Consider a andom variable X with probability density function
\$f(x) = ke^(_2x),0≤x≤\infty\$.Find the value of k,mean.

\$\int_-\infty^\infty\$f(x)dx = 1

Substitute f(x) in the formula

\$\int_0^\infty ke^(_2x)\$ dx = 1

Simplification

k \$\int_0^\infty e^(_2x)\$ dx = 1

Simplification

k \$(e^(_2x)/-2)_0^\infty = 1\$

After simplification

k = 2

E(X) = \$\int_-\infty^\infty\$x f(x)dx

Substitute f(x) in the formula

E(X) = \$\int_0^\infty x ke^(_2x)\$ dx

Simplification

E(X) = 2 \$\int_0^\infty xe^(_2x)\$ dx

E(X) = \$2((xe^(-2x)/-2)_0^\infty)-(\int_0^\infty (e^(-2x)/-2)dx\$

we know that \$\int udv = uv - \int v du\$

After simplification

\$E(X) = \int_0^\infty e^(-2x)dx = 1/2\$

Answer

A