What are the mean and standard deviation of the probability density function given by $p (x) = k (x - x^{2})$ $p(x)=k(x-x^2)$ for $x \in [0, 1]$ $x in [0,1]$ , in terms of k, with k being a constant such that the cumulative density across all x is equal to 1?

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Answer 1 · 2016-11-09T15:00:09

$E (X) = \frac{k}{12}$ $E(X)=k/12$

$s d = \sqrt{\frac{3 k - 5 k^{2}}{60}}$ $sd=sqrt((3k-5k^2)/60)$

the mean $E (X)$ $E(X)$ of a continuous pdf is given by

$E (X) = \int_{a l l x} x f (x) d x$ $E(X)=int_(all x)xf(x)dx$

so

$E (X) = \int_{0}^{1} k x (x - x^{2}) d x$ $E(X)=int_0^1kx(x-x^2)dx$

$E (X) = k \int_{0}^{1} (x^{2} - x^{3}) d x$ $E(X)=kint_0^1(x^2-x^3)dx$

$E(X)=k[1/3x^3-1/4x^4]_cancel(0)^1$

$E(X)=k(1/3-1/4)=k/12$

$Var(X)=E(X^2)-E^2(X)$

$E(X^2)=int_(all x)x^2f(x)dx$

$E(X^2)=int_0^1x^2k(x-x^2)dx$

$E(X^2)=kint_0^1(x^3-x^4)dx$

$E(X^2)=k[1/4x^4-1/5x^5]_cancel(0)^1$

$E(X^2)=k(1/4-1/5)=k/20$

$Var(X)=E(X^2)-E^2(X)$

gives

$Var(X)=k/20-(k/12)^2$

$Var(X)=(3k-5k^2)/60$

$sd=sqrt((3k-5k^2)/60)$

What are the mean and standard deviation of the probability density function given by p(x)=k(x-x^2) p(x)=k(x−x2)p(x)=k(x-x^2) for x in [0,1]x∈[0,1] x in [0,1], in terms of k, with k being a constant such that the cumulative density across all x is equal to 1?