CHAPTER 4
Section 4.1
1.
2.
1
1
f ( x)dx   12 xdx  14 x 2

a.
P(X  1) =
b.
P(.5  X  1.5) =
c.
P(X > 1.5) =
f(x) =
1
10


0

1.5

1
2
.5
f ( x)dx  
2
1
1.5 2
1.5

1.5
xdx  14 x 2

1
0
 .25
 .5
.5
xdx  14 x 2

2
1.5
 167  .438
for –5  x  5, and = 0 otherwise

0
1
dx  .5
a.
P(X < 0) =
b.
P(–2.5 < X < 2.5) =
c.
P(–2  X  3) =
d.
P(k < X < k + 4) =
5 10

3

2.5
1
 2.5 10
1
 2 10

dx  .5
k 4
k
dx  .5
1
10
dx  10x k
k 4
 101 [(k  4)  k ]  .4
3.
a.
0.4
f(x)
0.3
0.2
0.1
0.0
-3
b.
P(X > 0) =

2
-2
-1
0
x
1
2
2
.09375 (4  x 2 )dx  .09375 (4 x 
0
x3 
)  .5
3 
0
131
3
Chapter 4: Continuous Random Variables and Probability Distributions
1

.09375(4  x 2 )dx  .6875
c.
P(–1 < X < 1) =
d.
P(x < –.5 OR x > .5) = 1 – P(–.5  X  .5) = 1 –
1
.5

.5
.09375 (4  x 2 )dx
= 1 – .3672 = .6328
4.
a.
b.




f ( x; )dx 


e x
2
0
P(X  200) =
x
200
2
/ 2 2
f ( x; )dx 

dx   e  x

200
x
e x
2
0
2


/ 2 2
0
2
/ 2 2
 0  (1)  1
dx   e  x
2
/ 2 2

200
0
 .1353  1  .8647
P(X < 200) = P(X  200)  .8647, since x is continuous.
P(X  200) = 1 – P(X < 200)  .1353

c.
P(100  X  200) =
d.
For x > 0, P(X  x) =

x

200
100
f ( y; )dy  
f ( x; )dx   e  x
1=
0



/ 20, 000
2
f ( x)dx   kx2 dx  k
0
 
2
x3
3
0

200
 .4712
100
2
2
y  y 2 / 2 2
e
dx   e  y / 2
2
e
x
5.
a.
2

x
0
 1  ex
2
/ 2 2
 k 83   k  83
1.6
1.4
1.2
f(x)
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
x
1

b.
P(0  X  1) =
c.
P(1  X  1.5) =

d.
P(X  1.5) = 1 –

3
0 8
x 2 dx  18 x 3
1.5
1
1.5
0

1
0
 18  .125
3
8
x 2 dx  18 x 3

 18  32   18 1  19
64  .2969
3
8
x 2 dx  18 x 3

 1
1.5
1
1.5
0
132
3
3
    0  1 
1 3 3
8 2
27
64

37
64
 .5781
Chapter 4: Continuous Random Variables and Probability Distributions
6.
a.
0.8
0.7
0.6
f(x)
0.5
0.4
0.3
0.2
0.1
0.0
2.0
2.5
3.0
3.5
4.0
4.5
x

4
1
k[1  ( x  3) 2 ]dx   k[1  u 2 ]du 
4
3
k 
3
4
b.
1=
c.
P(X > 3) =
d.
P114  X  134   
e.
P( |X–3| > .5) = 1 – P( |X–3|  .5) = 1 – P( 2.5  X  3.5)
2
1

4
3
3 4
[1  ( x  3) 2 ]dx  .5 by symmetry of the p.d.f
13 / 4
3
11 / 4 4
1/ 4
3
4 1 / 4

[1  ( x  3) 2 ]dx 
=1–

.5
3
.5 4
[1  (u ) 2 ]du 
[1  (u ) 2 ]du 
7.
1
10
for 25  x  35 and = 0 otherwise
a.
f(x) =
b.
P(X > 33) =

35
1
33 10
dx  .2
35
x2 
c. E(X) =  x  dx 
  30
25
20  25
35
1
10
30  2 is from 28 to 32 minutes:
d.
P(28 < X < 32) =

P( a  x  a+2) =

32
1
28 10
a2
a
dx  101 x28  .4
1
10
32
dx  .2 , since the interval has length 2.
133
47
 .367
128
5
 .313
16
Chapter 4: Continuous Random Variables and Probability Distributions
8.
a.
0.20
f(y)
0.15
0.10
0.05
0.00
0
2
4
6
8
10
y
5
10
y2   2
1 2 
 y
y 
b.  f ( y)dy   ydy   (  y)dy =


5
50  0  5
50  5
1 
1  1 1
=  (4  2)  (2  )    1
2 
2  2 2

5
10
1
0 25
2
5
1
25
3

3
y2 
9
 .18
ydy 
 
50  0 50
c.
P(Y  3) =
d.
P(Y  8) = 
e.
P( 3  Y  8) = P(Y  8) – P(Y < 3) =
f.
P(Y < 2 or Y > 6) = 
a.
P(X  6) =
1
0 25

5
1
0 25
8
ydy   ( 52  251 y )dy 
5

2
1
0 25
23
 .92
25
46 9 37


 .74
50 50 50
10
ydy   ( 52  251 y )dy 
6
2
 .4
5
9.
6
5.5
  .15e .15( x5) dx  .15 e .15u du (after u = x – .5)
.5
=
e

.15u 5.5
0
0
 1 e
.825
 .562
b.
1 – .562 = .438; .438
c.
P( 5  Y  6) = P(Y  6) – P(Y  5)  .562 – .491 = .071
134
Chapter 4: Continuous Random Variables and Probability Distributions
10.
a.

b.
c.
d.



f ( x; k , )dx  


P(X  b) =
b

k k
 1
dx   k    k
k 1
x
 x
k k
dx   k
k 1
x

P(a  X  b) =
b
a



  k  1
  
b
 1 
 
   k   1   
 x  
b
k k
dx   k
k 1
x
b
k
k
 1 
   
   k       
 x  a  a   b 
k
k
Section 4.2
11.
 .25
a.
P(X  1) = F(1) =
b.
P(.5  X  1) = F(1) – F(.5) =
c.
P(X > .5) = 1 – P(X  .5) = 1 – F(.5) =
1
4
3
16
 .1875
15
16
 .9375
~ 2
~
d. .5 = F (  ) 
 ~ 2  2  ~  2  1.414
4
x
2
for 0  x < 2, and = 0 otherwise
e.
f(x) = F(x) =
f.
1
1 2 2
x3 
8
  1.333
E(X) =  x  f ( x)dx   x  xdx   x dx 


0
2
2 0
6 0 6
2

2
2
1
1 2
x4 
xdx   x 3 dx    2,
g. E(X ) =  x f ( x) dx   x

0
2
2 0
8 0
2

2
2
So Var(X) = E(X2) – [E(X)]2 =
h.
2
2   86   368  .222 , x  .471
2
From g , E(X2) = 2
135
Chapter 4: Continuous Random Variables and Probability Distributions
12.
a.
P(X < 0) = F(0) = .5
b.
P(–1  X  1) = F(1) – F(–1) =
c.
P(X > .5) = 1 – P(X  .5) = 1 – F(.5) = 1 – .6836 = .3164
d.
F(x) = F(x) =
e.
F ~  .5 by definition. F(0) = .5 from a above, which is as desired.
11
16
 .6875
2


d 1 3 
x3 
   4 x    = 0  3  4  3x   .09375 4  x 2



dx  2 32 
3  
32 
3 

13.
a.
b.
1 

1
k
dx  1   k
4
3
x
cdf: F(x)=

x

x
3

1
f ( y)dy  
x
1
k
k
 1  0  ( )(1)  1   k  3
3
3
x
1
3
3 y dy   3 y
  x 3  1  1  3 . So
3
x
1
4
 0, x  1
F x   
3
1  x , x  1
c.
1  18   18 or .125;
P(2  x  3)  F (3)  F (2)  1  271   1  18   .963  .875  .088
P(x > 2) = 1 – F(2) = 1 –

d.
  3 
 3 
3 3
2
E ( X )   x 4 dx    3 dx   3 x
 0 
1
1
2
2 2
x 
x 
1


 3 
 3
1
E ( X 2 )   x 2  4 dx    2 dx   3x
 033
1
1
x 
x 
1
2
9 3
3
V ( X )  E ( X )  [ E ( X )]  3     3   or .75
4 4
2
  V ( x)  3 4  .866
2
e.
2
P(1.5  .866  x  1.5  .866)  P( x  2.366)  F (2.366)
 1  (2.366 3 )  .9245
136

Chapter 4: Continuous Random Variables and Probability Distributions
14.
a.
If X is uniformly distributed on the interval from A to B, then
1
A B
A 2  AB  B 2
2
E( X )   x 
dx 
, E( X ) 
A
BA
2
3
2
 B  A .
V(X) = E(X2) – [E(X)]2 =
2
B
With A = 7.5 and B = 20, E(X) = 13.75, V(X) = 13.02
 0
 x  7 .5
b. F(x) = 
 12.5
 1
x  7 .5
7.5  x  20
x  20
c.
P(X  10) = F(10) = .200; P(10  X  15) = F(15) – F(10) = .4
d.
 = 3.61, so    = (10.14, 17.36)
Thus, P( –   X   + ) = F(17.36) – F(10.14) = .5776
Similarly, P( – 2  X   + 2) = P(6.53  X  20.97) = 1
a.
F(x) = 0 for x  0, = 1 for x  1, and for 0 < x < 1,
15.
x
x
x
F ( x)   f ( y)dy   90 y 8 (1  y)dy  90 ( y 8  y 9 )dy


0
90 y 
1
9
9
1
10
y
10

x
0
0
 10 x  9 x
9
4
10
1.0
0.8
3
F(x)
f(x)
0.6
2
0.4
1
0.2
0
0.0
0.0
0.2
0.4
0.6
0.8
1.0
0.0
0.2
0.4
x
0.6
x
b.
F(.5) = 10(.5)9 – 9(.5)10  .0107
c.
P(.25  X  .5) = F(.5) – F(.25)  .0107 – [10(.25)9 – 9(.25)10]
 .0107 – .0000  .0107
d.
The 75th percentile is the value of x for which F(x) = .75
 .75 = 10(x)9 – 9(x)10
 x  .9036
137
0.8
1.0
Chapter 4: Continuous Random Variables and Probability Distributions
e.

