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    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
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    What type of probability distribution is it? How do you calculate expectation of a random variable?
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    And think about how you sum up the first n natural numbers.
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    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 = \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 \sum xP(X=x) where \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:

    \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.
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    Latex is a clear nightmare, anyway good luck lol.
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    (Original post by Chittesh14)
    Latex is a clear nightmare, anyway good luck lol.
    LaTex is great.

    \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.
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    (Original post by RDKGames)
    LaTex is great.

    \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 .

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

    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.
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    (Original post by RDKGames)
    It's not the sum of all natural numbers.
    Why not?
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    (Original post by Chittesh14)
    Why not?
    Because you stop summing them at some point n lol...
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    (Original post by RDKGames)
    Because you stop summing them at some point n lol...
    Oh right lol fair enough.
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    rickyrossman TeamXO Chittesh14 RDKGames
    thanks everyone for your help!
    Posted on the TSR App. Download from Apple or Google Play
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    (Original post by MattDwyer)
    rickyrossman TeamXO Chittesh14 RDKGames
    thanks everyone for your help!
    No worries .
 
 
 
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