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Question 28 of 41

Q.Consider a random variable X with probability density function, f(x)={4x3,if 0<x<10,otherwisef(x)=\begin{cases}4x^3, & \text{if } 0<x<1 \\ 0, & \text{otherwise}\end{cases} Find E(X)E(X) and V(X)V(X).

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2023Subjective· 3mImportance★★★★★
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E(X)=45, E(X2)=23, V(X)=275≈0.0267E(X)=\dfrac45,\ E(X^2)=\dfrac23,\ V(X)=\dfrac{2}{75}\approx0.0267.

Density f(x)=4x3f(x)=4x^{3} for 0<x<10<x<1 (and 00 otherwise).

Step 1 — expectation.

E(X)=∫01x f(x) dx=∫01x⋅4x3 dx=∫014x4 dx=[4x55]01=45.E(X)=\int_0^1 x\,f(x)\,dx=\int_0^1 x\cdot 4x^{3}\,dx=\int_0^1 4x^{4}\,dx=\left[\frac{4x^{5}}{5}\right]_0^1=\frac45.

Step 2 — second moment.

E(X2)=∫01x2⋅4x3 dx=∫014x5 dx=[4x66]01=46=23.E(X^{2})=\int_0^1 x^{2}\cdot 4x^{3}\,dx=\int_0^1 4x^{5}\,dx=\left[\frac{4x^{6}}{6}\right]_0^1=\frac46=\frac23.

Step 3 — variance. …

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