E(X) =


90
 9 x10  11
x
E(X2) =




90
11
1
1
0
0
x  f ( x)dx  x  90 x 8 (1  x)dx  90 x 9 (1  x)dx

11 1
0
 119  .8182
1
1
0
0
x 2  f ( x)dx  x 2  90 x 8 (1  x)dx  90 x10 (1  x)dx
x 
11
90
12
x

12 1
0
 .6818
V(X)  .6818 – (.8182)2 = .0124,
x = .11134.
f.
   = (.7068, .9295). Thus, P( –   X   + ) = F(.9295) – F(.7068) = .8465 – .1602
= .6863, so the probability X is more than 1 sd from its mean equals 1–.6863 = 3137.
a.
F(x) = 0 for x < 0 and F(x) = 1 for x > 2. For 0  x  2,
16.
F(x) =

x
3
0 8
y 2 dy  18 y 3

x
0
 18 x 3
1.0
0.8
F(x)
0.6
0.4
0.2
0.0
0.0
0.5

1 1 3
8 2

b.
P(X  .5) = F(.5) =
c.
P(.25  X  .5) = F(.5) – F(.25)
d.
.75 = F(x) =
e.
E(X) =
f.

2.0
1
64
1
64
=
2

2
0
12
5
1.5
 18  14  
3
7
512
 .0137
x 3  x3 = 6  x  1.8171


E(X2) =
V(X) =
1
8
1.0
x
x  f ( x)dx  x 
x 
  32  
2
0
 x dx   x dx   x 
3
8
1
3
8 0
2
3
 x dx   x dx   x5
3
8
3
20
2
1
3
8 0
4
3 1
8 5
2
0
3 1
8 4
4
2
0
 32  1.5
 125  2.4
 .15 x = .3873
   = (1.1127, 1.8873). Thus, P( –   X   + ) = F(1.8873) – F(1.1127) = .8403 –
.1722 = .6681
138
Chapter 4: Continuous Random Variables and Probability Distributions
17.
xA
=p 
BA
x = (100p)th percentile = A + (B – A)p
a.
F(X) =
b.
1
1
x2 
1
1
A B
E( X )   x 
dx 
   
 B 2  A2 
A
BA
B  A 2 A 2 B  A
2
B

B

1
1
A 2  AB  B 2
E( X 2 )  
 B 3  A3 
3 B A
3


2
 A 2  AB  B 2    A  B  

B  A
( B  A)
  
V ( X )  
, x 
 
3
12
12

  2 
2
B
c.
A
f ( x) 
1
B n1  An1
dx 
B A
(n  1)( B  A)
1
1
 for –1 ≤ x ≤ 1
1  (1) 2
1

1
dx = .25
.5 2
a.
P(Y = .5) = P(X ≥ .5) =
b.
P(Y = –.5) = .25 as well, due to symmetry. For –.5 < y < .5, F(y) = .25 +

y
1
dx = .25 +
2
.5(y + .5) = .5 + .5y. Since Y ≤ .5, F(.5) = 1 and F(y) = 1 for y > .5 as well. That is,
y  .5
 0

F ( y)  .5  .5 y  .5  y  .5
 1
.5  y

1.0
0.8
0.6
F(y)
18.
E( X n )   x n 
0.4
0.2
0.0
-1.0
-0.5
0.0
y
139
0.5
1.0
.5
Chapter 4: Continuous Random Variables and Probability Distributions
19.
a.
P(X  1) = F(1) = .25[1 + ln(4)]  .597
b.
P(1  X  3) = F(3) – F(1)  .966 – .597  .369
c.
f(x) = F(x) = .25 ln(4) – .25 ln(x) for 0 < x < 4
a.
For 0  y  5, F(y) =
20.

1
y2
udu 
25
50
y
0
For 5  y  10, F(y) =

y
0

5
y
f (u)du   f (u)du   f (u)du
0
5
y 2
1
u 
2
y2
    du  y 
1
2 0  5 25 
5
50
1.0
0.8
F(y)
0.6
0.4
0.2
0.0
0
2
4
6
8
10
y
y 2p
 y p  50 p 
1/ 2
b.
For 0 < p  .5, p = F(yp) =
c.
E(Y) = 5 by straightforward integration (or by symmetry of f(y)), and similarly V(Y)=
50
y 2p
2
For .5 < p  1, p = y p 
 1  y p  10  5 2(1  p)
5
50
50
 4.1667 . For the waiting time X for a single bus,
12
25
E(X) = 2.5 and V(X) =
12
21.
E(area) = E(R2) =
r 2 f (r )dr   r 2  1  (10  r ) 2 dr 

9


11
140
3
4
501
  314.79
5
Chapter 4: Continuous Random Variables and Probability Distributions
22.
x


1 
1 
1

a. For 1  x  2, F(x) =  21  2 dy  2 y    2 x    4, so
1
y  1
x
y 



0
x 1


1 x  2
F(x) = 2 x  1x   4

1
x2

x
b.

1 
2 x p 
 4  p  2xp2 – (4 – p)xp + 2 = 0  xp = 14 [4  p 


xp 

~
~ = 1.64
find  , set p = .5  
p 2  8 p ] To
2
c.
E(X) =
E(X2) =
d.

2
1
2
 x2

1 
1

x  21  2 dx  2  x  dx 2  ln( x)   1.614
1
x
 x 

 2
 1
2
2
1


Amount left = max(1.5 – X, 0), so
E(amount left) =
23.
2
 x3

8
x  1 dx 2  x    Var(X) = .0626
 3
 1 3
2

2
1
1.5
1 

max( 1.5  x,0) f ( x)dx 2 (1.5  x)1  2 dx  .061
1
x 

With X = temperature in C, temperature in F =
9
 9
E  X  32  (120)  32  248,
5
 5
so  = 3.6
141
9
X  32, so
5
2
9
 9
Var  X  32     (2) 2  12.96 ,
5
 5
Chapter 4: Continuous Random Variables and Probability Distributions
24.


x
a.
E(X) =
b.
E(X) = 
c.
E(X2) = k k
k k
x k 1


dx  k
1
x k 1
dx 
k



k k x  k 1 
k
dx

 
k
 k  1 
k 1
x

1
k 2
, so
k 2
 k 2   k  2
k 2

Var(X) = 


 k  2   k 1
k  2k  12


d.
Var(X) = , since E(X2) = .
e.
E(Xn) = k k
a.
~ + 32) = P(1.8X + 32  1.8 ~ + 32) = P( X  ~ ) = .5
P(Y  1.8 
25.


x n( k 1) dx , which will be finite if n – (k+1) < –1, i.e. if n < k.
b.
90th for Y = 1.8(.9) + 32 where (.9) is the 90th percentile for X, since
P(Y  1.8(.9) + 32) = P(1.8X + 32  1.8(.9) + 32)
= (X  (.9) ) = .9 as desired.
c.
The (100p)th percentile for Y is 1.8(p) + 32, verified by substituting p for .9 in the
argument of b. When Y = aX + b, (i.e. a linear transformation of X), and the (100p)th
percentile of the X distribution is (p), then the corresponding (100p)th percentile of the
Y distribution is a(p) + b. (same linear transformation applied to X’s percentile)
142
Chapter 4: Continuous Random Variables and Probability Distributions
26.
a.

1=

k (1  x / 2.5) 7 dx =
0
2.5k
(1  x / 2.5) 6
6

=
0
k
 k = 2.4
2 .4
b.
2.5
2.0
f(x)
1.5
1.0
0.5
0.0
0

c.

