome of your sample space and write down the minimum number. For example, one outcome should be 32 (you selected 3 for your first card and 2 for your second card). In that case X=2 since 2 is the lowest number you selected. Another outcome is 44, so X=4 in that case.
Question 3 is very challenging. Hint: Let X = the times in the freestyle and Y = the times in the butterfly. Then, the total time is X+Y. You can use the properties of mean and variance I teach in Lesson 4 to work out the mean and variance of X+Y. Since X and Y are both normal distributions, X+Y is also normal. Part (b) wants X<Y which can also be written X - Y < 0. Again, you can find the mean and variance of X - Y.
Finally, the usual kind of stuff occurs for the rest of this assignment. I teach and discuss all the concepts covered in questions 4 through 7 of the assignment in my Lesson 6.
I assume your prof has written out the 4 conditions of a binomial distribution in class. Here is the way I would number and describe them:
1. Each trial has only two possible results: Success or Failure.
2. Each trial is independent.
3. There are a fixed number of trials, n.
4. The probability of success on each trial is constant, p.
There is no distance/online course for Stat 2000 this term.