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# S1 question watch

1. I've worked out the answers to the following questions but I don't know how I should set them out

1a) (It's a table)

When x=1, P(X=x) is 0.3-a,
When x=2, P)X=x) is 2a,
When x=3, P(X=x) is 0.7-a

"Find the possible values of a" - 0<a<0.3 (thanks to paronomase )

"Find the value of a when var(X)=0.64" - I think it's 0.1, but I found it by trial and error

Can anyone help me with setting the answers out correctly, please?
2. First of all,
do you know that 0< (or equal) P(X=x) < or = 1
Therefore, here you have to solve:

0< 0,3-a < 1 (again it is inferior or equal)
0 < 2a < 1
0 < 0,7-a < 1

this gives you

-0,7<a<0,3
0 < a < 0,5
-0,3 < a < 0,7

Now, you are looking for a value of a that satisfies all of these inequalities (the intersection of the intervals that satisfy the three inequalities)
With the 2nd one, you are imposed a>0
With the 1st one, you are imposed a<0,3

Therefore, those two conditions have to be satisfied, therefore 0<a<0,3
3. a is between zero and 0.3, as p(x=1,2,3) can not be less than zero. Sorted, right that

so 0.3-a + 2a +0.7-a=1, (1(0.3-a)+4(2a) +9(0.7-a))-(1(0.3-a) + 2(2a) +3(0.7-a)) squared=0.64 (for variance)

algebra it out and you get whatever the answer is
4. Thank you both for your help

I still can't do it though

I thought I had an answer, but when I used it to work out the next part, P(X1 + X2=4) = 0.28, I didn't get 0.28

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