Expected value problem and normal distribution

laythen

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Oct 21, 2011
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Hello, I would appreciate any help with the following problems:

The discrete random variable X is the number of students that show up for Professor Smith's office hours on Monday afternoons. The table below shows the probability distribution for X. What is the probability that the total number of students showing up on Monday afternoons in 2 weeks (assuming independence of results in the two Mondays) is less than 3?

X
0
1
2
3
P(X)
.45
.35
.15
.05

I had a similar problem like this one but it did not ask for two weeks of probability and it said "at least 1". I have no idea what to do for this one because they threw that "2 weeks" in there. What I initially tried to do was: (0)(.45)+(1)(.35)+(2)(.15) / (.45+.35+.15) and this gives me 68.42. The answer, however, is 0.775. As I said, I am completely lost and have no idea how to get 0.775.


My second question is this:

What percent of cases are likely to be between 86 and 93 in a normal distribution with mean 87 and variance 4?

What I did was 86-87/4 and I got 0.0987. Then I did 93-87/4 and I got 0.4332. Added those together I get 0.5319 but apparently the answer is .6902.

Any help appreciated please. Thank you,

Laythen
 
Hello, I would appreciate any help with the following problems:

The discrete random variable X is the number of students that show up for Professor Smith's office hours on Monday afternoons. The table below shows the probability distribution for X. What is the probability that the total number of students showing up on Monday afternoons in 2 weeks (assuming independence of results in the two Mondays) is less than 3?

X0123
P(X).45.35.15.05

I had a similar problem like this one but it did not ask for two weeks of probability and it said "at least 1". I have no idea what to do for this one because they threw that "2 weeks" in there. What I initially tried to do was: (0)(.45)+(1)(.35)+(2)(.15) / (.45+.35+.15) and this gives me 68.42. The answer, however, is 0.775. As I said, I am completely lost and have no idea how to get 0.775.
Let Y be the random variable for the number of students in week 1. Let Z be the random variable for the number of students in week 2. Let W be the random variable for the number in both weeks combined.

What is the distribution for W (given that Y and Z are independent and W = Y + Z)? Tell us what you get and how so we can see where, if anywhere, you are getting stuck?

As for at "least 1", is it not clear that less than 3 means at most 2 and that the processes for "at least" and "at most" must be quite analogous?
 
Hi, I apologize but I am still unclear on what to do. I have been trying to work it out since I posted it and since I read your reply, but I'm afraid I simply don't understand. If I look at it as only one week, then wouldn't I be able to say that the probability that less than 3 students shows up is .45+.35+.15? Then couldn't I square that to apply it to 2 weeks? Other than that I really can't think of any way to solve it besides what I previously posted. I will continue brainstorming.. Thank you for the reply.
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Hello, I believe I solved it. Here is what I did:

Assume my brackets are first week/second week and the numbers indicate the maximum number of students in those two weeks that are under 3.

(0,0) (1,0) (0,1) (1,1) (2,0) (0,2)

So essentially I may get 0 students both weeks, or 1 student the first week then 0 the second, or 0 the first then 1 the second, etc. So I did (0,0) which = (.45)(.45) and I did that for every bracket and added it up for 0.775. Is there another way of doing this? Or did I do the standard way? Thank you.
 
Last edited:
Hello, I believe I solved it. Here is what I did:

Assume my brackets are first week/second week and the numbers indicate the maximum number of students in those two weeks that are under 3.

(0,0) (1,0) (0,1) (1,1) (2,0) (0,2)

So essentially I may get 0 students both weeks, or 1 student the first week then 0 the second, or 0 the first then 1 the second, etc. So I did (0,0) which = (.45)(.45) and I did that for every bracket and added it up for 0.775. Is there another way of doing this? Or did I do the standard way? Thank you.
What you did was fine, and any method is essentially following the same logic. If you have a similar problem with multiple questions, it's probably safest to construct two two-way tables (one for sums of Y and Z and one for the probabilities of P(Y)P(Z) in the cells) and then summarize them into a table for values of W and the associated probabilities. That way you do not forget anything and reduce the chances of making a careless error. I find constructing tables at this site really cumbersome, but if you do not understand the tables that I am talking about, please let me know.
 
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