Math331, Fall 2008
Instructor: David Anderson
Section 11.2 Homework Answers
Homework: pg. 474, #’s 2, 6, 8.
2. From number 9 of Section 11.1 (which you did), the moment generating function of a
geometric RV, Xi, with parameter pis
MXi(t) = pet
1−qet,
where q= 1 −p, and the above function is only valid if t < −ln(q). Therefore, if
X1, X2,...,Xnare all independent geometric RVs with parameter p, we may use Theorem
11.3 to find the moment generating function of X1+X2+···+Xn.
MX1+···+Xn(t) = MX1(t)···MXn(t)
=pet
1−qet× · · · × pet
1−qet(ntimes )
=pet
1−qetn
.
Using Table 3 in the Appendix shows that this is the moment generating function for a
negative binomial RV with parameters nand p. The uniqueness Theorem 11.2 then shows
that X1+···+Xnmust be a negative binomial RV.
6. Let X1be the number of bottles under-filled by machine 1 from 10am to 11am tomorrow,
and X2be the number of bottles under-filled by machine 2 from 10am to 11am tomorrow.
Then, X1, X2are binomial RVs with parameters (100, .15),(80, .15). Therefore, by Theorem
11.4 (which we proved and discussed in class), X1+X2is a binomial RV with parameters
(180, .15). We are interested in finding
P{X1+X2= 27}=180
27 .1527.85180−27 =.083.
8. X, Y, Z are independent Poisson RVs with parameters λ1, λ2, λ3. Therefore, by Theorem
11.5 (which we proved and discussed in class), X+Y+Zis Poisson with parameter λ1+λ2+λ3
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