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$$\begin{equation}F(x)=\begin{cases}0 &\quad x<-10 \\ 0.25 & \quad -10\leqslant x <30 \\ 0.75 &\quad 30 \leqslant x <50 \\ 1 &\quad 50 \leqslant x \end{cases}\end{equation}$$ $a)\,\, P(X \leqslant 50)$

$b)\,\, P(X \leqslant 40)$

$c)\,\, P(40 \leqslant X \leqslant 60)$

$d)\,\, P(X < 0)$

$e)\,\, P(0\leqslant X < 10)$

$f)\,\, P(-10<X < 10)$

I'm having trouble wrapping my head about how the answers in the back of the book are obtained:

$a)\,\, 1$

$b)\,\, .75$

$c)\,\, .25$

$d)\,\, .25$

$e)\,\, 0$

$f)\,\, 0$

What I do not understand is how to correctly plug in the different values of $X$ and determine the probability of each...in particular when $X$ is between a range of two integers, yet the jumps in probabilities are at different values.

Would anyone care to explain the methodology behind finding these solutions? Thank you

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3 Answers 3

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Note that, by definition of a c.d.f. $P(X\le x)=F(x)$. Hence,
$P(X\le 50)=F(50)=1$
$P(X \le 40)=F(40)=0.75$
$P(40 \le X \le 60)=P(X\le 60)-P(X < 40)=F(60)-F(40^{-})=1-0.75=0.25$
For the strict inequality, since as $X$ gets close to $0$ from below, it satisfies the $-10<X<30$ condition:
$P(X<0)=0.25$
$P(0\le X < 10)=P(X < 10) - P(X < 0)=0.25-0.25=0$
$P(-10 < X < 10)=P(X<10)-P(X<-10)=0.25-0.25=0$

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Thank you Daniel. I notice that these are all involving less than operators... what if there was another part that asked something like P(X>25)? –  user62873 Feb 18 '13 at 21:48
    
Well, we know that exactly one of the following is true: $X\le 25$ or $X>25$. As a result, $P(X\le 25)+P(X>25)=1$. –  Daniel Littlewood Feb 18 '13 at 21:50
    
I see...so one could solve for it by evaluating 1-P(x≤25) yes? In the case of this problem it would be 1-.25 = .75 correct? –  user62873 Feb 18 '13 at 22:03
    
That is correct. :-) –  Daniel Littlewood Feb 18 '13 at 22:26
2  
Although the numerical answer for $P(40\leq X \leq 60)$ happens to be correct in this instance, it is highly misleading to tell a beginner that $$P(40\leq X \leq 60) = P(X\leq 60) - P(X\leq 40) = F(60)-F(40)$$ instead of $$P(40\leq X \leq 60) = P(X\leq 60) - P(X< 40) = F(60) - F(40^-).$$ It is inappropriate to stimulate the development of bad habits.... –  Dilip Sarwate Feb 18 '13 at 22:29

The $F(x)$ given to you is the cumulative distribution function. Before $-10$, it is $0$. At $-10$, it suddenly jumps to $0.25$, and stays there for a while. So there must be a weight of $0.25$ at $x=-10$. That is, $\Pr(X=-10)=0.25$.

Similarly, there is a sudden jump from $0.25$ to $0.75$ at $30$. Thus $\Pr(X=30)=0.50$.

Finally, $\Pr(X=50)=0.25$.

So we have the complete probability distribution function of $X$. The random variable $X$ takes on the values $-10$, $30$, and $50$, with probabilities $0.25$, $0.50$, and $0.25$. Now you can answer any question.

We could also answer the questions "algebraically." The only nuisance is that, because of the jumps, we need to carefully distinguish between $\lt$ and $\le$, also between $\gt$ and $\ge$.

Here is a sample. For (e), we want $\Pr(0\le x\lt 10)$. This is $\Pr(X\lt 10)-\Pr(X\lt 0)$.

Look at the cumulative distribution function $F(x)$ of $X$. Just short of $10$, it has value $0.25$. Just short of $0$, it has value $0.25$. Subtract: the result is $0$.

But it is intuitively clearer to observe that $X$ "takes on" no values between $0$ and $10$.

Remark: Occasionally, one even bumps into random variables that have a character which is a hybrid of the discrete and the continuous. For example, after the weight of $0.25$ at $x=-10$, the cumulative distribution function could after that climb smoothly to $1$.

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It's a step function that jumps at $-10$, $30$, and $50$. So at these steps you gain the increments indicated by $F$: $0.25, 0.75-0.25 = 0.5,$ and $1 - 0.75 = 0.25$ (the difference between before and after). Note also that if you hit any of the thresholds, you will get the full increase, which matters to distinguish $<$ from $\leq$ in the questions (c., eg, (e) below as you have a right end of the interval of $< 10$, not $\leq 10$). For instance,

(a) $\mathbf{P}(X \leq 50)=1$ as you jumped $50$ to reach the final $1$
(c) $\mathbf{P}(40 \leq X < 60)= 0.25$ as you jumped at 50 by 0.25
(e) $\mathbf{P}(0 \leq X < 10) = 0$ as you last jumped at $-10$, and will again at $10$.

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