Question

Let X be a random variable with pdf

𝑓(𝑥) = {

2(1 + 𝑥)

27 𝑖𝑓 2 ≤ 𝑥 ≤ 5

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Find (i) 𝑃(𝑋 < 4) (ii) 𝑃(3 < 𝑥 ≤ 4)

The density function of sheer strength of spot welds is given by

𝑓(𝑥) = {

𝑥/ 160000 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 400

800 − 𝑥/ 160000 𝑓𝑜𝑟 400 ≤ 𝑥 ≤ 800

Find the number a such that 𝑃(𝑋 < 𝑎) = 0.50

Answer 1

$i:\\P\left( X<4 \right) =\int_4^5{f\left( x \right) dx}=\int_4^5{\frac{2\left( 1+x \right)}{27}dx}=\frac{\left( 1+x \right) ^2}{27}|_{4}^{5}=\frac{11}{27}\\ii:\\P\left( 3<X\leqslant 4 \right) =\int_3^4{\frac{2\left( 1+x \right)}{27}dx}=\frac{\left( 1+x \right) ^2}{27}|_{3}^{4}=\frac{1}{3}\\\\a\leqslant 400:\\P\left( X<a \right) =\int_0^a{f\left( x \right) dx}=\int_0^a{\frac{x}{160000}dx}=\frac{a^2}{320000}\\\frac{a^2}{320000}=0.5\Rightarrow a^2=160000\Rightarrow a=400\\a=400 satisfies\,\,P\left( X<a \right) =0.5$