Will Landau
September 26, 2011
STAT 231 Homework 5 Solutions
Exercise 5.1 (Devore 4.11).
R∞
R∞
R2
R2
f. E(X) = −∞ xf (x)dx = −∞ xF 0 (x)dx = 0 x · x2 dx = (1/2) 0 x2 dx =
h 3 i2
x
= 86 ≈ 1.333
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0
R∞
R∞
R2
R2
g. E(X 2 ) = −∞ x2 f (x)dx = −∞ x2 F 0 (x)dx = 0 x2 · x2 dx = (1/2) 0 x3 dx =
h 4 i2
2
x
8
= 2. Hence V (X) = E(X 2 ) − E 2 (X) = 2 − 68 = 36
≈ .222 . Also,
8
0p
σX = V (X) = .471
h. E(h(X)) = E(X 2 ) = 2 from part g.
Exercise 5.2 (Devore 4.13).
First, we should get part a:
Rx
a. cdf: for x > 1, FX (x) =
We need limx→∞ F (x) =
if x ≤ 1.
d. E(X) =
E(X 2 ) =
R∞
1
R∞
1
x·
x2 ·
k
x4 dx
k
x4 dx
f (t)dt =
Rx
1
x
kt−4 dt = − k3 t−3 1 = − 3xk3 + k3 .
= 1, so k=3 and F (X) = 1 − x−3 if x > 1 and 0
k
3
=
=
−∞
R∞
1
R∞
1
k
x3 dx
k
x2 dx
V (X) = E(X 2 ) − E 2 (X) = k −
=
−k
2
x−2
∞
1
= 0 + k/2 = k/2=1.5 .
∞
= −kx−1 1 = 0 + k = k=3
k2
4 ,
σX =
√
V =
q
k−
k2
4
= 0.866
e. P (X ∈ (E(X) − σX , E(X) + σX )) = F (E(X) + σX ) − F (E(X) − σX ) =
F (1.5 + .866) − F (1.5 − .866) = F (2.366) − F (.634) = (1 − 2.366−3 ) − 0 (part
a) = .9245
Exercise 5.3 (Devore 4.28).
a. Φ(2.17) − Φ(0) = .4850
b. Φ(1) − Φ(0) = .3413
c. Φ(0) − Φ(−2.5) = .4938
d. Φ(2.5) − Φ(−2.5) = .9876
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e. Φ(1.37) = .9147
f. 1 − Φ−1.75 = .9599
g. Φ(2) − Φ(−1.5) = .9104
h. Φ(2.5) − Φ(1.37) = .0791
i. 1 − Φ(1.5) = .0668
j. P (|Z| ≤ 2.5) = P (−2.5 ≤ Z ≤ 2.5) = Φ(2.5) − Φ(−2.5) = .9876
Exercise 5.4 (Devore 4.29).
a. You can find .9838 is found in the 2.1 row and the .04 column of the
standard normal table (Table A-3 page A-7) so c = 2.14 .
b. Φ(c) − Φ(0) = .291 ⇒ Φ(c) = .791 ⇒ c =.81
c. P (c ≤ Z) = .121 ⇒ 1 − P (clZ) = P (Z < c) = Φ(c) = 1 − .121 = .8790 ⇒
c=1.17
d. Φ(c) − Φ(−c) = Φ(c) − (1 − Φ(c)) = 2Φ(c) − 1 ⇒ Φ(c) = .9920 ⇒ c=.97 .
e. .016 = P (c ≤ |Z|) ⇒ 1 − .016 = 1 − P (c ≥ |Z|) = P (−c ≤ Z ≤ c) =
Φ(c) − Φ(−c) = 2Φ(c) − 1 ⇒ Φ(c) = .9920 ⇒ c=2.41
Exercise 5.5 (Devore 4.30).
a. Φ(c) = .9100 ⇒ c ≈ 1.34 (Use the table. 0.9099 is the entry in the 1.3
row, 0.04 column)
b. Since the normal distribution is symmetric with mean 0, 9th percentile
= -91st percentile = -1.34
c. Φ(c) = .7500 ⇒ c ≈ .675 since .7486 and .7517 are in the .67 and .68
entries of the table, respectively.
d. 25th percentile = -75th percentile = -.675
e. Φ(c) = .06 ⇒ c -1.555 (.0594 and .0606 appear as the -1.56 and -1.55
entries of the table, respectively).
