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STAT/MA 416 Answers
Homework 5
October 16, 2007
Solutions by Mark Daniel Ward
PROBLEMS
We always let “Z” denote a standard normal random variable in the computations below. We
simply use the chart from page 222 of the Ross book, when working with a standard normal
random variable; of course, a more accurate computation is possible using the computer.
1a. We compute
Z
1
2
c(1 − x ) dx = c
1=
−1
1
1 3 x− x = c(4/3)
3
x=−1
and thus c = 3/4.
1b. We compute F (a) = P (X ≤ a) = 0 for a ≤ −1, and F (a) = P (X ≤ a) = 1 for a ≥ 1.
For −1 < a < 1, we have
a
Z a
3
3
1 3 3
1 3 2
2
F (a) = P (X ≤ a) =
(1 − x ) =
x− x =
a− a +
4
3
4
3
3
−1 4
x=−1
So the cumulative distribution function of X is


if a ≤ −1
0
3
2
1 3
F (a) = 4 a − 3 a + 3
if −1 < a < 1

1
if a ≥ 1
2. We first compute
∞
Z ∞ −x/2 e
e−x/2 −
dx
1=
Cxe
dx = C x
−1/2 x=0
−1/2
0
0
Z ∞
∞
= 2C
e−x/2 dx = −4Ce−x/2 x=0 = 4C
Z
∞
−x/2
0
Thus C = 1/4. Now we compute the probability that the system functions for at least 5
months:
−x/2 ∞
Z ∞
Z ∞ −x/2 1 −x/2
1
e
e
P (X ≥ 5) =
xe
dx =
x
−
dx
4
4
−1/2 x=5
−1/2
5
5
Z ∞
∞ 7
1
1 −5/2
−5/2
−x/2
=
10e
+2
e
dx =
10e
− 4 e−x/2 x=5 = e−5/2
4
4
2
5
3a. The function
(
C(2x − x3 ) if 0 < x < 5/2
f (x) =
0
otherwise
1
2
√
√
3
cannot be a probability density function.
R ∞ To see this, note 2x − x = −x(x − 2)(x + 2).
If C = 0, then f (x) = 0 for all x, so −∞ f (x) = 0, but every probability density function
√
R∞
has −∞ f (x) = 1. If C > 0, then f (x) < 0 on the range x ∈ (0, 2), but every probability
√
density function has f (x) ≥ 0 for all x. If C < 0, then f (x) < 0 on the range x ∈ ( 2, 5/2),
but every probability density function has f (x) ≥ 0 for all x. So no value of C will satisfy
all of the properties needed for a probability density function.
3b. The function
(
C(2x − x2 ) if 0 < x < 5/2
f (x) =
0
otherwise
2
cannot be a probability
R ∞density function. To see this, note 2x−x = −x(x−2). RIf∞C = 0, then
f (x) = 0 for all x, so −∞ f (x) = 0, but every probability density function has −∞ f (x) = 1.
If C > 0, then f (x) < 0 on the range x ∈ (0, 2), but every probability density function has
f (x) ≥ 0 for all x. If C < 0, then f (x) < 0 on the range x ∈ (2, 5/2), but every probability
density function has f (x) ≥ 0 for all x. So no value of C will satisfy all of the properties
needed for a probability density function.
4a. We compute
Z
∞
P (X > 20) =
20
10
−1 ∞
dx
=
10(−1)x
= 1/2
x=20
x2
4b. We have F (a) = P (X ≤ a) = 0 for a ≤ 10. For a > 10, we compute
Z a
Z 10
Z a
a
10
10
dx = 10(−1)x−1 x=10 = 1 −
F (a) = P (X ≤ a) =
f (x) dx =
0 dx +
2
a
−∞
−∞
10 x
So the cumulative distribution function of X is
(
1 − 10
a
F (a) =
0
if a > 10
otherwise
4c. The probability that one such device will function for at least 15 hours is P (X ≥ 15) =
10
1 − P (X < 15) = 1 − F (15) = 1 − 1 − 15
= 10/15 = 2/3. If all 6 devices are independent,
and Y denotes the number of the 6 devices that function for at least 15 hours each, then
P (Y ≥ 3) = P (Y = 3) + P (Y = 4) + P (Y = 5) + P (Y = 6)
6
6
6
6
3
3
4
2
5
1
(2/3) (1/3) +
(2/3) (1/3) +
(2/3) (1/3) +
(2/3)6 (1/3)0
=
3
4
5
6
≈ .8999
5. Let X denote the volume of sales in a week, given in thousands of gallons. We have the
density of X, and we want a with P (X > a) = .01. Note that we need 0 < a < 1. We have
1
Z ∞
Z 1
Z ∞
(1 − x)5
4
.01 = P (X > a) =
f (x) dx =
5(1 − x) dx +
0 dx = 5
(−1)
= (1 − a)5
5
a
a
1
x=a
√
√
5
5
5
So .01 = (1 − a) , and thus .01 = 1 − a, so a = 1 − .01 ≈ .6019.
