The cumulative distribution function (c.d.f.) | vedclass

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(A) The cumulative distribution function $F(x)$ is defined as the integral of the probability density function $f(x)$.$F(x) = \int_{0}^{x} f(t) dt = \

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$3$

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(A) The cumulative distribution function $F(x)$ is defined as the integral of the probability density function $f(x)$.$F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} 3(1 - 2t^2) dt$.Evaluating the integral:$F(x) = 3 \left[ t - \frac{2t^3}{3} \right]_{0}^{x} = 3 \left( x - \frac{2x^3}{3} \right)$.Comparing this with the given form $F(x) = k(x - \frac{2x^3}{k})$,we observe that $k = 3$ is not directly matching the structure unless we rewrite the expression.Wait,let us re-evaluate the expression: $F(x) = 3x - 2x^3$.If $F(x) = k(x - \frac{2x^3}{k}) = kx - 2x^3$,then by comparing coefficients,$k = 3$ and $2 = 2$. Thus,$k = 3$.

$X=x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$K$	$2K$	$2K$	$3K$	$K^2$	$2K^2$	$7K^2+K$

The cumulative distribution function (c.d.f.) $F(x)$ associated with the probability density function (p.d.f.) $f(x) = 3(1 - x^2)$ for $0 < x < 1$ and $f(x) = 0$ otherwise,is given by $F(x) = k(x - \frac{2x^3}{k})$. Find the value of $k$.

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