E(X ) = 
E(X) =
1
2
3
4
x

x  2.4(1  x / 2.5) 7 dx =
0
2

6
7
8
2.5(u  1)  2.4u 7  2.5du = 0.5, or $500. Similarly,
1


x 2  2.4(1  x / 2.5) 7 dx =
0

(2.5(u  1)) 2  2.4u 7  2.5du = 0.625, so V(X) =
1
0.625 – (0.5)2 = 0.375, and σX =
d.
5
0.375 = 0.612, or $612.
The maximum out-of-pocket expense, $2500, occurs when $500 + 20%(X – $500) equals
$2500; this accounts for the $500 deductible and the 20% of costs above $500 not paid by
the insurance plan. Solve: $2,500 = $500 + 20%(X – $500)  X = $10,500. At that
point, the insurance plan has already paid $8,000, and the plan will pay all expenses
thereafter.
Recall that the units on X are thousands of dollars. If Y denotes the expenses paid by the
company (also in $1000s), Y = 0 for X ≤ 0.5; Y = .8(X – 0.5) for 0.5 ≤ X ≤ 10.5; and Y =
(X – 10.5) + 8 for X > 10.5. From this,
E(Y) =


y  2.4(1  x / 2.5) 7 dx =
0



10.5
0.5
0 dx +
0

.8( x  0.5)  2.4(1  x / 2.5) 7 dx +
0.5
( x  10 .5)  2.4(1  x / 2.5) 7 dx = 0 + 0.16024 + .00013, or $160.37.
10.5
27.
0  360
360  0
= 180 and σX =
= 30 12 . Using the
2
12
linear representation of Y, E(Y) = (2π/360)E(X) – π = (2π/360)(180) – π = 0, and σY =
Since X is uniform on [0,360], E(X) =
(2π/360)σX = (2π/360)( 30 12 ) =
 12
6
≈ 1.814. (In fact, Y is uniform on [–π, π].)
143
Chapter 4: Continuous Random Variables and Probability Distributions
Section 4.3
28.
a.
P(0  Z  2.17) = (2.17) – (0) = .4850
b.
(1) – (0) = .3413
c.
(0) – (–2.50) = .4938
d.
(2.50) – (–2.50) = .9876
e.
(1.37) = .9147
f.
P( –1.75 < Z) + [1 – P(Z < –1.75)] = 1 – (–1.75) = .9599
g.
(2) – (–1.50) = .9104
h.
(2.50) – (1.37) = .0791
i.
1 – (1.50) = .0668
j.
P( |Z|  2.50 ) = P( –2.50  Z  2.50) = (2.50) – (–2.50) = .9876
a.
.9838 is found in the 2.1 row and the .04 column of the standard normal table so c = 2.14.
b.
P(0  Z  c) = .291  (c) = .7910  c = .81
c.
P(c  Z) = .121  1 – P(c  Z) = P(Z < c) = (c) = 1 – .121 = .8790 
d.
P(–c  Z  c) = (c) – (–c) = (c) – (1 – (c)) = 2(c) – 1
 (c) = .9920  c = .97
e.
P( c  | Z | ) = .016  1 – .016 = .9840 = 1 – P(c  | Z | ) = P( | Z | < c )
= P(–c < Z < c) = (c) – (–c) = 2(c) – 1
 (c) = .9920  c = 2.41
29.
144
c = 1.17
Chapter 4: Continuous Random Variables and Probability Distributions
30.
a.
(c) = .9100  c  1.34 (.9099 is the entry in the 1.3 row, .04 column)
b.
9th percentile = –91st percentile = –1.34
c.
(c) = .7500  c  .675 since .7486 and .7517 are in the .67 and .68 entries,
respectively.
d.
25th = –75th = –.675
e.
(c) = .06  c  .–1.555 (.0594 and .0606 appear as the –1.56 and –1.55 entries,
respectively).
a.
Area under Z curve above z.0055 is .0055, which implies that
( z.0055) = 1 – .0055 = .9945, so z.0055 = 2.54
b.
( z.09) = .9100  z = 1.34 (since .9099 appears as the 1.34 entry).
c.
( z.663) = area below z.663 = .3370  z.633  –.42
a.
P(X  100) = P z 
b.
P(X  80) = P z 
c.
P(65  X  100) = P
31.
32.




100  80 
 = P(Z  2) = (2.00) = .9772
10 
80  80 
 = P(Z  0) = (0.00) = .5
10 
100  80 
 65  80
z
 = P(–1.50  Z  2)
10 
 10
= (2.00) – (–1.50) = .9772 – .0668 = .9104
d.
P(70  X) = P(–1.00  Z) = 1 – (–1.00) = .8413
e.
P(85  X  95) = P(.50  Z  1.50) = (1.50) – (.50) = .2417
f.
P(|X – 80|  10) = P(–10  X – 80  10) = P(70  X  90)
P(–1.00  Z  1.00) = .6826
145
Chapter 4: Continuous Random Variables and Probability Distributions
33.
a.
18  15 

P(X  18) = P Z 
 = P(Z  2.4) = (2.4) = .9452
1.25 

b.
P(10  X  12) = P(–4.00  Z  –2.40)  P(Z  –2.40) = (–2.40) = .0082
c.
P( |X – 15|  1.5(1.25) ) = P( |Z| ≤ 1.5) = P(–1.5  Z  1.5) = (1.5) – (–1.5) = .8664
a.
P(X > .25) = P(Z > –.83) = 1 – .2033 = .7967
b.
P(X  .10) = (–3.33) = .0004
c.
We want the value of the distribution, c, that is the 95 th percentile (5% of the values are
higher). The 95th percentile of the standard normal distribution = 1.645. So c = .30 +
(1.645)(.06) = .3987. The largest 5% of all concentration values are above .3987 mg/cm 3.
a.
P(X  10) = P(Z  .43) = 1 – (.43) = 1 – .6664 = .3336.
P(X > 10) = P(X  10) = .3336, since for any continuous distribution, P(x = a) = 0.
b.
P(X > 20) = P(Z > 4)  0
c.
P(5  X  10) = P(–1.36  Z  .43) = (.43) – (–1.36) = .6664 – .0869 = .5795
d.
P(8.8 – c  X  8.8 + c) = .98, so 8.8 – c and 8.8 + c are at the 1st and the 99th percentile
of the given distribution, respectively. The 1st percentile of the standard normal
distribution has the value –2.33, so
8.8 – c =  + (–2.33) = 8.8 – 2.33(2.8)  c = 2.33(2.8) = 6.524.
e.
From a, P(x > 10) = .3336. Define event A as {diameter > 10}, then P(at least one A i) =
1 – P(no Ai) = 1  P( A) 4  1  (1  .3336) 4  1  .1972  .8028
a.
P(X < 1500) = P(Z < 3) = (3) = .9987; P(X ≥ 1000) = P(Z ≥ –.33) = 1 – (–.33) = 1 –
.3707 = .6293
b.
P(1000 < X < 1500) = P(–.33 < Z < 3) = (3) – (–.33) = .9987 – .2707 = .7280
c.
From the table, (z) = .02  z ≈ –2.05  x = 1050 – 2.05(150) = 742.5 μm. The
smallest 2% of droplets are those smaller than 742.5 μm in size
d.
P(at least one droplet in 5 that exceeds 1500 μm) = 1 – P(all 5 are less than 1500 μm) = 1
– (.9987)5 = 1 – .9935 = .0065
34.
35.
36.
146
Chapter 4: Continuous Random Variables and Probability Distributions
37.
38.
a.
P(X = 105) = 0, since the normal distribution is continuous;
P(X < 105) = P(Z < 0.2) = P(Z ≤ 0.2) = Φ(0.2) = .5793;
P(X ≤ 105) = .5793 as well, since X is continuous
b.
No, the answer does not depend on μ or σ. For any normal rv, P(|X–μ| > σ) = P(|Z| > 1) =
P(Z < –1 or Z > 1) = 2Φ(–1) = 2(.1587) = .3174
c.
From the table, (z) = .1% = .001  z ≈ –3.09  x = 104 – 3.09(5) = 88.55 mmol/L.
The smallest .1% of chloride concentration values are those less than 88.55 mmol/L
Let X denote the diameter of a randomly selected cork made by the first machine, and let Y be
defined analogously for the second machine.
P(2.9  X  3.1) = P(–1.00  Z  1.00) = .6826
P(2.9  Y  3.1) = P(–7.00  Z  3.00) = .9987
So the second machine wins handily.
39.
40.
41.
42.
a.
 + (91st percentile from std normal) = 30 + 5(1.34) = 36.7
b.
30 + 5( –1.555) = 22.225
c.
 = 3.000 m;  = 0.140. We desire the 90th percentile: 30 + 1.28(0.14) = 3.179
 = 43;  = 4.5
40  43 
 = P(Z < –0.667) = .2514
4.5 
60  43 

P(X > 60) = P z 
 = P(Z > 3.778)  0
4.5 



a.
P(X < 40) = P z 
b.
43 + (–0.67)(4.5) = 39.985
100  200 

P(damage) = P(X < 100) = P z 
 = P(Z < –3.33) = .0004
300 

P(at least one among five is damaged)
= 1 – P(none damaged)
= 1 – (.9996)5 = 1 – .998 = .002
From Table A.3, P(–1.96  Z  1.96) = .95. Then P( – .1  X   + .1) =
.1
.1
.1 
  .1
 .0510
P
 z   implies that = 1.96, and thus that  
1.96


 
147
Chapter 4: Continuous Random Variables and Probability Distributions
43.
Since 1.28 is the 90th z percentile (z.1 = 1.28) and –1.645 is the 5th z percentile (z.05 = 1.645),
the given information implies that  + (1.28) = 10.256 and  + (–1.645) = 9.671, from
which (–2.925) = –.585,  = .2000, and  = 10.
44.
45.
a.
P( – 1.5  X   + 1.5) = P(–1.5  Z  1.5) = (1.50) – (–1.50) = .8664
b.
P( X <  – 2.5 or X >  + 2.5) = 1 – P( – 2.5  X   + 2.5)
= 1 – P(–2.5  Z  2.5) = 1 – .9876 = .0124
c.
P( – 2  X   –  or  +   X   + 2) = P(within 2 sd’s) – P(within 1 sd) = P( –
2  X   + 2) – P( –   X   + )
= .9544 – .6826 = .2718
With  = .500 inches, the acceptable range for the diameter is between .496 and .504 inches,
so unacceptable bearings will have diameters smaller than .496 or larger than .504. The new
distribution has  = .499 and  =.002. P(X < .496 or X >.504) =
.496  .499 
.504  .499 


