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Introduction Statistics questions. I hope my question isn't too basic for this platform. I am given the CDF

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I've found $P(X \leq 1)$. I've found $P(X > 1)$ by $1-P(X \leq 1)$.

Now I'm asked for $P(X\geq 1)$.

Is this the same as $P(X > 1)$? I know that $P(X=1)=0$ so does $P(X\geq 1)=P(X > 1)+P(X=1)$?

I'm also asked to find, $P(−1.5 < X < 0.5)$. I've previously only seen problems of the form $P(−1.5 < X \leq 0.5)$ and know to solve them by subtraction.

Intuitively, it seems that by including the possibility that $X=1$ would make $P(X\geq 1) > P(X > 1)$ but since $P(X=1)=0$, I suppose it doesn't.

For the purposes of my homework, how do I treat problems that are inclusive or exclusive of the number being asked about?

For my curiosity, there most be some difference in practice, right? $P(X\leq 1)$ would be close to $P(X\leq .99999999)$ which would be different than $P(x\leq 1)$. Or am I overthinking things?

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  • $\begingroup$ Your idea seems OK. Distribution is continuous between -2 and 2. $\endgroup$
    – BruceET
    Sep 6 '20 at 15:44
  • $\begingroup$ See stats.stackexchange.com/…. $\endgroup$
    – whuber
    Sep 6 '20 at 15:57
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If $A$ and $B$ are disjoint, measurable sets, then $P(A \cup B) = P(A) + P(B).$ Disjoint means that the two sets don't overlap, i.e. the intersection is the empty set, or symbolically $A \cap B = \emptyset$. Measurable is a more technical concept, but (probably) every set you've ever seen is measurable, and any set like $\{ X < a \}$, $\{ X \le a \}$, $\{ X = a \}$, $\{ X \ge a \}$, or $\{ X > a \}$ is measurable.

This means that $$P( X \le a ) = P(\{ X < a \} \cup \{ X = a \}) = P(X < a) + P(X = a)$$ and $$P( X \ge a ) = P(\{ X > a \} \cup \{ X = a \}) = P(X > a) + P(X = a)$$

Since you've correctly noticed that because your cdf $F$ is continuous, $P(X = a) = 0$ for any number $a \in \mathbb{R}$, you can conclude that $$ P(X \le a) = P(X < a) \quad \text{and} \quad P(X \ge a) = P(X > a). $$

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