Question

Find k if the following is the p.d.f. of a r.v. X:
$f(x) = \left\{ \begin{array}{ll} kx^{2}(1-x) & \quad ,\ \ \ 0\ \textless \ x\ \textless \ 1; \\ 0 & \quad ,\ \ \ otherwise. \end{array} \right$

Answer 1

that's the answer for your question

Answer 2

Given : if the following is the p.d.f. of a r.v. X:

To Find : k

Solution:

$\int\limits^1_0 {(kx^2(1-x))} \, dx$

$\int\limits^1_0 {(kx^2-kx^3))} \, dx$

= kx³/3 - kx⁴/4  + c

$\left[\frac{kx^3}{3}-{kx^4}{4}}\right]^1_0$

= k(1³/3) - k(1⁴/4)  - (0 - 0)

= k/3 - k/4

= k/12

k/12 = 1   (as sum of probabilities = 1)

k = 12

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