3
6a. We compute
E[X] =
=
=
=
∞
∞
Z
1
1 ∞ 2 −x/2
−x/2
(x)
(x)(0) dx +
xf (x) dx =
(x)(e
) dx =
xe
dx
4
4 0
0
−∞
−∞
Z ∞
Z ∞
−x/2 ∞
−x/2
−x/2 ∞
1
e
1
e
2e
2
−x/2
x
2x
dx =
x
xe
dx
−
+4
4
−1/2 x=0
−1/2
4
−1/2 x=0
0
0
∞
Z ∞ −x/2 −x/2 ∞
1
e
e−x/2 2e
x
dx
+ 4x
−4
4
−1/2 x=0
−1/2 x=0
−1/2
0
−x/2 ∞
−x/2 ∞
−x/2 ∞
e
1
e
e
1
x2
+ 4x
+8
= (16) = 4
4
−1/2 x=0
−1/2 x=0
−1/2 x=0
4
Z
Z
0
Z
6b. We compute
∞
Z
−1
Z
c(1 − x ) dx +
−1
−∞
−∞
∞
Z
2
0 dx +
f (x) dx =
1=
1
Z
1
1
x3 0 dx = c x −
= c(4/3)
3 x=−1
and thus c = 3/4. Now we compute
Z
∞
−1
Z
E[X] =
Z
xf (x) dx =
−∞
x2 x
−
2
4
= (3/4)
x(3/4)(1 − x ) dx +
−1
= (3/4)
x=−1
∞
Z
2
(x)(0) dx +
−∞
1
4
1
(x)(0) dx
1
1 1
−
4 4
=0
6c. We compute
Z
∞
5
Z
x
(x)(0) dx +
xf (x) dx =
E[X] =
∞
Z
−∞
−∞
5
5
dx = 5 ln x|∞
x=5 = +∞
x2
7. We compute
Z
∞
1=
Z
0
Z
f (x) dx =
0 dx +
−∞
−∞
1
2
Z
∞
1
x3 0 dx = ax + b
= a + b/3
3 x=0
Z
1
(a + bx ) dx +
0
1
Also
Z
∞
3/5 = E[X] =
Z
0
xf (x) dx =
−∞
(x)(0) dx +
−∞
2
Z
x(a + bx ) dx +
0
∞
(x)(0) dx
1
1
2
x4 x
= a/2 + b/4
= a +b
2
4 x=0
We solve the system 1 = a + b/3 and 3/5 = a/2 + b/4 to obtain a = 3/5 and b = 6/5.
4
8. We compute
Z
∞
Z
0
Z
∞
Z
∞
xf (x) dx =
(x)(0) dx +
(x)(xe ) dx =
x2 e−x dx
−∞
−∞
0
0
Z ∞
Z ∞
−x
−x ∞
−x ∞
e
−x
2e
2e
2x
xe dx
= x
−
dx = x
+2
−1 x=0
−1
−1 x=0
0
0
Z ∞ −x −x ∞
−x ∞
e
e
2e
= x
+
2x
−
2
dx
−1 x=0
−1 x=0
−1
0
∞
∞ −x ∞
e−x e−x 2e
= x
+ 2x
+2
=2
−1 x=0
−1/2 x=0
−1 x=0
E[X] =
−x
10a. The times in which the passenger would go to destination A are 7:05–7:15, 7:20–7:30,
7:35–7:45, 7:50–8:00, i.e., a total of 40 out of 60 minutes. So 2/3 of the time, the passenger
goes to destination A.
