Continuous random variables — practice questions

(i) For $f(x)$ to be a valid PDF, the total area under the curve must be 1. We integrate $f(x)$ over its defined range and set the result to 1.

$\int_{0}^{2} k(4-x^2) dx = 1$

$k \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} = 1$

$k \left( (4(2) - \frac{2^3}{3}) - (0) \right) = 1$

$k \left( 8 - \frac{8}{3} \right) = 1$

$k \left( \frac{24-8}{3} \right) = 1$

$k \left( \frac{16}{3} \right) = 1$

$k = \frac{3}{16}$ (as required)

(ii) To find $P(X > 1)$, we integrate the PDF from 1 to the upper limit of the range, which is 2.

$P(X > 1) = \int_{1}^{2} \frac{3}{16}(4-x^2) dx$

$= \frac{3}{16} \left[ 4x - \frac{x^3}{3} \right]_{1}^{2}$

$= \frac{3}{16} \left[ (4(2) - \frac{2^3}{3}) - (4(1) - \frac{1^3}{3}) \right]$

$= \frac{3}{16} \left[ (8 - \frac{8}{3}) - (4 - \frac{1}{3}) \right]$

$= \frac{3}{16} \left[ \frac{16}{3} - \frac{11}{3} \right]$

$= \frac{3}{16} \left( \frac{5}{3} \right)$

$= \frac{5}{16}$

First, find the expectation, $E(X)$.

$E(X) = \int_{0}^{2} x f(x) dx = \int_{0}^{2} x \cdot \frac{3}{16}(4-x^2) dx$

$= \frac{3}{16} \int_{0}^{2} (4x - x^3) dx$

$= \frac{3}{16} \left[ 2x^2 - \frac{x^4}{4} \right]_{0}^{2}$

$= \frac{3}{16} \left[ (2(2^2) - \frac{2^4}{4}) - (0) \right]$

$= \frac{3}{16} \left[ (8 - \frac{16}{4}) \right] = \frac{3}{16} (8 - 4) = \frac{3}{16}(4) = \frac{3}{4}$

So, $E(X) = 0.75$.

Next, find $E(X^2)$ to calculate the variance.

$E(X^2) = \int_{0}^{2} x^2 f(x) dx = \int_{0}^{2} x^2 \cdot \frac{3}{16}(4-x^2) dx$

$= \frac{3}{16} \int_{0}^{2} (4x^2 - x^4) dx$

$= \frac{3}{16} \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_{0}^{2}$

$= \frac{3}{16} \left[ (\frac{4(2^3)}{3} - \frac{2^5}{5}) - (0) \right]$

$= \frac{3}{16} \left[ \frac{32}{3} - \frac{32}{5} \right] = \frac{3}{16} \cdot 32 \left[ \frac{1}{3} - \frac{1}{5} \right]$

$= 2 \left[ \frac{5-3}{15} \right] = 2 \left( \frac{2}{15} \right) = \frac{4}{15}$

Now, use the variance formula: $Var(X) = E(X^2) - [E(X)]^2$.

$Var(X) = \frac{4}{15} - (\frac{3}{4})^2 = \frac{4}{15} - \frac{9}{16}$

$= \frac{64 - 135}{240} = -\frac{71}{240}$. Wait, variance cannot be negative! Let me recheck the calculation for $E(X^2)$.

Ah, $E(X^2) = \frac{3}{16} [\frac{32}{3} - \frac{32}{5}] = \frac{3}{16} [\frac{160-96}{15}] = \frac{3}{16} [\frac{64}{15}] = \frac{3 \times 4}{15} = \frac{12}{15} = \frac{4}{5}$.

Let's try the variance calculation again with the corrected $E(X^2)$.

$Var(X) = E(X^2) - [E(X)]^2 = \frac{4}{5} - (\frac{3}{4})^2$

$= \frac{4}{5} - \frac{9}{16} = \frac{64 - 45}{80} = \frac{19}{80}$

So, $Var(X) = 0.2375$. This is positive, which is a good sign.