Applied Statistics and Probability for Engineers, 7th edition
CHAPTER 4
Section 4.1

4.1.1

a) P (1  X ) =  e − x dx = (−e − x )
1
= e −1 = 0.3679
1
2.5
b) P(1  X  2.5) =
e
−x
dx = (−e − x )
2.5
1
= e −1 − e − 2.5 = 0.2858
1
3
c) P( X = 3) =  e− x dx = 0
3
4
d) P( X  4) =  e− x dx = (−e− x ) = 1 − e− 4 = 0.9817
4
0
0

e) P(3  X ) =  e− x dx = (−e− x )

3
= e−3 = 0.0498
3

f) P( x  X ) =  e − x dx = (−e − x )

x
= e − x = 0.10 .
x
Then, x = −ln(0.10) = 2.3
x
g) P( X  x) =  e − x dx = (−e − x ) = 1 − e − x = 0.10 .
x
0
0
Then, x = −ln(0.9) = 0.1054
2
−1
−1
dx = ( 2 ) = ( ) − ( −1) = 0.75
3
x
x 1
4
2
2
4.1.2
a) P( X  2) =

1

b) P( X  5) =

5

2
−1
−1
dx = ( 2 ) = 0 − ( ) = 0.04
3
x
x 5
25
2
−1
−1
−1
4 x 3 dx = ( x 2 ) 4 = ( 64 ) − ( 16 ) = 0.0469
d) P( X  4 or X  8) = 1 − P(4  X  8) . From part (c), P(4  X  8) = 0.0469. Therefore,
P( X  4 or X  8) = 1 – 0.0469 = 0.9531
8
8
c) P( 4  X  8) =
2
−1
−1
dx = ( 2 ) = ( 2 ) − (−1) = 0.95
3
x 1
x
1 x
Then, x2 = 20, and x = 4.4721
x
x
e) P( X  x) = 
0
4.1.3
a) P( X  0) =
 0.5 cos xdx = (0.5 sin x)

− /2
b) P( X  − / 4) =
− / 4
0
− / 2
= 0 − ( −0.5) = 0.5
 0.5 cos xdx = (0.5 sin x)

− /2
− / 4
− / 2
= −0.3536 − ( −0.5) = 0.1464
Applied Statistics and Probability for Engineers, 7th edition
c) P( − / 4  X   / 4) =
 /4
 0.5 cos xdx = (0.5 sin x)

− /4
 /2
 0.5 cos xdx = (0.5 sin x)

d) P( X  − / 4) =
− /4
 /2
− / 4
 /4
− / 4
= 0.3536 − ( −0.3536) = 0.7072
= 0.5 − ( −0.3536) = 0.8536
x
 0.5 cos xdx = (0.5 sin x)

e) P( X  x ) =
− /2
x
− / 2
= (0.5 sin x ) − ( −0.5) = 0.95
Then, sin x = 0.9, and x = 1.1198 radians
4
x
x2
4 2 − 32
P
(
X

