Question 6 For f(x)=6x(1−x)f(x)=6x(1-x), 0≤x≤10\le x\le1, find Var(X)\operatorname{Var}(X). Use symmetry for the mean. See the diagram in the original worksheet below. Show solutionHide solution+Question 6 – Solution See the diagram in the original worksheet below. Step 1: Use symmetry to find the mean. The density satisfies f(1−x)=6(1−x)x=f(x),f(1-x)=6(1-x)x=f(x), so it is symmetric about x=1/2x=1/2. Therefore, μ=E[X]=12.\mu=E[X]=\frac12. Step 2: Compute the second moment. E[X2]=∫01x2f(x)dx=∫01x2(6x−6x2)dx=∫01(6x3−6x4)dx=[32x4−65x5]01=32−65=310.\begin{align*} E[X^2] &=\int_0^1x^2f(x)\,dx\\ &=\int_0^1x^2(6x-6x^2)\,dx\\ &=\int_0^1(6x^3-6x^4)\,dx\\ &=\left[\frac32x^4-\frac65x^5\right]_0^1 =\frac32-\frac65=\frac3{10}. \end{align*} Step 3: Apply the variance formula. Var(X)=E[X2]−(E[X])2=310−(12)2=310−14=120.\begin{align*} \operatorname{Var}(X) &=E[X^2]-(E[X])^2\\ &=\frac3{10}-\left(\frac12\right)^2\\ &=\frac3{10}-\frac14=\frac1{20}. \end{align*} Var(X)=120\boxed{\operatorname{Var}(X)=\frac1{20}}