If the probability density function (p.d.f.) | vedclass

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(B) For a probability density function, the total area under the curve must be $1$, so $\int_{-1}^{4} f(x) \, dx = 1$.$\int_{-1}^{4} k(9 + 8x - x^2) \

Vedclass · Accepted Answer

$3/250$

Vedclass · Answer

(B) For a probability density function, the total area under the curve must be $1$, so $\int_{-1}^{4} f(x) \, dx = 1$.$\int_{-1}^{4} k(9 + 8x - x^2) \, dx = 1$.$k \left[ 9x + 4x^2 - \frac{x^3}{3} \right]_{-1}^{4} = 1$.$k \left[ (9(4) + 4(4^2) - \frac{4^3}{3}) - (9(-1) + 4(-1)^2 - \frac{(-1)^3}{3}) \right] = 1$.$k \left[ (36 + 64 - \frac{64}{3}) - (-9 + 4 + \frac{1}{3}) \right] = 1$.$k \left[ (100 - \frac{64}{3}) - (-5 + \frac{1}{3}) \right] = 1$.$k \left[ \frac{300 - 64}{3} - \frac{-15 + 1}{3} \right] = 1$.$k \left[ \frac{236}{3} + \frac{14}{3} \right] = 1$.$k \left[ \frac{250}{3} \right] = 1$.$k = \frac{3}{250}$.

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

If the probability density function (p.d.f.) of a continuous random variable $X$ is given by $f(x) = \begin{cases} k(9 + 8x - x^2), & \text{for } -1 \leq x \leq 4 \\ 0, & \text{otherwise} \end{cases}$, then the value of $k$ is:

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