The continuous random variable $Y$ has probability density function $f(y)=k,y$ for $0leyle2$ and $f(y)=0$ elsewhere. Find the value of $k$ and the expected value $E[Y]$.

Step-by-step solution

1. Normalize the density: ∫₀² k y dy = 1.
2. Compute the integral: k ∫₀² y dy = k [y²/2]₀² = k·(4/2) = 2k.
3. Set 2k = 1, so k = 1/2.
4. Expected value: E[Y] = ∫₀² y f(y) dy = ∫₀² y (½ y) dy = ½ ∫₀² y² dy.
5. Integrate: ∫₀² y² dy = [y³/3]₀² = 8/3.
6. Multiply by ½: E[Y] = ½·(8/3) = 4/3.

Concept tested: Normalizing a continuous density and finding expected value.

Published 29 August 2026 · How this solution was prepared and checked

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