10b. The times in which the passenger would go to destination A are 7:10–7:15, 7:20–7:30,
7:35–7:45, 7:50–8:00, 8:05–8:10, i.e., a total of 40 out of 60 minutes. So 2/3 of the time, the
passenger goes to destination A.
11. To interpret this statement, we say that the location of the point is X inches from the
left-hand-edge of the line, with 0 ≤ X ≤ L. Since the point is chosen at random on the line,
with no further clarification, it is fair to assume that the distribution is uniform, i.e., X has
density function f (x) = L1 for 0 ≤ x ≤ L, and f (x) = 0 otherwise.
The ratio of the shorter to the longer segment is less than 1/4 if X ≤ 15 L or X ≥ 54 L. So
the ratio of the shorter to the longer segment is less than 1/4 with probability
Z 1L
Z L
5
1
1
dx +
dx = 1/5 + 1/5 = 2/5
4
L
L
0
L
5
1
13a. Let X denote the time (in minutes) until the bus arrives. Then X has density f (x) = 30
for 0 ≤ x ≤ 30, and f (x) = 0 otherwise.
R∞
R 30 1
The probability that we wait longer than 10 minutes is 10 f (x) dx = 10 30
dx = 2/3.
13b. Let E denote the event that the bus arrives after 10:25; let F denote the event that
the bus arrives at 10:15. Then the desired probability is
R 30 1
R∞
f
(x)
dx
dx
P (E ∩ F )
1/6
30
= R25
=
= 1/3
P (E | F ) =
= R25
∞
30 1
P (F )
1/2
f (x) dx
dx
15
15 30
14. Using proposition 2.1, we compute
Z ∞
Z 0
Z 1
Z
n
n
n
n
E[X ] =
x fX (x) dx =
(x )(0) dx +
(x )(1) dx +
−∞
=
−∞
0
1
∞
n
Z
(x )(0) dx =
1
xn dx
0
n+1 1
x
1
=
n + 1 x=0 n + 1
To use the definition of expectation, we need to find the density of the random variable Y =
X n . We first find the cumulative distribution function of Y . We know that P (Y ≤ a) = 0
5
for a ≤ 0, and P (Y
≤ a) = 1 for a ≥ 1. For 0 < a < 1, we have P (Y ≤ a) = P (X n ≤ a) =
√
R
na
√
√
P (X ≤ n a) = 0 1 dx = n a. Therefore, the cumulative distribution function of Y is
(√
n
a if 0 < a < 1
FY (a) =
0
otherwise
So the density of Y is
(
fY (x) =
1
1 n
a −1
n
0
if 0 < x < 1
otherwise
So the expected value of Y is
Z ∞
Z 0
Z 1
Z ∞
Z 1
1 1 −1
1 1/n
E[Y ] =
xfY (x) dx =
(x)(0) dx +
(x) x n dx +
x dx
(x)(0) dx =
n
−∞
−∞
0
1
0 n
1
1
1 x n +1 1
= 1
=
n n + 1
n+1
x=0
15a. We compute
5 − 10
X − 10
>
P (X > 5) = P
= P (Z > −5/6) = P (Z < 5/6) ≈ Φ(.83) ≈ .7967
6
6
15b. We compute
X − 10
16 − 10
4 − 10
<
<
P (4 < X < 16) = P
6
6
6
= P (−1 < Z < 1) = P (Z < 1) − P (Z < −1) = Φ(1) − (1 − Φ(1)) ≈ .6826
15c. We compute
X − 10
8 − 10
P (X < 8) = P
<
= P (Z < −1/3) = 1−P (Z < 1/3) ≈ 1−Φ(.33) ≈ .3707
6
6
15d. We compute
20 − 10
X − 20
<
= P (Z < 5/3) ≈ Φ(1.67) ≈ .9525
P (X < 20) = P
6
6
15e. We compute
X − 20
16 − 10
P (X > 16) = P
>
= P (Z > 1) = 1 − P (Z < 1) = 1 − Φ(1) ≈ .1587
6
6
16. Let X denote the rainfall in a given year.
Then X has a uniform (µ = 40, σ = 4)
50−40
distribution, so P (X ≤ 50) = P X−40
≤
= P (Z ≤ 2.5) = Φ(2.5) ≈ .9938, where Z
4
4
has a standard normal distribution. If the rainfall in each year is independent of all other
years, then it follows that the desired probability is (P (X ≤ 50))10 ≈ (.9938)10 ≈ .9397.
