| Author |
Message |
    
Moe (Moe)
| | Posted on Monday, June 10, 2002 - 04:06 pm: | 
|
I'm looking for some help today and tomorrow with some basic statistics--right now there isn't any available on campus. Believe me I would only be posting stat questions here if I was a little panic-stricken 
If anyone has a good working knowledge of Splus (or R language) and is willing to walk me through a couple of problems, I would gladly return a favor or pay $ for the time. Email here if you can help
Of course, for others with some basic stats here are three problems dealing with probability.
a) Two discrete random variables, X and Y, are independent. X can take values 1, 2, 3, while Y can take values 0 and 1. The probability function of X is the following: f(1)= 0.2; f(2)= 0.25; f(3)= 0.55. The probability function for Y is f(0)= 0.1; f(1)= 0.9. What is the joint probability of X = 1 and Y = 1?
b) The probability that a randomly selected US citizen is female is 0.51. Suppose you take a random sample of size 30 from the US population. What is the probability that 15 individuals in the sample will be female? (You can assume individuals have been sampled independently from the US population)
c) The probability that a randomly selected US citizen is female is 0.51. Suppose you take a random sample of size 30 from the US population. What is the probability that 13 or few individuals in the sample will be female? (You can assume individuals have been sampled independently from the US population) |
Ho Chung (Ho)
| | Posted on Monday, June 10, 2002 - 04:32 pm: |
|
moe, have you considered playing lotto? |
Moe (Moe)
| | Posted on Monday, June 10, 2002 - 05:06 pm: |
|
Ho, given that I'm ignorant regarding probabilities, I play Lotto all the time |
niall forbes (Forbesn)
| | Posted on Monday, June 10, 2002 - 09:36 pm: |
|
Whoa!! Moe, your posts were exactly an hour apart. Spooky. I got an A in Stats last term but I've forgotten it all already ;-) Sorry I can't help. My head hurts just reading the questions. LOL. |
Moe (Moe)
| | Posted on Monday, June 10, 2002 - 10:58 pm: |
|
Whoa!! Niall what is the probability of that
I 'think' these are the answers. Now onto the next sections.
a) Joint probabilities of independent events is just the product. 0.2*0.9 = 0.18
b) Choose(30,15)*.51^15*.49^15
= 0.1436001. That is a 14% chance
c) Choose(30,13)*.51^13*.49^17
= 0.1023427 That is a 10% chance
I hope this is right |
|
Closed: New threads not accepted on this page