Exercise 5.6 (Devore 4.32).
a. P (X ≤ 15) = Φ( 15−15.0
1.25 ) = Φ(0) = .5
b. P (X ≤ 17.5) = Φ( 17.5−f rm[o]−−5.0
) = Φ(2) = .9772
1.25
2
c. P (X ≥ 10) = 1 − P (X < 10) = 1 − Φ( 10−15
1.25 ) = 1 − Φ(−4) =
1 − .00003 = .99997≈ 1
14−15
d. P (14 ≤ X ≤ 18) = P (X ≤ 18) − P (X < 14) = Φ( 18−15
1.25 ) − Φ( 1.25 ) =
Φ() − Φ(2.4) − Φ(−.8) = .9918 − .2119 = .7799
e. P (|X − 15| ≤ 3) = P (−3 ≤ X − 15 ≤ 3) = P (12 ≤ X ≤ 18) = P (X ≤
12−15
18) − P (X < 12) = Φ( 18−15
1.25 ) − Φ( 1.25 ) = Φ(2.4) − Φ(−2.4) =
.9918 − .0082 = .9836
Exercise 5.7 (Devore 4.34).
a. P (X > .25) = P (Z > −.83) = 1 − Φ(.83) = 1 − .2033 = .7967
b. P (X.10) = Φ(−3.33) = .0004
c. The larges 5% of concentration values fall at or above the 95th
percentile. Hence, we need to find the 95th percentile of the normal
distribution with mean .30 and standard deviation .06. From the table,
the 95th percentile of the standard normal distribution is 1.6449. Thus,
if Z ∼ N (0, 1), then
.95 = P (Z ≤ 1.6449) = P (.06·Z +.3 ≤ .06·1.6449 +.3) = P (X ≤ 1.2869).
Hence, the 95th percentile is .3987, and
the largest 5% of concentration values are the ones at or above .3987.
Exercise 5.8 (Devore 4.42).
Let X be the temperature reading and let Z =
P (µ − .1 ≤ X ≤ µ + .1) = .95, which means:
X−µ
σ
∼ N (0, 1). We want
.95 = P (−.1 ≤ X − µ ≤ .1)
−.1
X −µ
.1
= P(
≤
≤ )
σ
σ
σ
−.1
.1
)
= P (Z ≤ ) − P (Z <
σ
σ
.1
−.1
= Φ( ) − Φ(
)
σ
σ
.1
.1
= Φ( ) − [1 − Φ )] (since the normal density is symmetric about 0)
σ
σ
.1
= 2Φ( ) − 1
σ
.1
With .95 = 2Φ( .1
σ ) − 1, we get Φ( σ ) = .975. From the table,
means σ = 0.0510
.1
σ
= 1.96, which
Exercise 5.9 (Devore 4.43).
If 10% of all resistors exceed 10.256 ohms, then 10.256 ohms is the 90th
percentile on the normal distribution of resistances: N(µ, σ 2 ). Hence
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Φ( 10.256−µ
) = .9. From the normal tables, this means
σ
10.256 − µ = 1.28σ ⇒ µ + 1.28σ = 10.256.
10.256−µ
σ
= z.9 = 1.28, so
If 5 % or resistors have a resistance smaller than 9.671 ohms, then 9.671 ohms
is the 5th percentile on the normal distribution of resistances: N(µ, σ 2 ). Hence
) = .05. From the normal tables, this means 9.671−µ
= z.05 = −1.64,
Φ( 9.671−µ
σ
σ
so 9.671 − µ = −1.64σ ⇒ µ − 1.64σ = 9.671.
Solving the system of equations:
µ + 1.28σ = 10.256
µ − 1.64σ = 9.671
we get µ = 10 and σ = .2
Exercise 5.10 (Devore 4.46).
a. P (67 ≤ X ≤ 75) = P (−1.00 ≤ Z ≤ 1.67) = .7938
b. P (70c ≤ X ≤ 70 + c) = P (−c/3 ≤ Z ≤ c/3) = Φ(c/3) − Φ(−c/3) =
2Φ(c/3) − 1. Now, if .95 = P (70c ≤ X ≤ 70 + c), then
.95 = 2Φ(c/3) − 1 ⇒ Φ(c/3) = .975 ⇒ c/3 = 1.96 ⇒ c=5.88
c. E(# of 10 that are acceptable) = 10?P(a single one is acceptable) =
7.938
d. p = P (X < 73.84) = P (Z < 1.28) = .9, soP (Y ≤ 8) = B(8; 10, .9) =
.264
Exercise 5.11 (Devore 4.47).