19. We compute
.10 = P (X > c) = P
Thus Φ
c−12
2
= .90. So
X − 12
c − 12
>
2
2
c−12
2
=P
≈ 1.28. So c ≈ 14.56.
c − 12
Z>
2
=1−Φ
c − 12
2
6
20a. We have n = 100 people, each of which is in favor of a proposed rise in school taxes
with probability p = .65. The number of the 100 people who are in favor of the rise in taxes
is a Binomial (n = 100, p = .65) random variable, which is well-approximated by a normal
random variable X with mean np = 65 and variance np(1 − p) = 22.75. So the probability
that at least 50 are in favor of the proposition is approximately
X − 65
49.5 − 65
P (X > 49.5) = P √
> √
≈ P (Z > −3.25) = Φ(3.25) ≈ .9994
22.75
22.75
20b. The desired probability is approximately
59.5 − 65
X − 65
70.5 − 65
√
P (59.5 < X < 70.5) = P
≈ P (−1.15 < Z < 1.15)
<√
< √
22.75
22.75
22.75
= P (Z < 1.15) − P (Z < −1.15) = Φ(1.15) − (1 − Φ(1.15)) ≈ .7498
20c. The desired probability is approximately
74.5 − 65
X − 65
< √
≈ P (Z < 1.99) = P (Z < 1.99) ≈ .9767
P (X < 74.5) = P √
22.75
22.75
21. We write X for the height of a randomly-selected 25-year-old man. We compute
X − 71
74 − 71
P (X > 74) = P √
> √
= P (Z > 1.2) = 1 − Φ(1.2) ≈ 1 − .8849 = .1151
6.25
6.25
Also,
P (X > 77 | X > 72) =
=
X−71
√
6.25
P
>
P (X > 77)
P (X > 77 and X > 72)
=
= P (X > 72)
P (X > 72)
√
P X−71
>
6.25
77−71
√
6.25
72−71
√
6.25
P (Z > 2.4)
1 − Φ(2.4)
1 − .9918
=
≈
≈ .0238
P (Z > .4)
1 − Φ(.4)
1 − .6554
23. Let X denote the number of times “6” shows during 1000 independent rolls of a fair die.
Then X is a Binomial random variable with n = 1000 and p = 1/6. So X is approximately
normal with mean np = 1000/6 and variance np(1 − p) = 1000(1/6)(5/6). Thus
P (150 ≤ X ≤ 200) = P (149.5 ≤ X ≤ 200.5)
=P
149.5 − (1000/6)
X − (1000/6)
200.5 − (1000/6)
p
≤p
≤p
1000(1/6)(5/6)
1000(1/6)(5/6)
1000(1/6)(5/6)
!
