Question #213481

A continuous random variable x has the pdf given by

f(x)={k(1-x) for 0<x<1 both inclusive

{o,elsewhere

find t,the value of the constant k. Also find the mean and the variance of x


1
Expert's answer
2021-07-15T11:25:51-0400
f(x)x=01k(1x)dx\displaystyle\int_{-\infin}^{\infin}f(x)x=\displaystyle\int_{0}^{1}k(1-x)dx

=k[xx22]10=12k=1=k[x-\dfrac{x^2}{2}]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{1}{2}k=1

k=2k=2

mean=μ=E[X]=xf(x)dxmean=\mu=E[X]=\displaystyle\int_{-\infin}^{\infin}xf(x)dx

=012x(1x)dx=[x2x33]10=13=\displaystyle\int_{0}^{1}2x(1-x)dx=[x-\dfrac{2x^3}{3}]\begin{matrix} 1\\ 0 \end{matrix}=\dfrac{1}{3}


E[X2]=x2f(x)dx=012x2(1x)dxE[X^2]=\displaystyle\int_{-\infin}^{\infin}x^2f(x)dx=\displaystyle\int_{0}^{1}2x^2(1-x)dx

=[2x33x42]10=16=[\dfrac{2x^3}{3}-\dfrac{x^4}{2}]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{1}{6}


Var(X)=E[X2](E[X])2Var(X)=E[X^2]-(E[X])^2

=16(13)2=118=\dfrac{1}{6}-(\dfrac{1}{3})^2=\dfrac{1}{18}


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