i think given your data, you can get the variance..
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
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So far, I have got 2b+3a = 1.4
Don't know what to do next..
Thanks!
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
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 ?
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