Let $X$ be a continuous random variable with probability density function $f(x)=k,x^2$ for $0lexle2$ and $f(x)=0$ otherwise.

Step-by-step solution

1. Step 1: Normalise the pdf: ${\int }_0^{2}k,x^2,dx=1$ gives $k{\int }_0^2{x}^{2}dx=k{\left[\frac{x^3}{3}\right]}_0^2=k\left(\frac{8}{3}\right)=1$ hence $k=\frac{3}{8}$.
2. Step 2: Compute $P(X>1.5)$: $P(X>1.5)={\int }_{1.5}^2\frac{3}{8}{x}^{2}dx=\frac{3}{8}{\left[\frac{x^3}{3}\right]}_{1.5}^2=\frac{1}{8}\left(8-3.375\right)=\frac{1}{8}(4.625)=\mathrm{0.578125.}$ (Rounded to 4 decimals: $0.5781$.)
3. Step 3: Compute the mean: $E[X]={\int }_0^{2}x\cdot \frac{3}{8}{x}^{2}dx=\frac{3}{8}{\int }_0^2{x}^{3}dx=\frac{3}{8}{\left[\frac{x^4}{4}\right]}_0^2=\frac{3}{8}\cdot \frac{16}{4}=\frac{3}{8}\cdot 4=\mathrm{1.5.}$
4. Step 4: Compute $E[X^2]$: $E[X^2]={\int }_0^2{x}^2\cdot \frac{3}{8}{x}^{2}dx=\frac{3}{8}{\int }_0^2{x}^{4}dx=\frac{3}{8}{\left[\frac{x^5}{5}\right]}_0^2=\frac{3}{8}\cdot \frac{32}{5}=\frac{96}{40}=\mathrm{2.4.}$
5. Step 5: Compute variance: $\mathrm{Var}(X)=E[X^2]-(E[X]{)}^2=2.4-1.5^2=2.4-2.25=\mathrm{0.15.}$
6. Step 6: Report the results: $k=3/8,;P(X>1.5)=0.5781,;E[X]=1.5,;\mathrm{Var}(X)=\mathrm{0.15.}$

Concept tested: Continuous probability distribution and moments

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked