Revision as of 08:45, 9 May 2023 by Admin (Created page with "'''Solution: E''' Using the formulas for the variance and mean of the uniform distribution: <math display = "block"> \begin{align*} \operatorname{E}(X^2) = \operatorname{Var...")
Exercise
ABy Admin
May 09'23
Answer
Solution: E
Using the formulas for the variance and mean of the uniform distribution:
[[math]]
\begin{align*}
\operatorname{E}(X^2) = \operatorname{Var}((X) + \operatorname{E}(X)^2 &= \frac{(100-a)^2}{12} + \left(\frac{100+a}{2}\right)^2 \\
&= \frac{100^2 -200a + a^2 +3(100)^2 + 600a + 3a^2}{12} \\
&= \frac{40000 + 400a + 4a^2}{12} \\
&= \frac{19600}{3} \\
0 &= 40000 -78400 + 400a + 4a^2 \\
0 &= a^2 + 100a - 9600 \\
0 &= (a-60)(a+160) \\
a &= 60.
\end{align*}
[[/math]]
Then, [math]Y[/math] is uniform on the interval 1.25(60) = 75 to 100. The 80th percentile is 75 + 0.8(25) = 95.