P Z 
  P Z 
  PZ  1.5  PZ  2.5 = Φ(–1.5) + [1 –
.002
.002




Φ(2.5)] = .073. 7.3% of the bearings will be unacceptable.
46.
a.
P(67  X  75) = P(–1.00  Z  1.67) = .7938
b.
P(70 – c  X  70 + c) = P
c
c
c
c
 Z    2( )  1  .95   ( )  .9750
3
3
3
 3
c
 1.96  c  5.88
3
47.
c.
10P(a single one is acceptable) = 7.938
d.
p = P(X < 73.84) = P(Z < 1.28) = .9, so P(Y  8) = B(8;10,.9) = .264
The stated condition implies that 99% of the area under the normal curve with  = 12 and  =
3.5 is to the left of c – 1, so c – 1 is the 99th percentile of the distribution. Thus c – 1 =  +
(2.33) = 20.155, and c = 21.155.
48.
a.
By symmetry, P(–1.72  Z  –.55) = P(.55  Z  1.72) = (1.72) – (.55)
b.
P(–1.72  Z  .55) = (.55) – (–1.72) = (.55) – [1 – (1.72)]
No, symmetry of the Z curve about 0.
148
Chapter 4: Continuous Random Variables and Probability Distributions
49.
X N(3432, 482)
a.
4000  3432 

Px  4000   P Z 
  Pz  1.18   1  (1.18)  1  .8810  .1190 ;
482


4000  3432 
 3000  3432
P3000  x  4000   P
Z 

482
482


 1.18    .90   .8810  .1841  .6969
b.
50.
c.
We will use the conversion 1 lb = 454 g, then 7 lbs = 3178 grams, and we wish to find
3178  3432 

Px  3178   P Z 
  1  (.53)  .7019
482


d.
We need the top .0005 and the bottom .0005 of the distribution. Using the Z table, both
.9995 and .0005 have multiple z values, so we will use a middle value, ±3.295. Then
3432±(482)3.295 = 1844 and 5020, or the most extreme .1% of all birth weights are less
than 1844 g and more than 5020 g.
e.
Converting to lbs yields mean 7.5595 and sd 1.0608. Then
7  7.5595 

P  X  7   P Z 
  1  (.53)  .7019 This yields the same answer as in
1.0608 

part c.
We use a Normal approximation to the Binomial distribution: X  b(x;1000,.03) ≈
N(30,5.394)
a.
b.
51.
2000  3432 
5000  3432 


P X  2000  X  5000   P Z 
  P Z 

482
482




  2.97   1  3.25   .0015  .0006  .0021
39.5  30 

Px  40   1  P x  39  1  P Z 

5.394 

 1  (1.76)  1  .9608  .0392


5% of 1000 = 50: P x  50   P Z 
50.5  30 
  (3.80)  1.00
5.394 
P(|X – |  ) = P(X   –  or X   + ) = 1 – P( –   X   + )
= 1 – P(–1  Z  1) = .3174.
Similarly, P(|X – |  2) = 1 – P(–2  Z  2) = .0456 and P(|X – |  3) = .0026.
These are considerably less than the bounds 1, .25, and .11 given by Chebyshev.
52.
a.
P(20 – .5  X  30 + .5) = P(19.5  X  30.5) = P(–1.1  Z  1.1) = .7286
b.
P(at most 30) = P(X  30 + .5) = P(Z  1.1) = .8643.
P(less than 30) = P(X < 30 – .5) = P(Z < .9) = .8159
149
Chapter 4: Continuous Random Variables and Probability Distributions
53.
p = .5  μ = 12.5 & σ2 = 6.25; p = .6  μ = 15 & σ2 = 6; p = .8  μ = 20 and σ2 = 4
a.
p
.5
.6
.8
P(15 ≤ X ≤ 20)
= .212
= .577
= .573
P(14.5 ≤ Normal ≤ 20.5)
= P(.80  Z  3.20) = .2112
= P(–.20  Z  2.24) = .5668
= P(–2.75  Z  .25) = .5957
p
.5
.6
.8
P(X ≤ 15)
= .885
= .575
= .017
P(Normal ≤ 15.5)
= P(Z  1.20) = .8849
= P(Z  .20) = .5793
= P(Z  –2.25) = .0122
p
.5
.6
.8
P(X ≥ 20)
= .002
= .029
= .617
P(Normal ≥ 19.5)
= P(Z ≥ 2.80) = .0026
= P(Z ≥ 1.84) = .0329
= P(Z ≥ –0.25) = .5987
b.
c.
54.
p = .10; n = 200; np = 20, npq = 18
 30  .5  20 

 = (2.47) = .9932
18


a.
P(X  30) =
b.
P(X < 30) =P(X  29) =
c.
P(15  X  25) = P(X  25) – P(X  14) =
 29  .5  20 

 = (2.24) = .9875
18


 25  .5  20 
 14  .5  20 

  

18
18




(1.30) – (–1.30) = .9032 – .0968 = .8064
55.
n = 500, p = .75   = 375, σ = 9.68246
a. P(360  X  400) = P(359.5  X  400.5) = P(–1.60  Z  2.58) = .9409
b.
56.
P(X < 400) = P(X  399.5) = P(Z  2.53) = .9943
P(X   + [(100p)th percentile for std normal])
X 

P
 ... = P(Z  […]) = p as desired
 

150
Chapter 4: Continuous Random Variables and Probability Distributions
57.
a.


Fy(y) = P(Y  y) = P(aX + b  y) = P X 
( y  b) 
 (for a > 0).
a 
Now differentiate with respect to y to obtain

1
fy(y) = Fy ( y ) 
2 a

e
1
2 a 2 2
[ y  ( a  b )]2
so Y is normal with mean a + b
and variance a22.
9
5
(115)  32  239 , variance = 12.96
b.
Normal, mean
a.
P(Z  1)  .5  exp 
b.
P(Z > 3)  .5  exp 
c.
P(Z > 4)  .5  exp 
58.
 83  351  562 
  .1587
 703  165 
  2362 
  .0013
 399.3333 
  3294 
  .0000317 , so
 340.75 
P(–4 < Z < 4)  1 – 2(.0000317) = .999937
d.
  4392 
  .00000029
 305.6 
P(Z > 5)  .5  exp 
151
Chapter 4: Continuous Random Variables and Probability Distributions
Section 4.4
59.
1
1
a.
E(X) =
b.

c.
P(X  4 ) = 1  e (1)( 4)  1  e 4  .982
d.
P(2  X  5) = 1  e (1)(5)  1  e (1)( 2)  e 2  e 5  .129
a.
P(X  100 ) = 1  e
1

1



60.
 (100)(.01386)
 ( 200)(.01386)
 1  e 1.386  .7499
 1  e 2.772  .9375
P(X  200 ) = 1  e
P(100  X  200) = P(X  200 ) – P(X  100 ) = .9375 – .7499 = .1876
b.
1
 72.15 ,  = 72.15
.01386
=
P(X >  + 2) = P(X > 72.15 + 2(72.15)) = P(X > 216.45) =
1  1  e ( 216.45)(.01386)  e 2.9999  .0498

c.
61.
~
~
.5 = P(X  ~ )  1  e (  )(.01386)  .5  e (  )(.01386)  .5
~(.01386 )  ln(. 5)  .693  ~  50
Mean =
a.

1

 25,000 implies  = .00004
P(X > 20,000) = 1 – P(X  20,000) = 1 – F(20,000; .00004)  e
1.2
P(X  30,000) = F(30,000; .00004)  1  e
P(20,000  X  30,000) = .699 – .551 = .148
b.
 
1

 .699
 25,000 , so P(X >  + 2) = P( x > 75,000) =
1 – F(75,000;.00004) = .05.
Similarly, P(X >  + 3) = P( x > 100,000) = .018
152
 (.00004)( 20, 000)
 .449
Chapter 4: Continuous Random Variables and Probability Distributions
62.
a.
Clearly E(X) = 0 by symmetry, so V(X) = E(X 2) =

b.
(3)

3

2

2
. Solving
P(|X – 0| ≤ 40.9) =
2


 40.9
2


x2

2
e  | x| dx = 


x 2 e x dx =
0
= (40.9)2 yields λ = 0.034577
2
40.9


e  | x| dx =

40.9
e x dx = 1 – e–40.9 λ = .75688
0
63.
a.
If a customer’s calls are typically short, the first calling plan makes more sense. If a
customer’s calls are somewhat longer, then the second plan makes more sense, viz. 99¢ is
less than 20min(10¢/min) = $2 for the first 20 minutes under the first (flat–rate) plan.
b.
h1(X) = 10X, while h2(X) = 99 for X ≤ 20 and 99 + 10(X – 20) for X > 20. With μ = 1/λ
for the exponential distribution, it’s obvious that E[h1(X)] = 10E[X] = 10μ. On the other
hand,

10
E[h2(X)] = 99 + 10 ( x  20 )e x dx = 99 + e  20 = 99 + 10μe–20/μ.
20

When μ = 10, E[h1(X)] = 100¢ = $1.00 while E[h2(X)] = 99 + 100e–2 ≈ $1.13.
When μ = 15, E[h1(X)] = 150¢ = $1.50 while E[h2(X)] = 99 + 150e–4/3 ≈ $1.39.
As predicted, the first plan is better when expected call length is lower, and the second
plan is better when expected call length is somewhat higher.

