Suppose that the cumulative distribution function of the random variable X is https://i.imgur.com/taHTeeH.png apparently the answer to a), b), c), e) are 0.56, 0.7, 0 and 0 respectively. I have no idea how they got these answer anyone able to help me to understand?
lets draw the distribution graph maybe? or take the derivatives to find the probability density function ....
the derivative of F(x) is simply: f(x) = .25 from 0 to 5
What do I do after I find the f(x)
integrate from x=0 to x=X
plug in the values of x given. the CDF gives P(X<x)... F(a) = P(X<a) = P(X\(\le\)a)
Okay, so here is the skinny. When it says \(\Pr(X<x)\) or \(\Pr(X\leq x)\), all you gotta do is plug \(x\) into the CDF. For continuous functions like this \(<\) and \(\leq\) are equivalent. When it says \(\Pr(X>x)\), remember that this is the same as \(1-\Pr(X\leq x)\).
my quandry is that we might need to modify it some to be a true probability function .... since .25 integrated from 0 to 5 is not equal to 1 ... but thats just a thought in my head at the moment
that would just give me 0.25(2.8) - 0
for b and d use the complement... 1-F(a) = P(X>a) = P(X\(\ge\)a)
@Australopithecus Does what I said make any sense?
I think so wio still I have no idea where 0.56 comes from
Actually,your cdf doesn't make any sense... F(5)=P(X<5) = 1.25. Either the cdf is off or the domain is.
it should probably be from 0 to 4.
@pgpilot326 The PDF is \(0.25\). The CDF is \(0.25x\).
yes, and ?
\[F(x)=\begin{cases}0&x<0\\.25x&x:(0,5)\\1&5\le x\end{cases}\] \[f(x)=\begin{cases}0&x<0\\.25&x:(0,5)\\0&5\le x\end{cases}\] now the issue is:\[\int_0^{5}.25~dx\ne1\]so this doesnt conform to a proper probability density function to me
so there is something wrong with the question and not my brain?
pfft, others have already stated this while i was trying to figure out the latex :)
yep
This text book trolls me all the time I think
\[ 0.20(2.8) = 0.56 \]
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