- #1
probhelp150
Homework Statement
X is a geometric random variable with p = 0.1. Find:
##a. F_X(5)##
##b. Pr(5 < X \leq 11)##
##c. Pr(X=7|5<X\leq11)##
##d. E(X|3<X\leq11)##
##e. E(X^2|3<X\leq11)##
##f. Var(X|3<X\leq11)##
Homework Equations
The Attempt at a Solution
Can someone check my work and help me?
a.
##F_X(5) = P(X\leq5)=1-(1-p)^x##
##=1-(0.9)^5##
##=0.40951##b.
##Pr(5<X\leq11)=F_X(11)-F_X(5)##
##=0.2766794039##c. Not too sure about this one.
##Pr(X=7) = F_X(7) - F_X(6) = 0.0531441##
##Pr(5<X\leq11)=0.2766794039##
##\frac{Pr(X=7)}{Pr(5<X<\leq11)} = \frac{0.0531441}{0.2766794039}=0.192078##
d.
##Pr(3<X\leq11) = F_X(11)-F_X(3) = 0.41518940391##
##E(X|3<X\leq11) = \sum_{x=4}^{11} \frac{x*0.1*0.9^{x-1}}{Pr(3<X\leq11)}##
##=\sum_{x=4}^{11} \frac{x*0.1*0.9^{x-1}}{0.41518940391)}##
##=\frac{0.1}{0.41518940391}\sum_{x=4}^{11} {x*0.9^{x-1}}##
##=6.9533987498770702912469710479707##
e. Don't know how to do this one.
##Pr(3<X\leq11) = F_X(11)-F_X(3) = 0.41518940391##
##E(X^2|3<X\leq11) = \sum_{x=4}^{11} \frac{x^2*0.1*0.9^{x-1}}{Pr(3<X\leq11)}##
##=\sum_{x=4}^{11} \frac{x^2*0.1*0.9^{x-1}}{0.41518940391)}##
##=\frac{0.1}{0.41518940391}\sum_{x=4}^{11} {x^2*0.9^{x-1}}##
##=53.415557495820389902397015631005##
f. Variance = ##E[X^2] - E[X]^2 = 5.0658033210283859687442550652196##
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