Answer to Question #203888 in Statistics and Probability for Ali

Question

Q(1) [10 Marks] [CLO2,C3] (a) The time required, in hours, to repair a car is a r.v. X with pdf fX(x) = c(4x − x 2 ), where 0 ≤ x ≤ 4. (i) Find the value of the constant c. (ii) Find the probability that for a car which arrives now at the garage, the amount of time needed to get repaired will be (a) at least one but less than three hours; (b) more than two hours.

(b) Let X be a r.v. whose pdf is: fX(x) = √ 1 2π exp ( − (x ^2) /2 ) , where −∞ < x < ∞. Determine the conditional pdf fX(x||x| ≤ 1). 

Answer 1

i.

$\int_0^4c(4x-x^2)dx=1$

$c(2x^2-\frac{1}{3}x^3)|_0^4=1$

$c(32-\frac{64}{3})=1$

$32-21\frac{1}{3}=\frac{1}{c}$

$\frac{32}{3}=\frac{1}{c}$

$32c=3$

$c=\frac{3}{32}$

iia. $P(1\le x<3)$

$P(1\le x<3)=\int_1^3\frac{3}{32}(4x-x^2)dx$

$=\frac{3}{32}(2x^2-\frac{1}{3}x^3)|_1^3$

$=\frac{3}{32}[(18-\frac{27}{3})-(2-\frac{1}{3})]$

$=\frac{3}{32}(9-\frac{5}{3})$

$=\frac{3}{32}\times\frac{22}{3}$

$=\frac{11}{16}$

iib. $P(x>2)$

$P(x>2)=\int_2^4\frac{3}{32}(4x-x^2)dx$

$=\frac{3}{32}(2x^2-\frac{1}{3}x^3)|_2^4$

$=\frac{3}{32}[(32-\frac{64}{3})-(8-\frac{8}{3})]$

$=\frac{3}{32}(\frac{32}{3}-\frac{16}{3})$

$=\frac{3}{32}\times\frac{16}{3}$

$=\frac{1}{2}$

b. The pdf is not clear but it looks like a standard normal distribution

$fX(x||x| ≤ 1)=\frac{f(X(x)}{f(x)}$ . |x|≤ 1

$-\infin<x<\infin\cap|x|\le 1=-1\le x\le 1$

$f(x)=\int_{-1}^1\frac{1}{\sqrt{2\pi}}e^{(-\frac{x^2}{2})} dx$

Which is equivalent to $P(-1\le Z\le 1)$

From Z tables or excel formula =NORM.S.DIST(1,TRUE)-NORM.S.DIST(-1,TRUE)),$P(Z\le1)-P(Z\le -1)=0.682689492$

Thus,

$fX(x||x| ≤ 1)=\frac{\frac{1}{\sqrt{2\pi}}e^{(-\frac{x^2}{2})}}{0.682689492}$ ,$-1\le x \le 1$

Simplify by converting 0.682689492 as a fraction

$0.682689492=\frac{170672373}{250000000}$

$fX(x||x| ≤ 1)=\dfrac{1953125{\cdot}2^\frac{13}{2}\mathrm{e}^{-\frac{x^2}{2}}}{170672373\sqrt{{\pi}}},-1\le x \le 1$