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(D) Since $f(x)$ is a probability density function,the total area under the curve must be $1$: $\int_{-2}^{2} k(4 - x^2) dx = 1$ Since the function is

Vedclass · Accepted Answer

$\frac{11}{16}$

Vedclass · Answer

(D) Since $f(x)$ is a probability density function,the total area under the curve must be $1$: $\int_{-2}^{2} k(4 - x^2) dx = 1$ Since the function is even,we have $2k \int_{0}^{2} (4 - x^2) dx = 1$. $2k [4x - \frac{x^3}{3}]_0^2 = 1$ $2k (8 - \frac{8}{3}) = 1 \Rightarrow 2k(\frac{16}{3}) = 1 \Rightarrow k = \frac{3}{32}$. Now,we calculate $P[-1 $P[-1 $= \frac{6}{32} [4x - \frac{x^3}{3}]_0^1 = \frac{3}{16} (4 - \frac{1}{3}) = \frac{3}{16} 	imes \frac{11}{3} = \frac{11}{16}$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

If the error involved in making a certain measurement is a continuous random variable $X$ with probability density function $f(x) = k(4 - x^2)$ for $-2 \leq x \leq 2$ and $f(x) = 0$ otherwise,then $P[-1 < X < 1] = $

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