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(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...") |
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<math display="block">\begin{aligned} | <math display="block">\begin{aligned} | ||
E | \operatorname{E}(X^{2})-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 | & =\int_{150}^{200}\left(x^{2}-150^{2}\right) f(x) d x | ||
\end{aligned}</math> | \end{aligned}</math> | ||
Latest revision as of 17:55, 13 May 2023
Key: C
In general,
[[math]]\begin{aligned}
\operatorname{E}(X^{2})-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]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.