MATH 511, Meade HW Solutions
3.4 – 2abdf, 7, 13, 21, **19, **20b 3/5/04
Notice that the inside front cover of the text book has a list of common distributions. It
gives formulas for the means and variances of each and the m.g.f. for most.
2. In this exercise one should try to recognize the distribution type by the given
m.g.f. and then use the formulas to solve the rest of the problem. Also,
remember that when specifying a distribution, it is equally important to specify
the parameters as well.
a. M(t) = (0.3 + 0.7et)5
i. binomial; n=5, p = 0.7
ii. µ = np = (5)(0.7) = 3.5
σ2 = np(1-p) = 1.05
iii. P(1 ≤ X ≤ 2) = P(X = 1) + P(X = 2) = 5C1 (0.7)1(0.3)4 + 5C2 (0.7)2(0.3)3
= 0.0284 + 0.1323 = 0.1607
b. M(t) = 0.3et . , t < -ln(0.7)
1 – 0.7et
i. geometric; p = 0.3
ii. µ = 1/p = 10/3
σ2 = (1 – p) / p2 = (0.7)/(0.09) = 7.7778
iii. (1 ≤ X ≤ 2) = P(X = 1) + P(X = 2) = (0.7)0(0.3)1 + (0.7)1(0.3)1
= 0.3 + 0.21 = 0.51
d. M(t) = 0.3et + 0.4e2t + 0.2e3t + 0.1e4t
i. f(x) = { 0.3, x = 1
0.4, x = 2
0.2, x = 3
0.1, x = 4 } , this is not a common distribtion.
ii. µ = E(X) = M’(0) = 0.3e(o) + 0.8e2(0) + 0.6e3(0) + 0.4e4(0) = 2.1
σ2 = E(X2) - µ2 = M’’(0) - µ2 = 0.3e(o) + 1.6e2(0) + 1.8e3(0) + 1.6e4(0) - µ2 =
= 5.3 – 4.41 = 0.89
iii. P(1 ≤ X ≤ 2) = P(X = 1) + P(X = 2) = 0.3 + 0.4 = 0.7
f. M(t) = x=1∑10 (0.1) etx
i. uniform; m = 10
ii. µ = (m + 1)/2 = 11/2 = 5.5
σ2 = (m2 – 1) / 12 = 99/12 = 8.25
iii. P(1 ≤ X ≤ 2) = P(X = 1) + P(X = 2) = 0.1 + 0.1 = 0.2
7. Let X = the number of bags under three pounds
X = geometric; p = 0.04
a. P(X ≥ 20) = P(X > 19) = (1 – p)19 = (0.96)19 =0.4604
b. P(X ≤ 20) = 1 - (1 – p)20 = 1 - (0.96)20 = 0.5580
c. P(X = 20) = (0.96)19(0.04) = 0.0184
