Question

A continuous random variable has a density function�(�) = 2(5 – x)
5
, where 2<x<3. Calculate
the following probability correct up to 3 decimal places, and make the graph for part (a) only in
Answer sheet:
P (x < 2.5)
P (x > 2.2)
P (2.1 ≤ � ≤ 2.7)

Answer 1

$$P(x<2.5)=\displaystyle\int_{2}^{2.5}\dfrac{2(5-x)}{5}dx$$

$$=\dfrac{2}{5}\big[5x-\dfrac{x^2}{2}\big]\begin{matrix} 2.5 \\ 2 \end{matrix}$$

$$=0.4\bigg(5(2.5)-\dfrac{(2.5)^2}{2}-(5(2)-\dfrac{(2)^2}{2})\bigg)$$

$$=0.55$$

$P(x<2.5)=0.55$

$$P(x>2.2)=1-P(x\leq2.2)=1-\displaystyle\int_{2}^{2.2}\dfrac{2(5-x)}{5}dx$$

$$=1-\dfrac{2}{5}\big[5x-\dfrac{x^2}{2}\big]\begin{matrix} 2.2 \\ 2 \end{matrix}$$

$$=1-0.4\bigg(5(2.2)-\dfrac{(2.2)^2}{2}-(5(2)-\dfrac{(2)^2}{2})\bigg)$$

$$=0.768$$

$P(x>2.2)=0.768$

$$P(2.1\leq x\leq 2.7)=\displaystyle\int_{2.1}^{2.7}\dfrac{2(5-x)}{5}dx$$

$$=\dfrac{2}{5}\big[5x-\dfrac{x^2}{2}\big]\begin{matrix} 2.7 \\ 2.1 \end{matrix}$$ $$=0.4\bigg(5(2.7)-\dfrac{(2.7)^2}{2}-(5(2.1)-\dfrac{(2.1)^2}{2})\bigg)$$

$$=0.624$$

$P(2.1\leq x\leq2.7)=0.624$