S1 random varibable help!!!!!!! Watch

Smashingdude
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#1
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The discrete random variable X has probability function

k(2 –x), x= 0, 1, 2,

P(X = x) = k(x–2), x= 3,

0, otherwise,


where k is a positive constant.

(a) Show that k = 0.25.
(b) Find E(X) and show that E(X ) = 2.5.
(c) Find Var(3X – 2).

Two independent observations X1 and X2 are made of X.

(d) Show that P(X1 + X2 = 5) = 0.
(e) Find the complete probability function for X1 + X2.
(f) Find P(1.3 = X1 + X2 = 3.2).

Please can anyone tell me how to do the parts d,e,f? i have done the first 3 parts but these last three are confusing. Ive got the answers for this but in the MS they havent explained anything.
anyways this is wht written in the MS.
How and wht the hell they did? and wht is the logic behind it?

(d) P(X1 + X2) = P(X1 = 3 X2 = 2) + P(X1 = 2 X2 = 3) = 0 + 0 = 0
Let Y = X1 + X2 y 0 1 2 3 4 5 6
(e)
P(Y = y) 0.25 0.25 0.0625 0.25 0.125 0 0.0625

(f) P(1.3 X1 + X2 3.2) = P(X1 + X2 = 2) + P(X1 + X2 = 3)
= 0.0625 + 0.25 = 0.3125

Please help me out here.
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gorgeousguy
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k(2 –x), x= 0, 1, 2,

P(X = x) = k(x–2), x= 3,
This is quite confusing. do you work out "k" using k(2-x) or k(x-2), also can you confirm exaclt what values x can take, as im not sure whether it is "0,1,2 or 0,1,2,3.

When you do this ill try to help you.
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Smashingdude
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Its actually :

............ { k(2 –x), x= 0, 1, 2,
..P(X=x) { k(x–2), x= 3
............ { 0, otherwise,

Forget the dots. The program wont allow me to leave spaces. so i just covered it with dots. I hope it makes sense now. X can take values of 0,1,2 usinf k(2-X) and can take 3 using k(X-2) .
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Paulus
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What syllabus is this?
because I hope i dont have to know about "observations"!

I got the first three parts also, i think, but I'm hoping there is a letter missing in part b - E(X[here])

x............0...1...2...3....To tal
P(X=x)....2k..k...0...k......4k
xP(X=x)...0...k...0..3k.....4k

a)

4k = 1 (because its the total)
therefore:
k=1/4

b) therefore I got E(X)=1

...just showed this so other people are clear...and to improve my confidence for tomorrow. Feel free to tell me if thats right...god i hope it is
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Jean
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for d, the probability of event x=2 and x=3 is 0 as one of the probability is 0. so when you times the probability of x=2(which is 0) and the probabilty of x=3, you'll get 0.

am I right?
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Jean
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for e, see attachment.
Attached files
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Jean
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#7
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for f,

P(1.3<X1=X2<3.2)= P(1.3<Y<3.2)
=P(2<Y<3)
=0.625+0.25
=0.3125
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Zaybraaa
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#8
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I'm stuck at e, the marking scheme isn't helpful and I'm on vacation can't even ask the teacher..
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BobbJo
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(Original post by Zaybraaa)
I'm stuck at e, the marking scheme isn't helpful and I'm on vacation can't even ask the teacher..
Solution in post #6
Do you understand it?
Consider the possible values of  X_1 + X_2
Consider the possible ways to obtain the values
Find the probabilities
Write the probability distribution function
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Zaybraaa
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#10
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#10
just seen the attached file, thank you sm <3

(Original post by BobbJo)Solution in post #6
Do you understand it?
Consider the possible values of  X_1 + X_2
Consider the possible ways to obtain the values
Find the probabilities
Write the probability distribution function
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Zaybraaa
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#11
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#11
hey can ik why is x till 6 for e?
(Original post by BobbJo)Solution in post #6
Do you understand it?
Consider the possible values of  X_1 + X_2
Consider the possible ways to obtain the values
Find the probabilities
Write the probability distribution function
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