4
)
=
dx
=
=
= 0.4375 , because
a)
3 8 16
16
3
4
4.1.4
5
x
x2
P
(
X

3
.
5
)
=
dx
=
b) ,
 8 16
3.5
=
3.5
2 5
5 2 − 3.5 2
= 0.7969 because
16
f X ( x) = 0 for x > 5.
x
x
52 − 4 2
dx
=
=
= 0.5625
4 8 16
16
4
5
c) P(4  X  5) =
5
f X ( x) = 0 for x < 3.
4.5
x
x2
d) P ( X  4.5) =  dx =
8
16
3
4.5
3
4.5 2 − 3 2
=
= 0.7031
16
5
3.5
x
x
x2
x2
5 2 − 4.5 2 3.5 2 − 32
+
=
+
= 0.5 .
e) P( X  4.5) + P( X  3.5) =  dx +  dx =
8
8
16 4.5 16 3
16
16
4.5
3
5
4.1.5
3.5
a) P(0  X ) = 0.5 , by symmetry.
1
b) P(0.5  X ) = 1.5 x 2 dx = 0.5 x 3
1
0.5
= 0.5 − 0.0625 = 0.4375
0.5
0.5
1.5x dx = 0.5x
c) P(−0.5  X  0.5) =
2
− 0.5
3 0.5
− 0.5
= 0.125
d) P(X < −2) = 0
e) P(X < 0 or X > −0.5) = 1
1
f) P( x  X ) = 1.5 x 2 dx = 0.5 x3
1
x
= 0.5 − 0.5 x3 = 0.05
x
Then, x = 0.9655
50.25
4.1.6
a) P( X  50) =
 2.0dx = 2 x
50.25
50
= 0.5
50
50.25
b) P( X  x) = 0.90 =
 2.0dx = 2 x
x
Then, 2x = 99.6 and x = 49.8.
50.25
x
= 100.5 − 2 x
Applied Statistics and Probability for Engineers, 7th edition
4.1.7
a) P(X < 2.25 or X > 2.75) = P(X < 2.25) + P(X > 2.75) because the two events are
mutually exclusive. Then, P(X < 2.25) = 0 and
2.8
 2dx = 2(0.05) = 0.10 .
P(X > 2.75) =
2.75
b) If the probability density function is centered at 2.55 meters, then f X ( x) = 2 for
2.3 < x < 2.8 and all rods will meet specifications.
4.1.8
a) P( X  90) = 0 because the pdf is not defined in the range (−,90) .
b)
200
P(100  X  200) =
−4
−6
−4
−6 2
 (−5.56 10 + 5.56 10 x)dx = (−5.56 10 x + 2.78 10 x )
200
100
100
= (−5.56 10−4  200 + 2.78 10−6  2002 ) − (−5.56 10−4 100 + 2.78 10−6 1002 )
= 0.278
c)
1000
 (4.44  10
P( X  800) =
−3
− 4.44  10− 6 x)dx = (4.44  10− 3 x − 2.22  10− 6 x 2 )
1000
800
800
= (4.44  10− 3  103 − 2.22  10− 6  106 ) − (4.44  10− 3  800 − 2.22  10− 6  8002 )
= 0.0888  0.09
d) Find a such that P( X  a) = 0.1
1000
P( X  a) =
 (4.44  10
−3
− 4.44  10− 6 x)dx = (4.44  10− 3 x − 2.22  10− 6 x 2 )
1000
a
= 0.1
a
(4.44  10− 3  103 − 2.22  10− 6  106 ) − (4.44  10− 3  a − 2.22  10− 6  a 2 ) = 0.1
(2.22) − (4.44  10− 3  a − 2.22  10− 6  a 2 ) = 0.1
Then, a  787.76
0.5
4.1.9
a) P( X  0.5) =
 0.5 exp(−0.5x)dx = − exp(−0.5x)
0.5
0
= 0.221
0

b) P( X  2) =  0.5 exp(−0.5 x)dx = − exp(−0.5 x)

c) P( X  a) =  0.5 exp(−0.5 x)dx = − exp(−0.5 x)

2

2
a
= 0.368
= exp(−0.5a) = 0.05  a = 5.991
a
40
4.1.10 a) P( X  40) =  (0.0025x − 0.075)dx = (0.00125x 2 − 0.075x)
40
30
30
= (0.00125 402 − 0.075  40) − (0.00125 302 − 0.075  30) = 0.125
Applied Statistics and Probability for Engineers, 7th edition
50
60
b) P(40  X  60) =  (0.0025x − 0.075)dx +  (−0.0025x + 0.175)dx
40
50
50
= (0.00125x − 0.075 x) + (−0.00125x + 0.175 x)
2
2
40
60
50
= 0.375 + 0.375 = 0.75
c) Find a such that P( X  a) = 0.99 (i.e. P( X  a) = 1 − 0.99 = 0.01
a
P( X  a) =  (0.0025x − 0.075)dx = (0.00125x 2 − 0.075x)
a
30
= 0.01
30
= (0.00125  a 2 − 0.075  a) − (0.00125  302 − 0.075  30) = 0.01
Then, a = 32.828
Section 4.2
4.2.1
a) P(X<2.8) = P(X  2.8) because X is a continuous random variable.
Then, P(X<2.8) =F(2.8)=0.2(2.8) = 0.56.
b) P( X  1.5) = 1 − P( X  1.5) = 1 − 0.2(1.5) = 0.7
c) P( X  −2) = FX (−2) = 0
d) P( X  6) = 1 − FX (6) = 0
4.2.2
Now,
f ( x) = e − x
x
for 0 < x and FX ( x) =  e − x dx = − e − x
x
0
0
0, x  0