64.
a.
(6) = 5! = 120
b.
5 3 1 3 1 1 3
               1.329
2 2 2 2 2 2 4
c.
F(4;5) = .371 from row 4, column 5 of Table A.4
d.
F(5;4) = .735
e.
F(0;4) = P(X  0; = 4) = 0
a.
P(X  5) = F(5;7) = .238
b.
P(X < 5) = P(X  5) = .238
c.
P(X > 8) = 1 – P(X  8) = 1 – F(8;7) = .313
d.
P( 3  X  8 ) = F(8;7) – F(3;7) = .653
e.
P( 3 < X < 8 ) =.653
f.
P(X < 4 or X > 6) = 1 – P(4  X  6 ) = 1 – [F(6;7) – F(4;7)] = .713
65.
153
Chapter 4: Continuous Random Variables and Probability Distributions
66.
67.
a.
 = 20, 2 = 80   = 20, 2 = 80   =
b.
 24 
P(X  24) = F  ;5  = F(6;5) = .715
 4 
c.
P(20  X  40) = F(10;5) – F(5;5) = .411
80
20
,=5
 = 24, 2 = 144   = 24, 2 = 144   = 6,  = 4
a.
P(12  X  24) = F(4;4) – F(2;4) = .424
b.
P(X  24) = F(4;4) = .567, so while the mean is 24, the median is less than 24, since
P(X  ~ ) = .5. This is a result of the positive skew of the gamma distribution.
c.
We want a value for which F(x;4) = .99. In table A.4, we see F(10;4) = .990. So with  =
6, the 99th percentile = 6(10) = 60.
d.
We want a value for which F(t;4)=.995. In the table, F(11;4)=.995, so t = 6(11)=66. At
66 weeks, only .5% of all transistors would still be operating.
a.
E(X) =  = n
b.
 30 
P(X  30) = F  ;10  = F(15;10) = .930
 2

c.
P(X  t) = P(at least n events in time t) = P( Y  n) when Y  Poisson with parameter t .
68.
1


n

; for  = .5, n = 10, E(X) = 20
e t t k
.
k!
k 0
n 1
Thus P(X  t) = 1 – P( Y < n) = 1 – P( Y  n – 1)  1 

69.
a.
{X  t} = A1  A2  A3  A4  A5
b.
P(X  t) =P( A1 )  P( A2 )  P( A3 )  P( A4 )  P( A5 ) =
.05t
t) = 1 – e
, fx(t) = .05e
but with parameter  = .05.
c.
.05t
e 
 t 5
 e .05t , so Fx(t) = P(X 
for t  0. Thus X also has an exponential distribution ,
By the same reasoning, P(X  t) = 1 – e
parameter n.
154
 n t
, so X has an exponential distribution with
Chapter 4: Continuous Random Variables and Probability Distributions
70.
 x p

71.

yX y
a.
{X2  y} =
b.
FY(y) = P(X2  y) =
1 y / 2
e
2
e
 x p
 1 p ,
 ln( 1  p)
~  .693 .
 x p  ln( 1  p)  x p 
. For p = .5, x.5 = 
With xp = (100p)th percentile, p = F(xp) = 1 – e


1 z2 / 2
e
dz , so fY(y) =
 y
2
1 1 / 2
1  y / 2  1 1 / 2
1  y / 2 1 / 2
. We recognize this
y

e
y

e
y
2
2
2
2

y
as the chi–squared p.d.f. with  = 1.
155
Chapter 4: Continuous Random Variables and Probability Distributions
Section 4.5
72.
1 1
 1
  3      2.66 ,
2  2
 2

1 
2
Var(X) = 91  1   1    1.926
 2 

a.
E(X) = 31 
b.
P(X  6) = 1  e
c.
P(1.5  X  6) = 1  e
a.
P(X  250) = F(250;2.5, 200) = 1  e
P(X < 250) = P(X  250)  .8257
 ( 6 /  )
 1  e (6 / 3)  1  e 4  .982
 ( 6 / 3) 2
2


 1  e (1.5 / 3)  e .25  e 4  .760
2
73.
P(X > 300) = 1 – F(300; 2.5, 200) =
 ( 250 / 200) 2.5
e (1.5)
 1  e 1.75  .8257
 .0636
2.5
b.
P(100  X  250) = F(250;2.5, 200) – F(100;2.5, 200)  .8257 – .162 = .6637
c.
~ is requested. The equation F( ~ ) = .5 reduces to
The median 
.5 =
2.5
2.5
~
 ~ 
e  (  / 200) , i.e., ln(.5)   
 , so ~ = (.6931).4(200) = 172.727.
 200 
74.
a.
For x > 3.5, F(x) = P( X  x) = P(X – 3.5  x – 3.5) = 1 – e
b.
E(X – 3.5) = 1.5
c.
P(X > 5) = 1 – P(X  5) = 1 
d.
P(5  X  8) =



( x  3. 5 ) 2
1.5
 3
 = 1.329 so E(X) = 4.829
 2
2
2  3 
Var(X) = Var(X – 3.5) = 1.5 2      .483
 2 

1  e 9
1  e   e  .368
 1  e   e  e  .3679  .0001  .3678
1
1
1
1
156
9
Chapter 4: Continuous Random Variables and Probability Distributions

75.
 1
Using the substitution y =  x  and dy = x
dx ,  

 



0
76.
a.

0
x
  1  x  
x e
dx =

1
1

y  e  y dy    1   by definition of the gamma function
 
2
.5  F ~   1  e  / 3 
e   / 9  .5  ~ 2  9 ln(. 5)  6.2383  ~  2.50
b.
1  e  3.5  /1.5  .5 
c.
P = F(xp) = 1 – e
~
d.

2

 
xp
~  3.52 = –2.25 ln(.5) = 1.5596  ~
= 4.75


 (xp/) = –ln(1 – p)  xp = [ –ln(1–p)]1/
The desired value of t is the 90th percentile (since 90% will not be refused and 10% will
be). From c, the 90th percentile of the distribution of X – 3.5 is 1.5[ –ln(.1)]1/2 = 2.27661,
so t = 3.5 + 2.2761 = 5.7761
77.
a.
E( X ) 
  2 

2 

e

 e 4.82  123 .97
V ( X )  e 2( 4.5).8
2

 e
.8

 1  15,367 .34 .8964   13,776 .53   117 .373
b.
ln(100 )  4.5 

P( X  100 )  P Z 
  0.13   .5517
.8


c.
ln( 200 )  4.5 

P( X  200 )  P Z 
  1  1.00   1  .8413  .1587  P( X  200 )
.8


a.
P(X ≤ 0.5) = F(0.5) = 1 – exp[– (0.5/β)α] = .3099
78.
b.
1 
2 


Using a computer, 1 
 = Γ(1.55) = 0.889 and 1 
 = Γ(2.10) = 1.047.
1
.
817
1
.
817




From these we find μ = (.863)(0.889) = 0.785 and σ 2 = (.863)2{1.047 – [0.889]2} = .1911,
or σ = 0.437. Hence, P(X > μ + σ) = P(X > 1.222) = 1 – F(1.222) = exp[– (1.222/β)α] =
.1524
c.
F(x) = ½  ½ = 1 – exp[– (x/β)α]  exp[– (x/β)α] = ½  (x/β)α = ln 2  x = β(ln 2)1/α =
.863(ln 2)1/1.817 = .7054
d.
Using the same math as part c, η(p) = β(–ln(1 – p))1/α = .863(–ln(1 – p))1/1.817
157
Chapter 4: Continuous Random Variables and Probability Distributions
79.
23.5  1.2 2  1.2 2
e
 1  14907 .168 ;
= 68.0335; V(X) = e


x = 122.0949
a.
E(X) = e 3.51.2
b.
P(50  X  250) = P z 
2
/2


ln(250 )  3.5 
ln(50)  3.5 

  P z 

1.2
1.2



P(Z  1.68) – P(Z  .34) = .9535 – .6331 = .3204.
ln(68 .0335 )  3.5 

 = P(Z  .60) = .7257. The lognormal
1.2


distribution is not a symmetric distribution.
c.
P(X  68.0335) = P z 
a.
 ln( ~ )   
~


.5 = F( ) = 
 , (where ~



80.
refers to the lognormal distribution and  and
 to the normal distribution). Since the median of the standard normal distribution is 0,
ln( ~ )  

 0 , so ln( ~ ) =   ~ = e  . For the power distribution,
~ = e 3.5  33.12
b.
 ln( X )  

1 –  = (z) = P(Z  z) = 
 z   P(ln( X )    z )



= P( X
 e  z ) , so the 100(1 – )th percentile is e  z . For the power distribution,
the 95th percentile is
81.
e 3.5(1.645)(1.2)  e 5.474  238.41
e 5(.01) / 2  e 5.005  149.157 ; Var(X) = e10(.01)  e.01  1  223.594
a.
E(X) =
b.
P(X > 125) = 1 – P(X  125) =  1  P z 
c.
P(110  X  125)   1.72   
d.
~ = e 5  148 .41
e.
P(any particular one has X > 125) = .9573  expected # = 10(.9573) = 9.573
f.
We wish the 5th percentile, which is e 5( 1.645)(.1)  125 .90


ln(125 )  5 
  1   1.72   .9573
.1

 ln(110 )  5 
  .0427  .0013  .0414
.1


158
Chapter 4: Continuous Random Variables and Probability Distributions
82.
83.
e1.99
2
/2
 10.024 ; Var(X) = e 3.8(.81)  e.81  1  125.395 , x = 11.20
a.
E(X) =
b.
P(X  10) = P(ln(X)  2.3026) = P(Z  .45) = .6736
P(5  X  10)
= P(1.6094  ln(X) 2.3026)
= P(–.32  Z  .45) = .6736 – .3745 = .2991
The point of symmetry must be
1
2
, so we require that
f  12     f  12    , i.e.,
 12    1  12    1  12    1  12    1 , which in turn implies that   .
84.
a.
E(X) =
b.
f(x) =
10
5
5
 .0255
  .714 , V(X) =
5  2 7
(49)(8)
7 
 x 4  1  x   30 x 4  x 5 for 0  X  1,
52

so P(X  .2) =
.2

0
c.

