If the function f defined by f(x) = K(x - x^2 | vedclass

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(B) Since $f(x)$ is the p.d.f. of a random variable $X$,the total area under the curve must be $1$: $\int_{0}^{1} K(x - x^2) dx = 1$ $K \left[ \frac{x

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$\frac{1}{2}$

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(B) Since $f(x)$ is the p.d.f. of a random variable $X$,the total area under the curve must be $1$: $\int_{0}^{1} K(x - x^2) dx = 1$ $K \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 1$ $K \left( \frac{1}{2} - \frac{1}{3} \right) = 1 \Rightarrow K \left( \frac{1}{6} \right) = 1 \Rightarrow K = 6$ Now,we calculate $P(X $P(X $= 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1/2}$ $= 6 \left( \frac{(1/2)^2}{2} - \frac{(1/2)^3}{3} \right)$ $= 6 \left( \frac{1/4}{2} - \frac{1/8}{3} \right) = 6 \left( \frac{1}{8} - \frac{1}{24} \right)$ $= 6 \left( \frac{3 - 1}{24} \right) = 6 \left( \frac{2}{24} \right) = 6 \left( \frac{1}{12} \right) = \frac{1}{2}$

If the function $f$ defined by $f(x) = K(x - x^2)$ for $0 < x < 1$ and $f(x) = 0$ otherwise,is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(X < \frac{1}{2})$ is

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