≈ P (−1.46 ≤ Z ≤ 2.87) = P (Z ≤ 2.87) − P (Z ≤ −1.46)
= P (Z ≤ 2.87) − P (Z ≥ 1.46) = P (Z ≤ 2.87) − (1 − P (Z ≤ 1.46))
= Φ(2.87) − (1 − Φ(1.46)) ≈ .9979 − (1 − .9279) = .9258
Given that “6” shows exactly 200 times, then the remaining 800 rolls are all independent,
with possible outcomes 1, 2, 3, 4, 5, each appearing with probability 1/5 on each die. Let
Y denote the number of times “5” shows during the 800 rolls. Then Y is a Binomial
random variable with n = 800 and p = 1/5. So Y is approximately normal with mean
7
np = 800/5 = 160 and variance np(1 − p) = 800(1/5)(4/5) = 128. Thus
Y − 160
149.5 − 160
√
√
P (Y < 150) = P (Y ≤ 149.5) = P
≤
128
128
≈ P (Z ≤ −.93) = P (Z ≥ .93) = 1 − P (Z ≤ .93) = 1 − Φ(.93) ≈ 1 − .8238 = .1762
25. Let X denote the number of acceptable items. Then X is a Binomial random variable
with n = 150 and p = .95. So X is approximately normal with mean np = (150)(.95) = 142.5
and variance np(1 − p) = (150)(.95)(.05) = 7.125. Thus
139.5 − 142.5
X − 142.5
√
P (150 − X ≤ 10) = P (140 ≤ X) = P (139.5 ≤ X) = P
≤ √
7.125
7.125
≈ P (−1.12 ≤ Z) = P (Z ≤ 1.12) = Φ(1.12) = .8686
26. With a fair coin, let X denote the number of heads. Then X is a Binomial random
variable with n = 1000 and p = 1/2. So X is approximately normal with mean np = 500
and variance np(1 − p) = 250. Thus the probability we reach a false conclusion is
524.5 − 500
X − 500
√
P (525 ≤ X) = P (524.5 ≤ X) = P
≤ √
≈ P (1.55 ≤ Z)
250
250
= 1 − P (Z ≤ 1.55) = 1 − Φ(1.55) ≈ 1 − .9394 = .0606
With a biased coin, let Y denote the number of heads. Then Y is a Binomial random variable
with n = 1000 and p = .55. So Y is approximately normal with mean np = 550 and variance
np(1 − p) = 247.5. Thus the probability we reach a false conclusion is
524.5 − 550
Y − 550
< √
P (Y < 525) = P (Y < 524.5) = P √
≈ P (Z ≤ −1.62)
247.5
247.5
= P (Z ≥ 1.62) = 1 − P (Z ≤ 1.62) = 1 − Φ(1.62) ≈ 1 − .9474 = .0526
28. We assume that each person is left-handed, independent of all the other people. Let
X denote the number of the 200 people who are lefthanded. Then X is a Binomial random
variable with n = 200 and p = .12. So X is approximately normal with mean np = 200(.12) =
24 and variance np(1 − p) = 200(.12)(.88) = 21.12. Thus the probability that at least 20 of
the 200 are lefthanded is
19.5 − 24
X − 24
√
P (20 ≤ X) = P (19.5 ≤ X) = P
<√
≈ P (−.98 < Z)
21.12
21.12
= P (Z < .98) = Φ(.98) ≈ .8365
32a. Let X denote the time (in hours) neededR to repair a machine. Then X is exponentially
∞
distributed with λ = 1/2. Then P (X > 2) = 2 12 e−(1/2)x dx = e−1 ≈ .368.
32b. The conditional probability is
R ∞ 1 −(1/2)x
e
dx
P (X > 10 & X > 9)
P (X > 10)
e−5
2
P (X > 10 | X > 9) =
=
= R10
=
= e−1/2
∞ 1 −(1/2)x
−9/2
P (X > 9)
P (X > 9)
e
e
dx
9 2
We could also have simply computed the line above by writing
P (X > 10)
1 − P (X ≤ 10)
1 − F (10)
1 − (1 − e−(1/2)10 )
e−5
=
=
=
= −9/2 = e−1/2
−(1/2)9
P (X > 9)
1 − P (X ≤ 9)
1 − F (9)
1 − (1 − e
)
e
An alternative method is to simply compute that probability
that the waiting time would
R∞
be at least one additional hour, which is P (X > 1) = 1 12 e−(1/2)x dx = e−1/2 (or equivalently,
8
P (X > 1) = 1 − P (X ≤ 1) = 1 − F (1) = 1 − (1 − e−(1/2)1 ) = e−1/2 ). Either way, the answer
is e−1/2 ≈ .6065.
33. Let X denote the time in years that the radio continues to functino for Jones. The
desired probability is
∞
Z ∞
1 −(1/8)x
1 e−(1/8)x P (X > 8) =
e
dx =
= e−1 ≈ .368
8
8 −1/8 x=8
8
We could also have written P (X > 8) = 1 − P (X ≤ 8) = 1 − (1 − e−(1/8)8 ) = e−1 ≈ .368.
