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**[SOLVED] Probability (cdf question)**

Here is the problem from the book

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Let X be a random variable with distribution function (cdf)

[tex]

F(x)=\begin{cases}

0 &\text{for } x\geq 0\\

\frac{x}{8} & \text{for } 0 \leq x < 1\\

\frac{1}{4} + \frac{x}{8} & \text{for } 1 \leq x < 2\\

\frac{3}{4} + \frac{x}{12} & \text{for } 2 \leq x < 3\\

1 & \text{for } x \geq 3\end{cases}

[/tex]

Calculate [tex]P(1 \leq X \leq 2)[/tex].

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This should be a simple question, but I am getting a different answer than my book is, and I believe the book is wrong (how could I be wrong? :tongue2:).

Here is what I did:

[tex]

\begin{align*}

P(1 \leq X \leq 2) & = P(1 \leq X < 2) \\

& = P(X < 2) - P(X \leq 1) \\

& = \frac{1}{4} + \frac{2}{8} - \left(\frac{1}{4} + \frac{1}{8}\right) \\

& = \frac{1}{8}

\end{align*}

[/tex]

We could also get the same answer by finding the pdf and then integrating it over the interval (1,2).

The book I have gives an answer of [tex]\frac{19}{24}[/tex].

Here is what they did:

[tex]

\begin{align*}

P(1 \leq X \leq 2) & = P(X\leq 2) - P(X < 1) \\

& = F(2) - \lim_{x\to 1^-}F(x) \\

& = \frac{11}{12} - \frac{1}{8} \\

& = \frac{19}{24}\\

\end{align*}

[/tex]

In my opinion this seems doubly wrong. They used the wrong function on both, but at least they were consistent I suppose. Am I being silly here and missing something, or is the book wrong? Thanks!

edit... I am looking at my book, and they have [tex]P(a < X \leq b) = F(b) - F(a)[/tex]. Now I won't argue with this, but the way it is used in the above example seems completely counterintuitive to me. I will admit my use of the cdf may be dubious for P(X < 2), but it feels right. Maybe the book is right after all. Thoughts?

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