Revision as of 17:52, 13 May 2023 by Admin (Created page with "'''Key: C''' In general, <math display="block">\begin{aligned} E\left(X^{2}\right)-E\left[(X \wedge 150)^{2}\right] &=\int_{0}^{200} x^{2} f(x) d x-\int_{0}^{150} x^{2} f(x...")
Exercise
May 13'23
Answer
Key: C
In general,
[[math]]\begin{aligned}
E\left(X^{2}\right)-E\left[(X \wedge 150)^{2}\right] &=\int_{0}^{200} x^{2} f(x) d x-\int_{0}^{150} x^{2} f(x) d x-150^{2} \int_{150}^{200} f(x) d x \\
& =\int_{150}^{200}\left(x^{2}-150^{2}\right) f(x) d x
\end{aligned}[[/math]]
Assuming a uniform distribution, the density function over the interval from 100 to 200 is [math]6 / 7400[/math] (the probability of [math]6 / 74[/math] assigned to the interval divided by the width of the interval). The answer is
[[math]]\int_{150}^{200}\left(x^{2}-150^{2}\right) \frac{6}{7400} d x=\left.\left(\frac{x^{3}}{3}-150^{2} x\right) \frac{6}{7400}\right|_{150} ^{200}=337.84[[/math]]