Let $Y$ be a continuous random variable with probability density function $f (y) = k y^{2}$ for $0 \leq y \leq 1$ and $f (y) = 0$ elsewhere.

hard 4 marks Class 12 · Mathematics · Probability

Answer

k = 3

P (Y > 0.5) = 0.875

E [Y] = \frac{3}{4}

Var (Y) = \frac{3}{80}

The correct answer is $k = 3$ , $P (Y > 0.5) = 0.875$ , $E [Y] = \frac{3}{4}$ , $Var (Y) = \frac{3}{80}$ . — State the final results together.

Step-by-step solution

The total probability must be 1: $\int_{0}^{1} k y^{2} d y = 1$ . Integrate to get $k [\frac{y ^{3}}{3}]_{0}^{1} = k /3 = 1$ , so $k = 3$ .
For $P (Y > 0.5)$ , integrate the pdf from $0.5$ to $1$ : $\int_{0.5}^{1} 3 y^{2} d y = [y^{3}]_{0.5}^{1} = 1 - (0.5)^{3} = 1 - 0.125 = 0.875$ .
Compute $E [Y] = \int_{0}^{1} y f (y) d y = \int_{0}^{1} y \cdot 3 y^{2} d y = 3 \int_{0}^{1} y^{3} d y = 3 [\frac{y ^{4}}{4}]_{0}^{1} = 3/4$ .
Compute $E [Y^{2}] = \int_{0}^{1} y^{2} f (y) d y = \int_{0}^{1} y^{2} \cdot 3 y^{2} d y = 3 \int_{0}^{1} y^{4} d y = 3 [\frac{y ^{5}}{5}]_{0}^{1} = 3/5$ .
Variance $= E [Y^{2}] - (E [Y])^{2} = \frac{3}{5} - (\frac{3}{4})^{2} = \frac{3}{5} - \frac{9}{16} = \frac{48 - 45}{80} = \frac{3}{80}$ .
State the final results together.

Common mistake: Integrating

y^{2}

instead of

y^{3}

for the mean, or forgetting to square the mean when computing variance.

Concept tested: Finding constant, probability, expectation and variance for a continuous pdf