34a. We write X for the total mileage of the car, given in thousands of miles. If X is
1
exponentially distributed with λ = 20
, then the desired conditional probability is
R ∞ 1 −(1/20)x
e
dx
P (X > 30)
e−3/2
P (X > 30 & X > 10)
20
=
= R30
= −1/2 = e−1
P (X > 30 | X > 10) =
∞ 1 −(1/20)x
P (X > 10)
P (X > 10)
e
e
dx
10 20
We could also have simply computed the line above by writing
P (X > 30)
1 − P (X ≤ 30)
1 − F (30)
1 − (1 − e−(1/20)30 )
e−3/2
=
=
=
=
= e−1
P (X > 10)
1 − P (X ≤ 10)
1 − F (10)
1 − (1 − e−(1/20)10 )
e−1/2
An alternative method is to simply compute that
that the lifetime has at
R ∞ 1probability
least 20,000 more miles, which is P (X > 20) = 20 20
e−(1/20)x dx = e−1 (or equivalently,
P (X > 20) = 1 − P (X ≤ 20) = 1 − F (20) = 1 − (1 − e−(1/20)20 ) = e−1 ). Either way, the
answer is e−1 ≈ .368.
34b. If X is uniformly distributed on the interval (0, 40), then the desired probability is
R 40 1
dx
P (X > 30 & X > 10)
1/4
40
P (X > 30 | X > 10) =
= R30
=
= 1/3
40 1
P (X > 10)
3/4
dx
10 40
As an alternate method, we could think of the reduced lifetime (given that the lifetime is
over 10,000 miles) as between 20,000 and 40,000 miles, so the probability of the lifetime
being greater than 30,000 miles is 10/30 = 1/3.
37a. We compute P (|X| > 1/2) = P (X > 1/2 or X < −1/2) = P (X > 1/2) + P (X <
−1/2) = 1/2
+ 1/2
= 1/2.
2
2
37b. We first compute the cumulative distribution function of Y = |X|. For a ≤ 0, we have
P (Y ≤ a) = 0, since |X| is never less than a in this case. For a ≥ 1, we have P (Y ≤ a) = 1,
since |X| is always less than a in this case. For 0 < a < 1, we compute
2a
=a
2
or equivalently, P (Y ≤ x) = x for 0 < x < 1. In summary, the cumulative distribution
function of Y = |X| is


0 x ≤ 0
FY (x) = x 0 < x < 1

1 x ≥ 1
P (Y ≤ a) = P (|X| ≤ a) = P (−a ≤ X ≤ a) =
9
Differentiating throughout with respect to x yields the density of Y = |X| is
(
1 0<x<1
fY (x) =
0 otherwise
Intuitively, if X is uniformly distributed on (−1, 1), then Y = |X| is uniformly distribution
on [0, 1).
(
e−x
39. Since X is exponentially distributed with λ = 1, then X has density fX (x) =
0
and cumulative distribution function
(
1 − e−x x ≥ 0
FX (x) =
0
otherwise
x≥0
otherwise
Next, we compute the cumulative distribution function of Y = ln X. We have
a
P (Y ≤ a) = P (log X ≤ a) = P (X ≤ ea ) = FX (ea ) = 1 − e−e
x
or equivalently, P (Y ≤ x) = 1 − e−e . Differentiating throughout with respect to x yields
the density of Y = ln X is
x
fY (x) = ex e−e
40. Since X is uniformly distributed on (0, 1), then X has density
(
1 0<x<1
fX (x) =
0 otherwise
and cumulative distribution function


0 x ≤ 0
FX (x) = x 0 < x < 1

1 x ≥ 1
Next, we compute the cumulative distribution function of Y = eX . For a ≤ 1, we have
P (Y ≤ a) = 0, since Y = eX is never less than a in this case (i.e., X is never less than 0
in this case). For a ≥ e, we have P (Y ≤ a) = 1, since Y = eX is always less than a in this
case (i.e., X is always less than 0 in this case). For 1 < a < e (and thus 0 < ln a < 1), we
compute
P (Y ≤ a) = P (eX ≤ a) = P (X ≤ ln a) = FX (ln a) = ln a
or equivalently, P (Y ≤ x) = ln x for 1 < x < e. In summary, the cumulative distribution
function of Y = eX is


x≤1
0
FY (x) = ln x 1 < x < e

1
x≥e
Differentiating throughout with respect to x yields the density of Y = eX is
(
1
1<x<e
fY (x) = x
0 otherwise