STAT 418
HW12 Solutions
Sections 6.2-6.5
15, 20, 21, 23, 27, 28, 32, 33; TE 9
15.
a. Let A denote the area of region R, then A =
dydx. Because f (x, y) is a joint pdf,
(x,y)R
cdydx = c
f (x, y)dydx =
1=
(x,y)R
(x,y)R
(x,y)R
dydx = cA.
Therefore c = 1/A.
b
STAT 418
HW7 Solutions
Sections 4.7- 4.9
51, 53, 54, 59, 61, 77, 78, 79; TE 16, 17
51. Let X = cfw_the number of typographical errors on a page. It is reasonable to assume
that the number of words on a page, n, is large and that the probability of an erro
STAT 418
HW8 Solutions
Sections 5.1-5.4
2, 3, 5, 6, 8, 10, 13, 15, 16, TE1, TE7
2. First, we must nd C. By the substitution rule with y = x/2; dy = dx/2,
0
xex/2 dx =
4yey dy.
0
By integration by parts, with u = y; du = dy and dv = ey dy; v = ey ,
0
0
4y
STAT 418
HW4 Solutions
Sections 3.4-3.5
37, 39, 43, 50, 52, 55, 64, 65, 78, TE 1, TE 6
37. Let H = cfw_Heads on rst ip, F = cfw_Fair coin, and T = cfw_Two-headed coin. Note
that F and T form a partition.
a. By Bayes rule,
P (F |H ) =
P (H |F )P (F )
.5 .5
STAT 418
HW2 Solutions
Sections 2.1-2.5
4,10,12,18,23,27,36,39; TE 10,11,12
4.
a. The sample space consists of all the possible sequences of ips until a Head. The 0s
are Tails; the 1 is the rst Head.
b.
i Because A ips rst, A = cfw_1, 0001, 0000001, . . .
STAT 418
HW2 Solutions
Sections 3.1-2.4
2, 5, 6, 11, 12, 15, 18, 21, 23, 28
2. Let E = cfw_First die lands on a 6 and Fi = cfw_Sum of the dice = i.
P (E |F2 ) = P (E |F3 ) = P (E |F4 ) = P (E |F5 ) = P (E |F6 ) = 0.
P (E |F7 ) =
P (E F7 )
1/36
=
P (F7 )
6
STAT 418
HW6 Solutions
Sections 4.3- 4.6
30, 35, 38, 41, 42, 46, 49, TE 13, 14
30.
2n (1/2)n =
E (X ) =
n=1
a. No, E (X ) represents the average in the long run. In a single game, there is a 1/2 + 1/4
+ 1/8 + 1/16 + 1/32 = .96875 probability that X $32.
STAT 418
HW9 Solutions
(pp. 195-196)
63, 65, 71, 80
63. Let N (t) = cfw_the number of people who enter the casino in time t. It is reasonable to
assume that N (t) P oisson(). We are given that = 1/2 = .5, so we may use the Poisson
distribution to approxim
STAT 418
HW16 Solutions
Sections 8.1-8.4
1, 2, 7, 8, 13, 15, 17, 18, 19, 21, 22; TE 7
1. By Chebyshevs inequality with k =
20,
P (0 < X < 40) = 1 P (|X 20| > 20) 1 1/20 = 19/20
2.
a. By Markovs inequality,
P (X 85)
E(X)
= 15/17.
85
b. By Chebyshevs inequ
STAT 418
HW14 Solutions
Sections 7.1-7.4
1, 3, 6, 16, 18, 21, 26, 33, 36; TE 1, 2
1. Let X = 1 if the coin toss lands heads, and let it equal 0 otherwise. Also, let Y denote
the value that shows up on the die. Then, with p(i, j) = P (X = i, Y = j),
6
6
E(
STAT 418
HW13 Solutions
Sections 6.6-6.7
34, 37, 42, 43; TE 11
34. Let X be the number of men among the 200 who never eat breakfast. Then X
B (200, .252). Let Y be the number of women among the 200 who never eat breakfast. Then
Y B (200, .236).
a. Approx
STAT 418
HW10 Solutions
Sections 5.5-5.7
19, 21, 23, 27, 31, 33, 35; TE 9, 13
19. Let Z = (X 12)/2, then Z is a standard normal. Find c such that 0.10 = P [Z >
(c 12)/2]. From Table 5.1, P [Z < 1.28] = .90 and so
(c 12)/2 = 1.28 c = 14.56
21. Let X = cfw_
STAT 418
HW1 Solutions
Sections 1.1-1.5
3,5,7,11,14,21,22,27; TE 8,10,12
3. Consider each job as a place in a sequence. Because only one person can
be assigned to a job, it follows that the total number of possible assignments
is a permutation of the numb