# Verify that f(x) = 2x/( k(k + 1)) for x = 1, 2, 3, . . . ,k can serve as the probability distribution of a random variable in a given range

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If f(x) = 2x/k(k+1), x= 1, 2, 3, 4,....k.

To verify whether this is probability distribution function, we should prove that {Sum f(x) over x = 1, 2, 3, 4,...k} = 1.

x takes a discrete value. That is x takes values like x= 1, 2, 3, 4, ......, k.

Let X be the random variable

f(X =x) = 2x/k(k+1) = 2x/k(k+1), x= 1, 2, 3, 4....k.

Total frequency = {Sum f(X = x) , x= 1, 2, 3, ...n.} = 2*1/k(k+1)+2*2/k(k+1)+2*3/k9k+1)...2*k/k(k+1).

{Sum f(X = x) , x= 1, 2, 3, ...n.} = {2/k(k+1)}{1+2+3+..........k}

{Sum f(X = x) , x= 1, 2, 3, ...n.} = {2/k(k+1)}{Sum of the k natural numbers starting from 1}.

{Sum f(X = x) , x= 1, 2, 3, ...n.} = {2/(k(k+1)}{k(+1)/2}

{Sum f(X = x) , x= 1, 2, 3, ...n.} = 1..

Therefore f(x) is a frequency density function. Or f(x) is a probability distribution function.