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. m = np = (5)(0.7) = 3.5
s2 = 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. m = 1/p = 10/3
s2 = (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. m = E(X) = M(0) =
0.3e(o) + 0.8e2(0) + 0.6e3(0) + 0.4e4(0) = 2.1
s2 = E(X2) - m2 = M(0) - m2 = 0.3e(o)
+ 1.6e2(0) + 1.8e3(0) + 1.6e4(0) - m2 =
= 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. m = (m + 1)/2 = 11/2 = 5.5
s2 = (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
13. Let X = number of flips of a fair coin needed to
observe heads-tails in
consecutive flips.
This problem has a negative binomial distribution since this
is a sequence of
bernoulli trials and we wish to find the probability that in
x number of trials, we
will observe exactly two successes. The fact that one the
first success is a
heads and the other a tails does not complicate things since
they both have
the same probability.
a. X = negative binomial: r = 2, p = 0.5
f(x) = (x 1)C(2-1) (0.5)2 (0.5)x-
2
= (x 1)
(0.5)x = (x 1)/2x
, x = 2, 3, 4,
b. m = r/p = (2)/(1/2) = 4
s2 = r(1 p) / p2 = (2)(0.5)/(0.25) = 4
s = √s2 = 2
c.
i. P(X ≤ 3) = P(X = 2) + P(X = 3) = [(2 1)/4)] +
[(3 1)/ 8] = 0.5
ii. P(X ≥ 5) = 1 P(X ≤ 4) = 1 [P(X = 2) +
P(X = 3) + P(X = 4)]
= 1 [(1/4) +
(1/4) + (3/16)] = 1 (11/16)
= 5/16
iii. P(X = 3) = 1/4
21. M(t) = (44/120) + (45/120)et
+ (20/120)e2t + (10/120)e3t + (1/120)e5t
a. m = E(X) = M(0) = [0 + 45 + 40 + 30 + 5]/120 = 120/120 =
1
s2 = E(X2) - m2 = M(0) - m2 = [0 + 45 + 80 + 90 + 25]/120 1
= (240/120)
1 = 1
b. P(X ≥ 1) = 1 P(X=0) = 1 (44/120) = 76/120 = 0.63
Notice that from the m.g.f. one can rebuild the p.m.f.
of x.
f(x) = { (44/120) ,
x = 0
(45/120) ,
x = 1
(20/120) ,
x = 2
(10/120) ,
x = 3
(0/120) , x = 4
(1/120)
, x = 5}
c.
**Extra Credit
19. In this problem, it
suffices to show that M(0) = 1, since in that case
R(0) = M(0) = E(X) = m
R(0) = M(0) [M(0)]2 = E(X2)
[E(X)]2 = E(X2) - m2 = s2
M(t) = xєS∑
f(x)etx
M(0) = xєS∑
f(x)e(0)x = xєS∑ f(x) = 1 □
This makes sense because M(0) and M(0) give the first and
second
moments respectively. So M(0) would give the zeroth moment or
the sum of
the p.m.f. over the entire sample space. And of course, all
p.m.f. s by
definition sum to one over their sample space.
20b. Use the result from 19 to find the mean and
variance of the binomial
distribution.
M(t) = (1 p + petx)n, R(t) = ln[(1 p + petx)n] = n ln(1 p + petx)
R(t) = npetx . R(t) = (1 p + petx) npetx (npetx)(petx)
(1 p + petx) (1 p + petx)2
R(0) = np
. R(0) = (1 p + p) np (np)(p)
(1 p + p) (1 p + p)2
R(0) = np □ R(0) = np (np)(p) = np(1 p) □