.4
4
.2
d.
E(1 – X) = 1 – E(X) = 1 –
a.
E(X) =
b.
E[(1 – X)m] =
85.

30 x 4  x 5 dx  .0016
 30x
P(.2  X  .4) =


 x 5 dx  .03936
5 2
  .286
7 7
     1
    1 
 1
 1


x
1

x
dx

x 1  x  dx
0   

0
  
   
      1 
 

=



       1             
1
x
     1
 1
x 1  x  dx
  
1
   
     m   
m   1

x  1 1  x 
dx 

0
  
    m  
 1  x 
1
0
m

For m = 1, E(1 – X) =


.
159
Chapter 4: Continuous Random Variables and Probability Distributions
86.
a.
100
1

 Y  100
Y  1
; Var(Y) =
 Var  

 
7
 20  2800 28
 20  2   
E(Y) = 10  E 

   3,   3 , after some algebra.
        1
2
b.
 12

 8

;3,3  F  ;3,3 = F(.6;3,3) – F(.4; 3,3).
 20

 20

P(8  Y  12) = F 
The standard density function here is 30y2(1 – y)2,
so P(8  Y  12) =
 30 y 1  y 
.6
.4
c.
2
2
dy  .365 .
We expect it to snap at 10, so P( Y < 8 or Y > 12) = 1 – P(8  X  12)
= 1 – .365 = .665.
Section 4.6
87.
The given probability plot is quite linear, and thus it is quite plausible that the tension
distribution is normal.
88.
The z percentiles and observations are below, along with the probability plot. The plot is quite
straight except for the point corresponding to the largest observation. This observation is
clearly much larger than what would be expected in a normal random sample. Because of this
outlier, it would be inadvisable to analyze the data using any inferential method that depended
on assuming a normal population distribution.
Observation
152.7
172.0
172.5
173.3
193.0
204.7
216.5
234.9
262.6
422.6
400
lifetime
Percentile
–1.645
–1.040
–0.670
–0.390
–0.130
0.130
0.390
0.670
1.040
1.645
300
200
-2
-1
0
z %ile
160
1
2
Chapter 4: Continuous Random Variables and Probability Distributions
89.
The z percentile values are as follows: –1.86, –1.32, –1.01, –0.78, –0.58, –0.40, –0.24,–0.08,
0.08, 0.24, 0.40, 0.58, 0.78, 1.01, 1.30, and 1.86. The accompanying probability plot is
reasonably straight, and thus it would be reasonable to use estimating methods that assume a
normal population distribution.
thickness
1.8
1.3
0.8
-2
-1
0
1
2
z %ile
The Weibull plot uses ln(observations) and the extreme value percentiles of the pi values
given; i.e., η(p) = ln[–ln(1–p)]. The accompanying probability plot appears sufficiently
straight to lead us to agree with the argument that the distribution of fracture toughness in
concrete specimens could well be modeled by a Weibull distribution.
1.1
1.0
0.9
obsv
90.
0.8
0.7
0.6
0.5
0.4
-4
-3
-2
-1
Extreme value percentile
161
0
1
Chapter 4: Continuous Random Variables and Probability Distributions
91.
The (z percentile, observation) pairs are (–1.66, .736), (–1.32, .863), (–1.01, .865), (–.78,
.913), (–.58, .915), (–.40, .937), (–.24, .983), (–.08, 1.007), (.08, 1.011), (.24, 1.064), (.40,
1.109), (.58, 1.132), (.78, 1.140), (1.01, 1.153), (1.32, 1.253), (1.86, 1.394). The
accompanying probability plot is straight, suggesting that an assumption of population
normality is plausible.
1.4
1.3
obsvn
1.2
1.1
1.0
0.9
0.8
0.7
-2
-1
0
1
2
z %ile
92.
The 10 largest z percentiles are 1.96, 1.44, 1.15, .93, .76, .60, .45, .32, .19 and .06; the
remaining 10 are the negatives of these values. The accompanying normal probability
plot is reasonably straight. An assumption of population distribution normality is
plausible.
500
400
load life
a.
300
200
100
0
-2
-1
0
z %ile
162
1
2
Chapter 4: Continuous Random Variables and Probability Distributions
b.
For a Weibull probability plot, the natural logs of the observations are plotted against
extreme value percentiles; these percentiles are –3.68, –2.55, –2.01, –1.65, –1.37, –1.13,
–.93, –.76, –.59, –.44, –.30, –.16, –.02, .12, .26, .40, .56, .73, .95, and 1.31. The
accompanying probability plot is roughly as straight as the one for checking normality (a
plot of ln(x) versus the z percentiles, appropriate for checking the plausibility of a
lognormal distribution, is also reasonably straight – any of 3 different families of
population distributions seems plausible.)
ln(loadlife)
6
5
4
-4
-3
-2
-1
0
1
W %ile
93.
To check for plausibility of a lognormal population distribution for the rainfall data of
Exercise 81 in Chapter 1, take the natural logs and construct a normal probability plot. This
plot and a normal probability plot for the original data appear below. Clearly the log
transformation gives quite a straight plot, so lognormality is plausible. The curvature in the
plot for the original data implies a positively skewed population distribution – like the
lognormal distribution.
8
3000
7
6
rainfall
ln(rainfall)
2000
5
4
3
1000
2
1
0
-2
-2
-1
0
1
-1
0
z %ile
2
z %ile
163
1
2
Chapter 4: Continuous Random Variables and Probability Distributions
94.
a.
The plot of the original (untransformed) data appears somewhat curved.
5
precip
4
3
2
1
0
-2
-1
0
1
2
z %iles
b.
The square root transformation results in a very straight plot. It is reasonable that this
distribution is normally distributed.
sqrt
2.0
1.5
1.0
0.5
-2
-1
0
1
2
z %iles
The cube root transformation also results in a very straight plot. It is very reasonable that
the distribution is normally distributed.
1.6
cubert
c.
1.1
0.6
-2
-1
0
z %iles
164
1
2
Chapter 4: Continuous Random Variables and Probability Distributions
95.
The pattern in the plot (below, generated by Minitab) is reasonably linear. By visual
inspection alone, it is plausible that strength is normally distributed.
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
125
135
145
strength
Average: 134.902
StDev: 4.54186
N: 153
We use the data (table below) to create the desired plot. This half–normal plot reveals some
extreme values, without which the distribution may appear to be normal.
ordered absolute
values (w's)
0.89
1.15
1.27
1.44
2.34
3.78
3.96
12.38
30.84
43.4
probabilities
0.525
0.575
0.625
0.675
0.725
0.775
0.825
0.875
0.925
0.975
z values
0.063
0.19
0.32
0.454
0.6
0.755
0.935
1.15
1.44
1.96
2
z values
96.
Anderson-Darling Normality Test
A-Squared: 1.065
P-Value: 0.008
1
0
0
5
10
15
20
25
wi
165
30
35
40
45
Chapter 4: Continuous Random Variables and Probability Distributions
97.
The (100p)th percentile (p) for the exponential distribution with  = 1 satisfies F((p)) = 1 –
5 1.5
exp[–(p)] = p, i.e., (p) = –ln(1 – p). With n = 16, we need (p) for p = 16
, 16 ,..., 1516.5 .
These are .032, .398, .170, .247, .330, .421, .521, .633, .758, .901, 1.068, 1.269, 1.520, 1.856,
2.367, 3.466. this plot exhibits substantial curvature, casting doubt on the assumption of an
exponential population distribution. Because  is a scale parameter (as is  for the normal
family),  = 1 can be used to assess the plausibility of the entire exponential family.
600
500
failtime
400
300
200
100
0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
percentile
Supplementary Exercises
98.
10
 .4
25
a.
P(10  X  20) =
b.
P(X  10) = P(10  X  25) =
c.
For 0  X  25, F(x) =

x
0
d.
E(X) =
15
 .6
25
1
x
. F(x)=0 for x < 0 and = 1 for x > 25.
dy 
25
25
 A  B   0  25  12.5 ; Var(X) = B  A2
2
2
12
166

