# Statistics Probability Distribution..

• Apr 9th 2008, 05:50 AM
Danielisew
Statistics Probability Distribution..
Hey

I am stuck on this question in the June '04 Edexcel past paper.. I know how to find the answer but I can't get any further so I am stuck.. It's reasonably easy probably.. Here's the question:

3.
A discrete random variable X has a probability function as shown in the table below, where a and b are constants

x - 0, 1, 2, 3
P(X=x) - 0.2, 0.3, b, a

Given that E(X) = 1,7,

(a) Find the value of a and the value of b

------------------------

So far, I have got 2b+3a = 1.4

Don't know what to do next..(Worried)

Thanks!
• Apr 9th 2008, 05:58 AM
kalagota
i think given your data, you can get the variance.. \$\displaystyle Var \, X = E[X^2] - E[X]^2\$
• Apr 9th 2008, 06:20 AM
Danielisew
I know that, but how does that help me in finding a and b? I need to find them and they are constants..
• Apr 9th 2008, 06:24 AM
kalagota
of course your variance is in terms of a and b.. thus you have 2 equations in 2 unknowns.. Ü
• Apr 9th 2008, 06:25 AM
Oli
Writing down the equation for the variance doesn't help, because you don't know what the variance actually is, and so you end up with another unknown.

The sum of the probabilities must be equal to 1. That is how you get your 2nd equation.

0.2 + 0.3 + b + a = 1
So
b + a = 0.5

And:
2a + 3b = 1.4 from what you did.

These are linearly independent, so solve to find a and b.
b=0.4
a=0.1
• Apr 9th 2008, 06:32 AM
colby2152
Quote:

Originally Posted by Danielisew
Hey

I am stuck on this question in the June '04 Edexcel past paper.. I know how to find the answer but I can't get any further so I am stuck.. It's reasonably easy probably.. Here's the question:

3.
A discrete random variable X has a probability function as shown in the table below, where a and b are constants

x - 0, 1, 2, 3
P(X=x) - 0.2, 0.3, b, a

Given that E(X) = 1,7,

(a) Find the value of a and the value of b

------------------------

So far, I have got 2b+3a = 1.4

Don't know what to do next..(Worried)

Thanks!

You will have a system of two equations. The other equation is: \$\displaystyle b+a=0.5\$ because all the probabilities must add up to one.
• Apr 9th 2008, 06:38 AM
Danielisew
Oh yeah stupid me.. THanks for that.
I did the variance one whilst minusing the E(X) and got the equation:

2b + 12a = 1.1

Don't know if that would work too..(Worried)

I'll use the obvious one a+b = 0.5
• Apr 9th 2008, 06:39 AM
kalagota
(Rofl) haha.. yeah.. i forgot that thing!
• Apr 9th 2008, 06:55 AM
Danielisew
Wait... Guys I am even more stuck and confused now.
I got a = 0.4 and b = 0.1

I simultaneously solved :

2b + 3a = 1.4
2b + 2a = 1

Isn't that second one the same as a + b = 0.5 ?

Anyway.. A later question says show that the variance of X is 1.41, but I can't get that to work lol.. Is Oli right or me with values of a and b? Can you double the equation to give 2b + 2a = 1 ?
• Apr 9th 2008, 07:46 AM
Oli
You are right, I solved for 2a+3b=1.4. Should have solved for 3a+2b=1.4. My mistake.

The variance is:
E(X^2)-E(X)^2
This is:
0^2.P(X=0)+(1^2)P(X=1)+(2^2)P(X=2)+(3^2)P(X=3)-E(X)^2
=1*0.3+4*b+9*a-1.7^2
=0.3+0.4+3.6-1.7^2
=4.3-1.7^2
=1.41