Question

Suppose you choose at random a real number X from the interval [2, 10].
(a) Find the density function f(x) and the probability of an event E for this experiment, where E is a sub-interval [a, b] of [2, 10].
(b) From (a), find the probability that X > 5, that 5 < X < 7, and that X² − 12X + 35 > 0.

Answer 1

Answer:

$\mathbf{(a)} P(E) = \frac{1}{8}(b-a)\\ \mathbf{(b)} P(X>5) = \frac{5}{8}, P(5<X<7) = \frac{1}{4} , P(X^2 - 12X + 35>0) = \frac{3}{4}$

Step-by-step explanation:

$\mathbf{(a)}$

The objective is to find the density function $f(x)$ and the probability of the event E, which is choosing at random from the interval $[a,b] \subset [2,10].$

Let $X$ be a random variable that follows the uniform distribution on the interval [2,10]. Then, its density function is

                               $f_x(x) = \dfrac{1}{10-2} = \dfrac{1}{8}, \quad x \in [2,10]$

and 0 otherwise.

Now, we can find the probability of the event $E$.

         $P(E) = P(a\leq X\leq b) = \int \limits_{a}^{b} f_x(x)\, dx = \int \limits_{a}^{b} \frac{1}{8} \, dx = \frac{1}{8} x \Big|_{a}^{b} = \frac{1}{8} (b-a)$

$\mathbf{b)}$

Let's find the probability that $X>5$. By the definition, we obtain

        $P(X>5) = \int\limits^{10}_{5} f_x(x) \, dx = \int\limits^{10}_{5} \frac{1}{8} \, dx = \frac{1}{8} x \Big|_{5}^{10} = \frac{1}{8} (10 - 5) = \frac{5}{8}$

The probability that $5<X<7$ is

     $P(5<X<7) = \int\limits^{7}_{5} f_x(x) \, dx = \int\limits^{7}_{5} \frac{1}{8} \, dx = \frac{1}{8} x \Big|_{5}^{7} = \frac{1}{8} (7 - 5) = \frac{2}{8} = \frac{1}{4}$

To find the probability that $X^2-12X + 35 > 0$, first solve the inequality$x^2-12x+35 > 0$ .

The roots of the quadratic equation $x^2-12x+35 = 0$ are

$x_{1/2} = \frac{12 \pm \sqrt{144- 4 \cdot 35} }{2} = \frac{12 \pm 2 }{2} = 6 \pm 1 \implies x_1 = 6+1 = 7, x_2 = 6-1 = 5$  

Therefore, the solutions of the inequality are $x<5$ or $x>7$ (look at the graph below).

Now, we can calculate the probability as follows

                    $P(X^2-12X+ 35 > 0) = P(X<5 \; \text{or} \; \text X>7)$

It is easier to calculate the probability using the complementary event.

The complementary event of $X<5$ or $X>7$ is

       $\neg (X< 5 \; \vee \; X>7) \iff X>5 \;\wedge \; X<7 \iff 5 < X < 7$

and the probability of this event is

                      $P(5<X<7) = 1 - P (X <5 \; \text{or} \; X>7)$

Therefore, our probability is

         $P (X <5 \; \text{or} \; X>7) = 1- P(5<X<7) = 1 - \int\limits^7_5 f_x(x)\, dx \\\phantom{P (X <5 \; \text{or} \; X>7) }=1- \frac{1}{8} x \Big|_{5}^{7} = 1 -\frac{1}{8} (7 - 5) = 1-\frac{2}{8} = \frac{3}{4}$