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S1 discrete random variables Watch

1. for this question part (b) to get E(X) I did 60 + 2 /2 so = 31
but for the next part working out the variance the formula

1/12 (n^2 -1) doesn't work, for the value of n do I use 60?
2. What board are you doing?
3. (Original post by calla_lily)
What board are you doing?

Edexcel
4. Same as me! Which paper is this?
5. Forgive me if I'm wrong but this is how i see it:

1a) All probabilities are the same

1b) as each value is equally likely mean = E(x) = n+1/2 = 30+1/2 = 15.5 (using n=30 surely, not sure why you are using 60?)

variance = (n+1)(n-1) = ? but then, like you, I don't quite understand why this doesn't work using this method
12
6. Babe, it seems to me you are using the formulas to find the mean and variance of a discrete uniform distribution (all outcomes have the same probability), if this is not true for your question, which you have not told us, you have to use the other method for finding E(X) and Var(X) where var(x) is e(x^2)-(e(x))^2 etc...
7. (Original post by calla_lily)
Forgive me if I'm wrong but this is how i see it:

1a) All probabilities are the same

1b) as each value is equally likely mean = E(x) = n+1/2 = 30+1/2 = 15.5 (using n=30 surely, not sure why you are using 60?)

variance = (n+1)(n-1) = ? but then, like you, I don't quite understand why this doesn't work using this method
12
for E(X) it's 60+2/2 = 31 which is correct

(Original post by Bobsyourgrandma)
Babe, it seems to me you are using the formulas to find the mean and variance of a discrete uniform distribution (all outcomes have the same probability), if this is not true for your question, which you have not told us, you have to use the other method for finding E(X) and Var(X) where var(x) is e(x^2)-(e(x))^2 etc...
for part a the answer was that they are equally likely

8. now
so
all you do for is replace with .
you should also be able to sum the first series using a geometric series formula where .
9. (Original post by Rainingshame)

now
so
all you do for is replace with .
you should also be able to sum the first series using a geometric series formula where .

but this is what the mark scheme says: http://gyazo.com/3ff2dbffffcbd935eeea130809c76f1c.png

it looks like they used a version of the 1/12 (n^2 - 1) but I don't know what they did
10. (Original post by Secret.)
but this is what the mark scheme says: http://gyazo.com/3ff2dbffffcbd935eeea130809c76f1c.png

it looks like they used a version of the 1/12 (n^2 - 1) but I don't know what they did
I never learnt that method. I do AQA and so We only had one method taught. They should allow you to use any correct method and so this one will work every time.
11. (Original post by Secret.)
but this is what the mark scheme says: http://gyazo.com/3ff2dbffffcbd935eeea130809c76f1c.png

it looks like they used a version of the 1/12 (n^2 - 1) but I don't know what they did
For E(X) = (30+1)/2 = 15.5

For Var(X) = (29(31))/12 = 899/12

E(R) = E(2X)
= E(X) * 2
= 15.5 * 2
= 31

Var(R) = Var(2x)
=2^2 * (899/12)
= 299⅔
12. (Original post by Secret.)
but this is what the mark scheme says: http://gyazo.com/3ff2dbffffcbd935eeea130809c76f1c.png

it looks like they used a version of the 1/12 (n^2 - 1) but I don't know what they did
We've done this one before

http://www.thestudentroom.co.uk/show....php?t=2350711
13. (Original post by Air1337)
For E(X) = (30+1)/2 = 15.5

For Var(X) = (29(31))/12 = 899/12

E(R) = E(2X)
= E(X) * 2
= 15.5 * 2
= 31

Var(R) = Var(2x)
=2^2 * (899/12)
= 299⅔
on the mark scheme E(X) = 31

Aah right thanks so much !
14. (Original post by Secret.)
on the mark scheme E(X) = 31

Aah right thanks so much !
Yeah look further down into the calculations lol.

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