For p.d.f. $f(x) = k(4 - x^2)$ for $-2 \leq x \leq 2$ (and 0 elsewhere), the value of $k$ is:
A$3/16$
B$1/8$
C$1/16$
D$3/32$
Answer & Solution
Correct answer: D. $3/32$
$\int_{-2}^2 k(4 - x^2)\,dx = k[4x - x^3/3]_{-2}^2 = k\{(8 - 8/3) - (-8 + 8/3)\} = k \cdot 32/3 = 1$ ⇒ $k = 3/32$.
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