Question

If the r.v. X is X is N(0,2).find P(1≤x≤2│x≥1).

Answer 1

Answer on Question #48728 – Math – Statistics and Probability

If the r.v. $X$ is $X$ is $N(0,2)$. Find $P(1 \leq x \leq 2 | x \geq 1)$.

Solution

Using the definition of conditional probability,

$$P (A \mid B) = \frac {P (A \text{ and } B)}{P (B)},$$

It leads to

$$\begin{array}{l} P (1 \leq x \leq 2 | x \geq 1) = \frac {P (1 \leq x \leq 2 \text{ and } x \geq 1)}{P (x \geq 1)} = \frac {P (1 \leq x \leq 2)}{P (x \geq 1)} = \frac {P \left(\frac {1 - 0}{\sqrt {2}} \leq \frac {x - 0}{\sqrt {2}} \leq \frac {2 - 0}{\sqrt {2}}\right)}{P \left(\frac {x - 0}{\sqrt {2}} \geq \frac {1 - 0}{\sqrt {2}}\right)} \\ = \frac {P (0 . 7 1 \leq z \leq 1 . 4 1)}{P (z \geq 0 . 7 1)} = \frac {P (z \leq 1 . 4 1) - P (z \leq 0 . 7 1)}{1 - P (z \leq 0 . 7 1)}. \end{array}$$

From z-table:

$$P (z \leq 0. 7 1) = 0. 7 6 1 1, P (z \leq 1. 4 1) = 0. 9 2 0 7.$$

Thus

$$P (1 \leq x \leq 2 | x \geq 1) = \frac {0 . 9 2 0 7 - 0 . 7 6 1 1}{1 - 0 . 7 6 1 1} = 0. 6 6 8 1.$$

Answer: 0.6681.

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