Choose the correct option from the given alternative: P.d.f. of a.c.r.v X is f (x) = 6x (1 − x), for 0 $\leq$ x $\leq$ 1 and = 0, otherwise (elsewhere) If P (X a), then a = .....

Question

Choose the correct option from the given alternative: P.d.f. of a.c.r.v X is f (x) = 6x (1 − x), for 0 $\leq$ x $\leq$ 1 and = 0, otherwise (elsewhere) If P (X < a) = P (X > a), then a = .....

KollegeApply · Accepted Answer

$\frac{1}{2}$ Explanation: Given: f(x) = 6x(1 − x), 0 $\leq$ x $\leq$ 1 P(X < a) = P(X > a) Since total probability = 1, P(X < a) = P(X > a) = $\frac{1}{2}$ int_0^a 6x(1-x) dx = $\frac{1}{2}$ Integrate `int 6x(1-x) dx = int (6x-6x^2)dx = 3x2 − 2x3 3a^2 - 2a^3 = $\frac{1}{2}$ 2a^3 - 3a^2 + $\frac{1}{2}$ = 0 4a3 − 6a2 + 1 = 0 4($\frac{1}{8}$)-6($\frac{1}{4}$) + 1 = 0 =$\frac{1}{2}$ - $\frac{3}{2}$ + 1 = 0 a = $\frac{1}{2}$`

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