S1 random varibable help!!!!!!!

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.

Reply 1

gorgeousguy

7

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.

Reply 2

Smashingdude

OP

4

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) .

Reply 3

Paulus

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....Total
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 :s-smilie:

Reply 4

Jean

2

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?

Reply 5

Jean

2

for e, see attachment.

probability.doc28.7KB

Reply 6

Jean

2

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

Reply 7

Zaybraaa

1

I'm stuck at e, the marking scheme isn't helpful and I'm on vacation can't even ask the teacher..

Reply 8

username3249896

12

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

Reply 9

Zaybraaa

1

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

Reply 10

Zaybraaa

1

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

12 yrs late buddy

e.For example: when X1 X2=1 P{X1 X2}=P(x1=0)*P(X2=1)*2This is multiplied by two, because you can get X1=0 and X2=1 in any order, for example getting x1=0 and x2=1 or the other way round x1=1 and x2=0So, P{X1 X2}=P(x1=0)*P(X2=1)*2P{1}=P(x1=0)*P(X2=1)*2P(x1=0)=2*0.25=0.5P(X2=1)=0.25P{1}=0.25*0.5*2=0.25 :smile:

Reply 13

_zahra2608

3

damn i see no one knows the answer and its been 14 years

Reply 14

ghostwalker

17

Original post by _zahra2608

damn i see no one knows the answer and its been 14 years

Plenty of people could help, but this isn't a do your homework for you site. See the maths forum posting guidelines - sticky thread at top of forum.

If you want some assistance, post your working so far, and say where precisely you're stuck.

(edited 3 years ago)

Reply 15

_neeesarak

1

thank you !

S1 random varibable help!!!!!!!

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