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F(x)= \begin{cases} 0, & \text{if } x < 0, \\ x, & \text{if } 0 \leq x < \frac{1}{2} \\ 1, & \text{if } x \geq \frac{1}{2} \end{cases}

Find $P\{X<\frac{1}{2}\}$, $P\{0.2 \leq X<0.7\}$, $P\{X=\frac{1}{2}\}$.

This is what I have so far:

$P\{0.2 \leq X<0.7\} \rightarrow P(X < \frac{7}{10}) - P(X \leq \frac{1}{5})$.

Then $F(\frac{2}{3})-F(\frac{1}{5}) \rightarrow \frac{7}{10}-\frac{1}{5}=\frac{1}{2}.$

I need feedback if I did this portion correctly, I'll work on the rest in a while.

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    0.7 is not the same as 2/3. Other than that, it's fine. Oops, not so fine -- see stud_iisc's comment. – 2017-02-14
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    Adding to @quasi's comment, $F(2/3)$ isn't $2/3$. – 2017-02-14

1 Answers 1

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  1. $P\{X<\frac{1}{2}\} = F(\frac{1}{2})$, since $X$ is a continuous random variable.

  2. $P\{0.2 \leq X<0.7\} = F(0.7)-F(\frac{1}{5})$, since $X$ is a continuous random variable.

  3. $0$, since $X$ is a continuous random variable.

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    For the second part $F(0.7)-F(\frac{1}{5})$. Would it be $1-\frac{1}{5}=\frac{4}{5}$? – 2017-02-14
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    Yes. And also for 1, $F(0.5)=1$. – 2017-02-14