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cumulative distribution function of the random X is given by

$$ F(x) = \begin{cases} 1-(1+x)e^{-x} & \text{for } x\gt 0 \\ 0 & \text{elsewhere} \end{cases} $$

  1. Find $P(X \leq2 )$
  2. Find $P(1\lt X\lt 3)$

Hi, i am not sure of how to solve this problem. Do i just have to integrate the values in given cdf equation?

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    $\begingroup$ What was a cumulative distribution function again...? $\endgroup$ – Saucy O'Path Mar 24 at 19:59
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Hint

By definition of CDF for a continuous random variable we have:

$$F_X(x)\triangleq\Pr\{X<x\}=\Pr\{X\le x\}$$and $$\Pr\{a<x<b\}=\Pr\{x<b\}-\Pr\{x<a\}$$

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A cumulative distribution function shows the probability of $X$ being less than a given value $x$, ie $P(X<x)$. Therefore $P(X<2) = 1-(1+2)e^{-2} = 1-3e^{-2}$ and $P(1<X<3)$ = $P(X<3) - P(X<1) = 1-(1+3)e^{-3}-(1-(1+1)e^{-1}) = 2e^{-1}-4e^{-3}$

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