625
 52.083 , so σx = 7.22
12
Chapter 4: Continuous Random Variables and Probability Distributions
99.
a.
1
For 0  y  25, F(y) =
24

y
0
2
u  u

12

y
 1  u 2 u 3 


 . Thus
 24  2  36 


 0

0
y0
1 
3 

2
F(y) =   y  y  0  y  12

18 
 48 
y  12

1
b.
P(Y  4) = F(4) = .259, P(Y > 6) = 1 – F(6) = .5
P(4  X  6) = F(6) – F(4) = .5 – .259 = .241
c.
E(Y) =
1
24
12

0
1
24
E(Y2) =
12
y
1  y3 y4 

y 2 1  dy 


 6
24  3
48 
 12 
0
y

y 3 1  dy  43 .2 , so V(Y) = 43.2 – 36 = 7.2
 12 
12

0
d.
P(Y < 4 or Y > 8) = 1 – P(4  X  8) = .518
e.
the shorter segment has length min(Y, 12 – Y) so
12
6
 min( y,12  y)  f ( y)dy   min( y,12  y)  f ( y)dy
90
  min( y,12  y)  f ( y)dy   y  f ( y)dy   (12  y)  f ( y)dy =
 3.75
24
E[min(Y, 12 – Y)] =
0
0
12
6
12
6
0
6
100.
a.
Clearly f(x)  0. The c.d.f. is , for x > 0,
F ( x) 

x

f ( y)dy 

x
0
x
1
32 
16
dy   
 1
3
2
2  y  4 
 y  4
x  42
32
0
( F(x) = 0 for x  0.)
Since F() =



f ( y)dy  1, f(x) is a legitimate pdf.
b.
See above
c.
P(2  X  5) = F(5) – F(2) = 1 
d.
16  16 
 1    .247
81  36 



32
32
E ( x) 
x  f ( x)dx 
x
dx  ( x  4  4) 
dx
3


0
x  4
x  43




0
e.

32
x  4 
2
dx  4


0
E(salvage value) = 

32
x  43


0
dx  8  4  4
100
32

dx  3200
x  4 x  43
167


0
1
 x  4 4
dx 
3200
 16 .67
(3)( 64 )
Chapter 4: Continuous Random Variables and Probability Distributions
101.
a.
 x2
7 3
By differentiation, f(x) =   x
4 4
 0
0  x 1
7
1 x 
3
otherwise
1.0
0.8
f(x)
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
2.5
x
17
 7 3  .5 11
  .917
  2    2  
23
3
12
 4 4 
3
102.
b.
P(.5  X  2) = F(2) – F(.5) = 1 
c.
E(X) =
7
131
3
7 3 
2
x

x
dx

0
1 x   4  4 x dx  108  1.213
1
 = 40 V;  = 1.5 V
a.
 42  40 
 39  40 
  

 1.5 
 1.5 
P(39 < X < 42) = 
= (1.33) – (–.67) = .9082 – .2514 = .6568
b.
We desire the 85th percentile: 40 + (1.04)(1.5) = 41.56
c.
P(X > 42) = 1 – P(X  42) = 1  
 42  40 
 = 1 – (1.33) = .0918
 1 .5 
Let D represent the number of diodes out of 4 with voltage exceeding 42.
 4
 0
P(D  1 ) = 1 – P(D = 0) = 1   .0918 .9082 =1 – .6803 = .3197
0
168
4
Chapter 4: Continuous Random Variables and Probability Distributions
103.
 = 137.2 oz.;  = 1.6 oz
 135  137.2 
 = 1 – (–1.38) = 1 – .0838 = .9162
1. 6


a.
P(X > 135) = 1  
b.
With Y = the number among ten that contain more than 135 oz, Y ~ Bin(10, .9162). So,
P(Y  8) = b(8; 10, .9162) + b(9; 10, .9162) + b(10; 10, .9162) =.9549
c.
 = 137.2;
a.
Let S = defective. Then p = P(S) = .05; n = 250   = np = 12.5,  = 3.446. The
random variable X = the number of defectives in the batch of 250. X ~ Binomial. Since
np = 12.5  10, and nq = 237.5  10, we can use the normal approximation.
135  137.2

 1.65    1.33
104.
 24.5  12.5 
  1  3.48  1  .9997  .0003
 3.446 
P(Xbin  25)  1  
b.
P(Xbin = 10)  P(Xnorm  10.5) – P(Xnorm  9.5)
=  .58   .87  .2810  .1922  .0888
a.
P(X > 100) = 1  
b.
P(50 < X < 80) = 
105.
 100  96 
  1  .29  1  .6141  .3859
 14 
 80  96 
 50  96 
  

 14 
 14 
= (–1.5) – (–3.29) = .1271 – .0005 = .1266.
c.
a = 5th percentile = 96 + (–1.645)(14) = 72.97.
b = 95th percentile = 96 + (1.645)(14) = 119.03. The interval (72.97, 119.03) contains the
central 90% of all grain sizes.
169
Chapter 4: Continuous Random Variables and Probability Distributions
106.
a.
F(X) = 0 for x < 1 and = 1 for x > 3. For 1  x  3, F ( x) 
1
  0dy  

x
1
P(X  2.5) = F(2.5) = 1.5(1 – .4) = .9; P(1.5  x  2.5) =
F(2.5) – F(1.5) = .4
c.
E(X) = 
d.
E(X2) = 
3
1

x
3
1

f ( y)dy
3 1
3 31
3
 2 dx   dx  1.5 ln( x)1  1.648
2 x
2 1 x
x2 
 =.553
3 1
3 3
 2 dx   dx  3 , so V(X) = E(X2) – [E(X)]2 = .284,
2 x
2 1
1  x  1.5
 0

e. h(x) =  x  1.5
 1

so E[h(X)] = 
x
3 1
 1
 2 dy  1.51  
2 y
 x
b.


1 .5  x  2 .5
2 .5  x  3
3 1
 x  1.5  2  x
2.5
3
2
1.5
dx   1 
2.5
3 1

dx  .267
2 x2
107.
a.
0.5
0.4
f(x)
0.3
0.2
0.1
0.0
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
x
b.
F(x) = 0 for x < –1 or = 1 for x > 2. For –1  x  2,
x 1
1
x 3  11
F ( x) 
4  y 2 dy   4 x 


1 9
9
3  27
 
c.

The median is 0 iff F(0) = .5. Since F(0) =
11
27
, this is not the case. Because
median must be greater than 0.
d.
Y is a binomial r.v. with n = 10 and p = P(X > 1) = 1 – F(1) =
170
5
27
11
27
< .5, the
Chapter 4: Continuous Random Variables and Probability Distributions
108.
1
= 1.075,  
1
a.
E(X) =
b.
P(3.0 < X) = 1 – P(X  3.0) = 1 – F(3.0) = e–.93(3.0) = .0614
P(1.0  X  3.0) = F(3.0) – F(1.0) = .333
c.
The 90th percentile is requested; denoting it by c, we have


= 1.075
.9 = F(c) = 1 – e–(.93)c, whence c =
ln(. 1)
 2.476
(.93)
109.
a.

  (150  150) 
  exp[  exp( 0)]  exp( 1)  .368 , where
90



exp(u) = eu. P(X  300) = exp[  exp( 1.6667)]  .828 ,
P(X  150) = exp  exp 
and P(150  X  300) = .828 – .368 = .460.
b.
The desired value c is the 90th percentile, so c satisfies

  (c  150)  . Taking the natural log of each side twice in succession

90



 (c  150)
yields ln[–ln(.9)] =
, so c = 90(2.250367) + 150 = 352.53.
90
.9 = exp  exp 

   x    
  x    
  exp 

 exp  exp 









1
c.
f(x) = F(X) =
d.
We wish the value of x for which f(x) is a maximum; this is the same as the value of x for
which ln[f(x)] is a maximum. The equation of
d [ln  f ( x) ]
 0 gives
dx
  x    
 x   
  1 , so
 0 , which implies that x = . Thus the mode is .
exp 




e.
E(X) = .5772 +  = 201.95, whereas the mode is 150 and the median is
–(90)ln[–ln(.5)] + 150 = 182.99. The distribution is positively skewed.
a.
E(cX) = cE(X) =
b.
E[c(1 – .5eax)] =
110.
c

c[.5  a]

 c1  .5e   e dx    a

ax
 x
0
171
Chapter 4: Continuous Random Variables and Probability Distributions
111.
a.
From a graph of f(x; , ) or by differentiation, x* = .
b.
No; the density function has constant height for A  X  B.
c.
F(x;) is largest for x = 0 (the derivative at 0 does not exist since f is not continuous
there) so x* = 0.
d.
 
ln[ f  x;  ,  ]   ln    ln      1 ln( x) 
x
;
d
 1 1
ln[ f  x;  ,  ] 
  x  x*  (  1) 
dx
x



 12     2.
2 
e.
From d x*  
a.

112.


0


0
f ( x)dx   .1e .2 x dx   .1e .2 x dx  .5  .5  1
0.10
0.08
f(x)
0.06
0.04
0.02
0.00
-2
b.
c.
x
-1
0
x

For x  0, F(x) =
x
1
1
  .1e .2 y dy  1  e .2 x .
0
2
2
P(X < 0) = F(0) =
.1e .2 y dy 
2
1 .2 x
e .
2
For x < 0, F(x) =

1
1
 .5 , P(X < 2) = F(2) = 1 – .5e–.4 = .665,
2
P(–1  X  2) – F(2) – F(–1) = .256, 1 – (–2  X  2) = .670
172
Chapter 4: Continuous Random Variables and Probability Distributions
113.
a.
Clearly f(x; 1, 2, p)  0 for all x, and
=
 p e

1 x
1
0





f ( x; 1 , 2 , p)dx

 1  p 2 e 2 x dx  p  1e 1x dx  1  p  2 e 2 x dx
0
0
= p + (1 – p) = 1
b.
For x > 0, F(x; 1, 2, p) =

x
f ( y; 1 , 2 , p)dy  p(1  e 1x )  (1  p)(1  e 2 x ).
0
c.