We want .99 = P (W < c − 1) = .99, where W ∼ N (12, 3.53 ) is the weight of a
random parcel. Standardizing, we get .99 = P (Z < c−1−12
= Φ( c−13
3.5
3.5 ). From
c−13
the table, we get 3.5 = z.99 = 2.33 ⇒ c = 2.326 ∗ 3.5 + 13 ⇒ c=21.155
Exercise 5.12 (Devore 4.53).
p
Remember that for X ∼ Binomial(n, p), µX = np and σX = np(1 − p).
Hence, for p=.5, µX = 25 ∗ .5 = 12.5 and σX = 25 ∗ .5 ∗ (1 − .5) = 5.25. For
p=.6, µX = 25 ∗ .6 = 15 and σX = 25 ∗ .6 ∗ (1 − .6) = 6. For p=.8,
µX = 25 ∗ .8 = 20 and σX = 25 ∗ .8 ∗ (1 − .8) = 4.
The following tables show the results.
a.
4
b.
c.
The comparisons are pretty good, but they decrease as p goes from .5 to .8. In
general, the normal approximation improves with large n (for a smoother
binomial pdf) and p close to .5 (for a binomial pdf without much skew).
Exercise 5.13 (Devore 4.54).
p
p = .10 and n = 200, so µX = np = 20, and σX = np(1 − p) = 18 = 4.243.
Let Y ∼ N (20, 18). Then, Y approximates X. Using the normal
approximation:
a. P (X ≤ 30) ≈ P (Y ≤ 30.5) = Φ( 30.5−20
4.243 ) = Φ(2.47) = .9932
b. P (X < 30) ≈ P (Y ≤ 29.5) = Φ( 29.5−20
4.243 ) = Φ(2.24) = .9875
c. P (15 ≤ X ≤ 25) = P (X ≤ 25) − P (X ≤ 14) ≈ P (Y ≤ 25.5) − P (Y ≤
14.5−20
14.5) = Φ( 25.5−20
4.243 ) − Φ( 4.243 ) = Φ(1.30) − Φ(−1.30) = .9032 − .0968 =
.8064
Exercise 5.14 (Devore 4.59).
a. E(X) =
b. σX =
1
λ
1
λ
=1
=1
c. P (X ≤ 4) = FX (4) = 1 − e−4 = 0.9827
d. P (2 ≤ X ≤ 5) = P (X ≤ 5) − P (X < 2) = FX (5) − FX (2) =
1 − e−5 − (1 − e−2 ) = e−2 − e−5 = 0.1286
Exercise 5.15 (Devore 4.61).
Let X be the rainfall duration. X ∼ Exponential(λ = 1/µ = 1/2.725 = .367)
a. P(duration is at least 2 hours) =
P (X ≥ 2) = 1 − P (X < 2) = 1 − FX (2) = 1 − (1 − e−.367·2 ) = .480
P(at most 3 hours) = P (X ≤ 3) = FX (3) = 1 − e−.367·3 = .667
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P(between 2 and 3 hours) = P (2 ≤ X ≤ 3) = P (X ≥ 2) − P (X > 3) =
P (X ≥ 2) − (1 − P (X ≤ 3)) = .480 − (1 − 0.667) (from the previous 2
calculations) = .147
b. µX = 2.725. Hence, µ + 2σ = 3σ = 8.175.
P (X > µ + 2σ) = 1 − P (X ≤ µ + 2σ) = 1 − FX (8.175) =
1 − (1 − e−.367·8.175 ) = .046
P (X < µ − σ) = P (X < 0) = FX (0) = 1 − e0 = 0
Exercise 5.16 (Devore 4.69).
a. For X ≥ t, all components need to last at least t hours. Hence,
{X ≥ t} = A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5
b. P (X ≥ t) = P (A1 )P (A2 )P (A3 )P (A4 )P (A5 ) =
P (some component lasts at least t hours)5 = [1 −
P (some component lasts less than t hours)]5 = [1−(1−e−.01·t )]5 = e−.05t .
