Stat 311: HW 4, Chapters 4 & 5, Solutions

Fritz Scholz

Ch. 4, Problem 7. (a) Koko is more likely to catch the mouse outside than inside,
since 0.12 > 0.112, as the two tree diagrams show.

Koko waits outside

B
P(B) = P(B | A) P(A)
P(B | A) = 0.2
= 0.6 0.2 = 0.12
A

P(A) = 0.6

Bc
A: mouse takes outside exit
B: Koko catches mouse
P(B | A) = 0.8
c

B
Ac

P(Ac) = 0.4 P(B | Ac) = 0

Koko waits inside

A B

P(A) = 0.6 P(B | A) = 0

S B

P(S | A ) = 0.3
c
P(B | Ac S) = 0

Ac B
0.4 0.7 0.4 = 0.112
P(Ac) = 0.4 P(B | Ac Sc) = 0.4

Sc

P(Sc | Ac) = 0.7

Bc
A: mouse takes outside exit
S: Koko sleeps P(Bc | Ac Sc) = 0.6
B: Koko catches mouse
(b) Let C7 be the event that the mouse is caught within the next 7 days when Koko waits outside each night. It is
easier to contemplate the complementary event of missing the mouse 7 nights in a row, the misses each night being
independent events. Then
P(C7 ) = 1 P(C7c ) = 1 (1 0.12)7 = .5913
Ch. 4, Problem 11. View the n = 110 room reservations as so many independent Bernoulli trials. Then the number
Y of people actually showing up is binomial with parameters n = 110 and p = .96. Then

P(Y > 100) = 1 P(Y 100) = 1 pbinom(100, 110, .96) = 0.9870095

i.e., there is a 98.7% chance that the hotel will have to turn someone away.
Ch. 2, Problem 2. (a) see plot. (b) Clearly f (x) 0 for all x and the triangle area (1/2) height base width = 1.
2.5
2.0

f(x) = 2(x 1)
1.5
f(x)

1.0
0.5

area
= (1 2 ) 2 1 = 1
0.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0

(c) P(1.50 < X < 1.75) = P(X 1.75)P(X 1.5) = 0.56250.25 = 0.3125 since
1 1
P(X 1.5) = 2 (1.5 1) (1.5 1) = 0.25 and P(X 1.75) = 2 (1.75 1) (1.75 1) = 0.752 = 0.5625
2 2

Ch. 2, Problem 7. X N (5, 102 ), then

 
X 0 (5)
(a) P(X < 0) = P = pnorm(0.5) = 0.6914625
10
= pnorm(0, 5, 10) = 0.6914625

 
X 5 (5)
(b) P(X > 5) = 1 P(X 5) = 1 P = 1 pnorm(1) = 0.1586553
10
= 1 pnorm(5, 5, 10) = 0.1586553

(c) P(3 < X < 7) = P(X 7) P(X 3) = pnorm(1.2) pnorm(0.2) = 0.3056706

= pnorm(7, 5, 10) pnorm(3, 5, 10) = 0.3056706

(d) P(|X + 5| < 10) = P(10 < X + 5 < 10) = P(10 X + 5 10) = P(15 X 5)
= P(X 5) P(X 15) = pnorm(1) pnorm(1) = 0.6826895
= pnorm(5, 5, 10) pnorm(15, 5, 10) = 0.6826895
(e) P(|X 3| > 2) = P({X > 5} {X < 1}) = P(X < 1) + P(X > 5)
= P(X < 1) + 1 P(X 5) = pnorm(0.6) + 1 pnorm(1) = 0.8844021
= pnorm(1, 5, 10) + 1 pnorm(5, 5, 10) = 0.8844021

Ch. 2, Problem 8. (a) Since X1 N (1, 9) and X2 N (3, 16) are independent, we have

E(X1 + X2 ) = 1 + 3 = 4 and var(X1 + X2 ) = 9 + 16 = 25

(b) E(X2 ) = 3 and var(X2 ) = (1)2 var(X2 ) = 16.

(c) E(X1 X2 ) = 1 3 = 2 and var(X1 X2 ) = var(X1 + (1)X2 ) = varX1 + (1)2 varX2 = 9 + 16 = 25
(d) E(2X1 ) = 2EX1 = 2 1 = 2 and var(2X1 ) = 22 var(X1 ) = 4 9 = 36
(e) E(2X1 2X2 ) = E(2X1 ) E(2X2 ) = 2EX1 2EX2 = 2 1 2 3 = 4 and

var(2X1 2X2 ) = var(2X1 ) + var(2X2 ) = 22 var(X1 ) + 22 var(X2 ) = 4 (9 + 16) = 100