Revision as of 00:44, 7 May 2023 by Admin (Created page with "'''Solution: E''' Let <math>X_k</math> be the random change in month <math>k</math>. Then <math>\operatorname{E}(X_k) = (0.5)(1.1) + 0.5(−0.9) = 0.1</math> and <math>\opera...")
Exercise
ABy Admin
May 07'23
Answer
Solution: E
Let [math]X_k[/math] be the random change in month [math]k[/math]. Then [math]\operatorname{E}(X_k) = (0.5)(1.1) + 0.5(−0.9) = 0.1[/math] and [math]\operatorname{Var}(X_k) = 0.5(1.1)^2 + 0.5(−0.9)^2 − (0.1)^2 = 1.[/math] Let [math]S = \sum_{k=1}^{100}X_k [/math]. Then, [math]\operatorname{E}(S) = 1000(0.1) = 10[/math] and [math]\operatorname{Var}(S) = 100(1) = 100 [/math]. Finally,
[[math]]
\operatorname{P}(100 + S \gt 91) = \operatorname{P}(S \gt -9) = \operatorname{P}( Z \gt \frac{-9-10}{\sqrt{100}} = -1.9 ) = 0.9713.
[[/math]]