Hence, P (X ≤ t) = 1 − e−.05t , so that X ∼ Exp(λ = .05)
c. By the same reasoning, P (X ≤ t) = 1 − e−nλt , so
X has an exponential distribution with parameter nλ .
Exercise 5.17 (Devore 4.73).
Let X be the lifetime of a random specimen. Then, the cdf of X is
α
2.5
F (x) = F (x; α = 2.5, β = 200) = 1 − e−(x/β) = 1 − e−(x/200) for x ≥ 0 and
0 for x < 0.
a. At most 250:
2.5
P (X ≤ 250) = F (250) = 1 − e−(250/200) = 1 − e−1.7469 = .8257
Less than 250: since X is continuous,
P (X < 250) = P (X ≤ 250) = .8257
More than 300: P (X > 300) = 1 − P (X ≤ 300) = 1 − F (300) =
2.5
1 − (1 − e−(300/200) ) = .0636
b. Between 100 and 250:
P (100 ≤ X ≤ 250) = P (X ≤ 250) − P (X < 100) = .8257 − F (100) (from
2.5
part a) = .8257 − (1 − e(100/200) ) = .8257 − (.1620) = .6637
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c. This question asks for the median m of X. By definition, F (m) = .5, so:
α
1 − e−(m/β) = .5
α
.5 = e−(m/β)
α
m
log(.5) = −
β
α
m
log(2) =
β
m = β[log(2)]1/α = 200[log(2)]1/2.5 = 172.727
Exercise 5.18 (Devore 5.1).
a. P (X = 1, Y = 1) = p(1, 1) = .20
b. P (X ≤ 1 and Y ≤ 1) = p(0, 0) + p(0, 1) + p(1, 0) + p(1, 1) = .42
c. At least one hose is in use at both islands.
P (X 6= 0 and Y 6= 0) = p(1, 1) + p(1, 2) + p(2, 1) + p(2, 2) = .70
d. Sum the row probabilities to get pX (x) = .16, .34, .50 for x = 0, 1, 2. Sum
the column probabilities to get pY (y) = .24, .38, .38 for y = 0, 1, 2.
P (X ≤ 1) = pX (0) + pX (1) = .50
Exercise 5.19 (Devore 5.3).
a. p(1, 1) = .15 , the entry in the 1st row and 1st column of the joint
probability table.
b. P (X1 = X2 ) = p(0, 0) + p(1, 1) + p(2, 2) + p(3, 3) =
.08 + .15 + .10 + .07 = .40
c. A = {(x1 , x2 ) | x1 ≥ 2 + x2 } ∪ {(x1 , x2 ) | x2 ≥ 2 + x1 }
P (A) = p(2, 0) + p(3, 0) + p(4, 0) + p(3, 1) + p(4, 1) + p(4, 2) + p(0, 2) +
p(0, 3) + p(1, 3) = .22
d. exactly 4: P (exactly 4) = p(1, 3) + p(2, 2) + p(3, 1) + p(4, 0) = .17
P (at least 4) =
P (exactly 4) + p(4, 1) + p(4, 2) + p(4, 3) + p(3, 2) + p(3, 3) + p(2, 3) = .46
Exercise 5.20 (Devore 5.4).
a. P1 (0) = P (X1 = 0) = p(0, 0) + p(0, 1) + p(0, 2) + p(0, 3) = .19
P1 (1) = P (X1 = 1) = p(1, 0) + p(1, 1) + p(1, 2) + p(1, 3) = .30, etc.
7
E(X1 ) = 0 · .19 + 1 · .3 + 2 · .25 + 3 · .14 + 4 · .12 = 1.7
b. P2 (0) = P (X2 = 0) = p(0, 0) + p(1, 0) + p(2, 0) + p(3, 0) + p(4, 0) = .19,
etc.
c. p(4, 0) = 0 yet p1 (4) = .12 > 0 and p2 (0) = .19 > 0, so
p(x1 , x2 ) 6= p1 (x1 ) · p2 (x2 ) for some (x1 , x2 ). Hence, the two variables are
not independent.
8