E(X) =

0


x  p1e 1x  (1  p)2 e 2 x dx


0
0
 p  x1e 1x dx  (1  p)  x 2 e 2 x dx 
2
2p

21  p 
d.
E(X ) =
e.
For an exponential r.v., CV =
12
22


 2 p  2 1 p  
 2

2
2
 1


1
2
CV =  




p
1

p
 


  1

2 
= [2r – 1]1/2 where r =
1 provided
f.

n

,
1
, so Var(X) =
1

1

2p
12
p
1

1  p 
2
21  p 

22
 p 1  p  
 
 2 
 1
 1 . For X hyperexponential,
2


 2 p22  (1  p)12

= 
 1
2
  p 2  (1  p)1 

 p
2
2
 (1  p)12
1
2
 . But straightforward algebra shows that r >
 p2  (1  p)1 2
1  2 , so that CV > 1.
2

n

2
2
, so  
n

and CV =
1
n
173
 1 if n > 1.
Chapter 4: Continuous Random Variables and Probability Distributions
114.
a.
1 

5
b.
k
51
dx

k

 k    15 1 where we must have  > 1.

 1
x
For x  5, F(x) =

x
5
c.
E(X) =


5
x
k
1 
 1
 5
dy  51  1   1   1   

y
x 
5
 x
 1
.
 1
   15
k
5  1
, provided  > 2.
dx

dx 



1

5
  2
x
x
 X

X

 5 
d. P ln    y   P  e y   P X  5e y  F 5e y  1  
y 
5

 5e 
 5

1  e   1 y , the cdf of an exponential r.v. with parameter  – 1.

  
 1
115.
a.
b.
c.
I 
A lognormal distribution, since ln  o  is a normal r.v.
I 
 i
 I
I

PI o  2 I i   P o  2   P ln  o
 Ii

  Ii
 ln 2  1 
1  
  1   6.14  1
 .05 

 I

  ln 2   1  P ln  o




  Ii


  ln 2 



I 
I 
E o   e1.0025 / 2  2.72, Var o   e 2.0025  e .0025  1  .0185
 Ii 
 Ii 
174
Chapter 4: Continuous Random Variables and Probability Distributions
116.
a.
The Weibull pdf exists for x ≥ 0, but only a section is graphed below.
0.020
f(x)
0.015
0.010
0.005
0.000
150
117.
160
170
x

180
190
1759
b.
P(X > 175) = 1 – F(175; 9, 180) = e 180  .4602
P(150  X  175) = F(175; 9, 180) – F(150; 9, 180)
= .5398 – .1762 = .3636
c.
P(at least one) = 1 – P(none) = 1 – (1 – .3636)2 = .5950
d.
We want the 10th percentile: .10 = F( x; 9, 180) = 1  e 180 . A small bit of algebra
leads us to x = 140.178. Thus 10% of all tensile strengths will be less than 140.178 MPa.



F(y) = P(Y  y) = P(Z +   y) = P Z 
y    




 
 y 


x
9
1
2
e
 12 z 2
dz . Now
differentiate with respect to y to obtain a normal pdf with parameters  and .
118.
a.


FY(y) = P(Y  y) = P(60X  y) = P X 
 y

y
; . Thus fY(y)
  F 
60 
 60  
y
 y

1
y
 1 60 
e
= f
, which shows that Y has a gamma distribution

 60  ;   60  




60





with parameters  and 60.
b.
With c replacing 60 in a, the same argument shows that cX has a gamma distribution
with parameters  and c.
175
Chapter 4: Continuous Random Variables and Probability Distributions
119.
a.
Y = –ln(X)  x = e–y = k(y), so k(y) = –e–y. Thus since f(x) = 1,
g(y) = 1  | –e–y | = e–y for 0 < y < , so y has an exponential distribution with parameter 
= 1.
b.
y = Z +   y = h(z) = Z +   z = k(y) =
y   

and k(y) =
1
, from which the

result follows easily.
c.
y = h(x) = cx  x = k(y) =
a.
If we let
y
1
and k(y) = , from which the result follows easily.
c
c
120.
  2 and   2 , then we can manipulate f(v) as follows:

2
2

  
f ( )  2 e  / 2 
e  / 2 
 2 1 e   / 2      1 e  ,
2
2

2

2
2
2
2
2
2

2

which is in the Weibull family of distributions.
b.
F    
25
0


400




e 800 d ; cdf: F  ;2, 2  1  e
 


2


v2
 1  e 800 , so
625
F 25;2, 2  1  e 800  1  .458  .542
121.
a.
Assuming independence, P(all 3 births occur on March 11) =
b.
3651 3 (365)  .0000073
c.
3651 3  .00000002
Let X = deviation from due date. XN(0, 19.88). Then the baby due on March 15 was 4
days early. P(x = –4) ≈ P(–4.5 < x < –3.5)
  3.5 
  4.5 
 
  
   .18   .237   .4286  .4090  .0196 .
 19.88 
 19.88 
Similarly, the baby due on April 1 was 21 days early, and P(x = –21)
  20.5 
  21.5 
  
   1.03   1.08  .1515  .1401  .0114 .
 19.88 
 19.88 
≈ 
The baby due on April 4 was 24 days early, and P(x = –24) ≈ .0097
Again, assuming independence, P( all 3 births occurred on March 11) =
.0196.0114.0097  .00002145
d.
To calculate the probability of the three births happening on any day, we could make
similar calculations as in part c for each possible day, and then add the probabilities.
176
Chapter 4: Continuous Random Variables and Probability Distributions
122.
a.
F(x) = e
 x
and F(x) = 1  e
 x
, so r(x) =
e  x
e x
  , a constant; this is consistent
with the memoryless property of the exponential distribution.
b.
 
 

r(x) = 
  1
 x ; for  > 1 this is increasing, while for  < 1 it is a decreasing function.


c.
ln(1 – F(x)) = 

x2 

  x 
 2 


x
x2 


,
 1  dx    x 

F
(
x
)

1

e


2






x2 

x    x  2  
f(x) =  1  e 
 
0x
123.
a.


 1

ln 1  U   x   Pln( 1  U )  x   P 1  U  e x
 

 x
 x
 P U 1 e
 1  e since FU(u) = u (U is uniform on [0, 1]). Thus X has an
FX(x) = P 


exponential distribution with parameter .
b.
By taking successive random numbers u1, u2, u3, …and computing x i  
1
ln 1  u i  ,
10
… we obtain a sequence of values generated from an exponential distribution with
parameter  = 10.
124.
a.
b.
E(g(X))  E[g() + g()(X – )] = E(g()) + g()E(X – ), but E(X) –  = 0 and
E(g()) = g() ( since g() is constant), giving E(g(X))  g().
V(g(X))  V[g() + g()(X – )] = V[g()(X – )] = (g())2V(X – ) = (g())2V(X).
v
v
v
v
g ( I )  , g ( I )  2 , so E  g ( I )    R 

I
 I 20
I
2
v
v
v
V  g ( I )    2   V I ,  g ( I )  2   I 
800
20
 I 
125.
g() + g()(X – )  g(X) implies that E[g() + g()(X – )] = E(g()) = g()  E(g(X)), i.e.
that g(E(X))  E(g(X)).
177
Chapter 4: Continuous Random Variables and Probability Distributions
126.
 2X 2
For y > 0, F ( y )  P(Y  y )  P

2
 
take the cdf of X (Weibull), replace x by

 y 


2y
. Now
  P X 
 y   P X 2 


2 
2




 y
, and then differentiate with respect to y to
2
obtain the desired result fY(y).
127.
a.
(850  150 ) 2 (8)( 2)
8
= 710; V(X) =
 σX ≈ 84.423
82
(8  2) 2 (8  2  1)
P(|X – 710| ≤ 84.423) = P(625.577 ≤ X ≤ 794.423) =
E(X) = 150 + (850 – 150)
7
1
(10 )  x  150   850  x 
1

 
 dx = .684. The computation of this integral
625.577 700 (8)(2)  700   700 
requires a calculator or computer.

794.423
7
1
(10 )  x  150   850  x 
1

 
 dx = .376. Again, the computation
750 700 (8)( 2)  700   700 
of the requested integral requires a calculator or computer.

850
b.
P(X > 750) =
a.
For w < 0, F(w) = 0. F(0) = P(V ≤ vd) = 1 – exp(–λvd), since V is exponential. For w > 0,
F(w) = P(W ≤ w) = P(k(V–vd) ≤ w) = P(V ≤ vd + w/k) = 1 – exp(–λ[vd+w/k]). This can be
written more compactly as
128.
F(w) = 1 – exp(–λ[vd+w/k]) for w ≥ 0 and F(w) = 0 otherwise
b.

exp(–λ[vd+w/k]). Since the only other possible value of W is
k
zero, which would not contribute to the expected value, we can compute E(W) as the
appropriate integral:
For w > 0, f(w) = F′(w) =
E(W) =


0
w

k
exp( [v d  w / k ]) dw = exp( v d )
178


0
w

k
exp( w / k )dw =
k

e  v d