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sebastianzx6r
October 31st 2005, 07:10 PM
I'm having trouble with a couple of problems.

Let a chip be taken at random from a bowl that contains 6 white chips, 3 red chips, and 1 blue chip. Let the random variable X = 1 if the outcome is a white chip; let X = 5 if the outcome is a red chip; and let X = 10 if the outcome is a blue chip.
a. Find the p.m.f. of X
b. Find the mean and standard deviation of X.

For a I need a function where you can plug in X=1 and get 6/10, X=5 and get 3/10, etc..

Once I have that I can find the mean and standard deviation. I've been trying to figure out that equation.
I have (x^2+5)/10x
It gives the correct x and y values for the 1st 2 x's but not the 3rd.

Also

Suppose X is a random variable such that E(X+4)=10 and E((X+4)^2)=116

How do I find the mean and variance?
Any help would greatly be appreciated.

Thanks

.Godspeed.
October 31st 2005, 09:24 PM
I'm having trouble with a couple of problems.

Let a chip be taken at random from a bowl that contains 6 white chips, 3 red chips, and 1 blue chip. Let the random variable X = 1 if the outcome is a white chip; let X = 5 if the outcome is a red chip; and let X = 10 if the outcome is a blue chip.
a. Find the p.m.f. of X
b. Find the mean and standard deviation of X.

For a I need a function where you can plug in X=1 and get 6/10, X=5 and get 3/10, etc..

Once I have that I can find the mean and standard deviation. I've been trying to figure out that equation.
I have (x^2+5)/10x
It gives the correct x and y values for the 1st 2 x's but not the 3rd.

Also

Suppose X is a random variable such that E(X+4)=10 and E((X+4)^2)=116

How do I find the mean and variance?
Any help would greatly be appreciated.

Thanks

Why don't we try:
P(X=x) = 6/10, x = 1
3/10, x = 5
1/10, x = 10

Also, the fact that it is a probability mass function of a discrete (as opposed to a continuous) distribution implies that over all x, the sum of the probabilities must equal 1. This certainly is true for our P(X).

For the second question, since the expected value is a linear operator and the expected value of a constant is the constant itself, E(X+4)=E(X)+4=10-->E(X)=6.

Also, E((X+4)^2)=E(X^2+8X+16)=E(X^2)+E(8X)+E(16)
=E(X^2)+8*E(X)+16=E(X^2)+8*6+16=116-->E(X^2)=52
And, Var(X)=E(X^2)-E(X)^2=52-6^2=16.

Help?

Ken
October 31st 2005, 11:08 PM
I'd really appreciate if you don't make duplicate threads next time. You can post anywhere and I will find it. I see I just wasted time answering your question in your duplicate thread.

SirVLCIV
October 31st 2005, 11:21 PM
I'd really appreciate if you don't make duplicate threads next time. You can post anywhere and I will find it. I see I just wasted time answering your question in your duplicate thread.