Edexcel IAL Mathematics S2.2.4 mean and variance of continuous random variablesPractise continuous-variable moments by integrating density functions to find E(X), E(X²), variance and transformed expectations.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointsIntegrate xf(x) and x²f(x) over each support interval before applying Var(X).Use E(aX+bY+c) or a stated variance to form equations for unknown constants.
S2.2.4 - Mean and variance of continuous random variables question 1[Maximum number: 4]A random variable X has probability density function given byf(x)={14−12⩽x<122x−3412⩽x⩽k0 otherwise f(x)=\left\{\begin{array}{cc} \frac{1}{4} & -\frac{1}{2} \leqslant x<\frac{1}{2} \\ 2 x-\frac{3}{4} & \frac{1}{2} \leqslant x \leqslant k \\ 0 & \text { otherwise } \end{array}\right.f(x)=⎩⎨⎧412x−430−21⩽x<2121⩽x⩽k otherwise where k is a positive constant.Use calculus to find E(X)Show Answer[∫−0.50.514x dx]+∫0.51.252x2−34x dx=[0]+[2x33−38x2]0.51.25\left[\int_{-0.5}^{0.5} \frac{1}{4} x \mathrm{~d} x\right]+\int_{0.5}^{1.25} 2 x^{2}-\frac{3}{4} x \mathrm{~d} x=[0]+\left[\frac{2 x^{3}}{3}-\frac{3}{8} x^{2}\right]_{0.5}^{1.25}[∫−0.50.541x dx]+∫0.51.252x2−43x dx=[0]+[32x3−83x2]0.51.25M1A1=(2×1.2533−38×1.252)−(2×0.533−38×0.52)=\left(\frac{2 \times 1.25^{3}}{3}-\frac{3}{8} \times 1.25^{2}\right)-\left(\frac{2 \times 0.5^{3}}{3}-\frac{3}{8} \times 0.5^{2}\right)=(32×1.253−83×1.252)−(32×0.53−83×0.52)dM1=93128=\frac{93}{128}=12893A1(4)Add to Test