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How to find P(a < X ≤ b) from a distribution function

The rule: P(a < X ≤ b) = F(b) − F(a)

A distribution function (CDF) $F (x)$ already stores all the probability accumulated up to the point $x$ . So the probability of landing inside an interval is simply the difference of its values at the two ends:

P (a < X \leq b) = F (b) - F (a)

Nothing has to be integrated: take the piece of $F (x)$ that covers each endpoint and substitute. If an endpoint falls where $F$ is constant, that value is used as it is.

Problem

A random variable $X$ is given by the distribution function

F (x) = ⎩ ⎨ ⎧ 0, \frac{1}{26} (x^{3} - 1), 1, x \leq 1 1 < x \leq 3 x > 3

Find $P (1 < X \leq 2)$ .

Answer

P (1 < X \leq 2) = \frac{7}{26} \approx 0.269

Checked by hand: the solution is correct.

Step-by-step solution

1. Write the rule for the interval

For a continuous random variable the probability of falling into $(a, b]$ is the increment of the distribution function:

P (a < X \leq b) = F (b) - F (a)

Here $a = 1$ and $b = 2$ , so

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

2. Substitute both endpoints

Both endpoints sit in the middle piece $1 < x \leq 3$ , where $F (x) = \frac{1}{26} (x^{3} - 1)$ (the point $x = 1$ is the border, and both pieces give the same value there). For $x = 2$ :

F (2) = \frac{1}{26} (2^{3} - 1) = \frac{1}{26} (8 - 1) = \frac{7}{26}

For $x = 1$ :

F (1) = \frac{1}{26} (1^{3} - 1) = 0

3. Take the difference

P (1 < X \leq 2) = \frac{7}{26} - 0 = \frac{7}{26} \approx 0.269

A quick sanity check: the value is between $0$ and $1$ , and it is smaller than $F (3) = 1$ , as it must be for a part of the whole range.

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