Question

Let X be the percentage score on a college-entrance exam for students who did not participate in an exam-preparation seminar. X is modeled by a uniform distribution on [a,100]. Let Y be the percentage score on a college-entrance exam for students who did participate in an exam-preparation seminar. Y is modeled by a uniform distribution on [1.25a,100]. It is given that E[X^2] = 19,600/3.Calculate the 80th percentile of Y. A. 80 B. 85 С. 90 D. 92 E. 95

Answer 1

Answer:

E. 95

Step-by-step explanation:

$\text{Given that:} \\ \\ X \sim U(a,100) \\ \\ fx(x) = \dfrac{1}{100-a} \\ \\ E[X^2] = \int _x f_x (x) \ dx \\ \\ \implies \int^{100}_{a} \dfrac{x^2}{100-a}dx \\ \\ =\dfrac{100^3-a^3}{3(100-a)} \\ \\ =\dfrac{100^2+100a+a^2}{3} \\ \\ E[X^2] = \dfrac{19600}{3}=\dfrac{100^2+100a+a^2}{3} \\ \\\implies a^2 +100a+10000=19600 \\ \\ \implies a^2 +100a -9600 = 0 \\ \\ \implies a = \dfrac{-100 \pm \sqrt{100^2 -4(-9600)} }{2} \\ \\=\dfrac{-100 \pm \sqrt{48400}} {2} \\ \\ = \dfrac{-100 \pm 220}{2} \\= 60 \ OR \ 160$

$\text{SInce X is a percentage score on an exam }\ P(X \ge 0) = 1 \text{which means } \ a = 60\\ \\Y \sim U (1.25a,100) = U(75,100)\\ \\ f_y(y) = \dfrac{1}{100-75} = \dfrac{1}{25} \\ \\ \text{Let assume the 80th percentile of Y be} \ y_{80} } \\ \\ \\Then; \ P(Y \le y_{80}) =80\% \implies \int^{y_{80}}_{75} \ f_Y (y) \ dy = 0.8 \\ \\ \implies \int^{y_{80}}_{75}\dfrac{1}{25} \ dy = 0.8 \\ \\ \implies \dfrac{y_{80}-75}{25}= 0.8 \\ \\ \implies y_{80} = 95$