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further maths statistics help?

really stuck on these questions, would appreciate any help please

1.) a random number X is chosen from the fractions 1/n,2/n,3/n.....1
prove that the expectation of X >1/2, but the variance of x < 1/12

2.) X~U(n). prove that 6 Var(X) is always divisible by E(X)

Any help appreciated, thanks
What type of probability distribution is it? How do you calculate expectation of a random variable?
And think about how you sum up the first n natural numbers.
It'd be better if you post the actual questions tbh.
But regardless, you can rewrite the fractions as 1/n, 2/n, 3/n, ...., n/n (1).

So the expectation would be the mean: the sum of all the fractions divide by the number of them:

1/n+2/n+...+n/n=(1+2+...+n)/n)n=(1+2+...+n)n21/n + 2/n + ... + n/n = \frac{(1 + 2 + ... + n)/n)}{n} = \frac{(1 + 2 + ... + n)}{n^2}
Notice the link here is that 1 + 2 + ... + n is the sum of all the natural numbers.

You can also use the formula xP(X=x)\sum xP(X=x) where P(X=x)\sum P(X=x) is 1.
We know that the probability of choosing each fraction or value of X from the distribution is equal, (1/n).
So, using the formula:

Unparseable latex formula:

\begin{aligned} \sum xP(X=x) & = \sum x \cdot \frac{1}{n} \\ & = \frac{1}{n} \left( \frac{1}{n} + \frac{2}{n} + \ldots + \frac{n}{n} \right) \\ & = \frac{1}{n^2} (1+2+ \ldots + n)



Once you find that out, you should notice the clue: as n is a the number of terms in the series, it must be a positive integer.

I hope you can take it from there.

For the second part, I'm not sure waht you mean. X~U[n]. I'm assuming it's X~U[0,n].
Use the formula for the variance and mean of a continuous uniform distribution and you'll understand it from there.
(edited 6 years ago)
Latex is a clear nightmare, anyway good luck lol.
Original post by Chittesh14
Latex is a clear nightmare, anyway good luck lol.


LaTex is great.

Unparseable latex formula:

\begin{aligned} \sum xP(X=x) & = \sum x \cdot \frac{1}{n} \\ & = \frac{1}{n} \left( \frac{1}{n} + \frac{2}{n} + \ldots + \frac{n}{n} \right) \\ & = \frac{1}{n^2} (1+2+ \ldots + n)



You're welcome.

Also check this:

Original post by Chittesh14
Notice the link here is that 1 + 2 + ... + n is the sum of all the natural numbers.
(edited 6 years ago)
Original post by RDKGames
LaTex is great.

Unparseable latex formula:

\begin{aligned} \sum xP(X=x) & = \sum x \cdot \frac{1}{n} \\ & = \frac{1}{n} \left( \frac{1}{n} + \frac{2}{n} + \ldots + \frac{n}{n} \right) \\ & = \frac{1}{n^2} (1+2+ \ldots + n)



You're welcome.

Also check this:


Thanks buddy :smile:.

Btw, what is there to check? :O
I missed 0, but no difference to sum lol.
Original post by Chittesh14
Thanks buddy :smile:.

Btw, what is there to check? :O
I missed 0, but no difference to sum lol.


It's not the sum of all natural numbers.
Original post by RDKGames
It's not the sum of all natural numbers.


Why not?
Original post by Chittesh14
Why not?


Because you stop summing them at some point nn lol...
Original post by RDKGames
Because you stop summing them at some point nn lol...


Oh right lol fair enough.
@rickyrossman @TeamXO @Chittesh14 @RDKGames
thanks everyone for your help!
Original post by MattDwyer
@rickyrossman @TeamXO @Chittesh14 @RDKGames
thanks everyone for your help!


No worries :smile:.

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