for 0 < x. Then, FX ( x) = 
−x
1 − e , x  0
4.2.3
Now, f ( x ) = 0.5 cos x for -/2 < x < /2 and
x
FX ( x) =
 0.5 cos udu = (0.5 sin x)

− /2
x
− / 2
= 0.5 sin x + 0.5
0, x  − / 2


Then, FX ( x ) = 0.5 sin x + 0.5, −  / 2  x   / 2

1, x  /2

4.2.4
Now, f ( x ) =
2
for x > 1 and
x3
2
−1
−1
FX ( x) =  3 du = ( 2 ) = ( 2 ) + 1
u 1
x
1 u
x
x
0, x  1


1
1− 2 , x  1

 x
Then, FX ( x) = 
= 1− e −x
Applied Statistics and Probability for Engineers, 7th edition
4.2.5
x
u
u2
x2 − 9
=
Now, f ( x) = x / 8 for 3 < x < 5 and FX ( x) =  du =
8
16 3
16
3
x
0, x  3

 2
x −9
,3  x  5
for 0 < x. Then, F X ( x) = 
16

1, x  5

4.2.6
Now, f ( x) =
e− x /10
for 0 < x and
10
x
FX ( x) = 1/10  e − x /10 dx = −e − x /10
x
0
= 1 − e − x /10
0
for 0 < x.
0, x  0

Then, FX ( x) = 
− x /10
, x0
1 − e
a) P(X<60) = F(60) =1 - e-6 = 1 - 0.002479 = 0.9975
30

b) 1/10 e − x /10 dx = e-1.5 − e-3 =0.173343
15
c) P(X1>40)+ P(X1<40 and X2>40)= e-4+(1- e-4) e-4=0.0363
d) P(15<X<30) = F(30)-F(15) = e-1.5- e-3=0.173343
x
4.2.7
Now, f(x) = 2 for 2.3 < x < 2.8 and F ( x) =
 2dy = 2 x − 4.6
2.3
for 2.3 < x < 2.8. Then,
0,
x  2.3


F ( x) = 2 x − 4.6, 2.3  x  2.8

1,
2.8  x

P( X  2.7) = 1 − P( X  2.7) = 1 − F (2.7) = 1 − 0.8 = 0.2 because X is a continuous random
variable.
4.2.8
f ( x) = 2e−2 x , x  0
4.2.9

0.2, 0  x  4
f ( x) = 
0.04, 4  x  9

4.2.10 For 30  x  50 ,
Applied Statistics and Probability for Engineers, 7th edition
x
P( X  x) =  (0.0025x − 0.075)dx = (0.00125x 2 − 0.075x)
x
30
30
= (0.00125x 2 − 0.075x) − (0.00125  302 − 0.075  30)
= 0.00125x 2 − 0.075x − 1.125
For 50  x  70 ,
x
P( X  x) = F (50) +  (−0.0025x + 0.175)dx
50
= 0.5 + (−0.00125x 2 + 0.175 x)
x
50
= 0.5 + (−0.00125x 2 + 0.175 x) − (−0.00125  502 + 0.175  50)
= 0.5 − 0.0125 x 2 + 0.175 x − 5.625 = −0.0125 x 2 + 0.175 x − 5.125
0, x  30

2
0.00125x − 0.075 x − 1.125,30  x  50
F ( x) = 
2
− 0.00125x + 0.175 x − 5.125,50  x  70
1, x  70

P( X  55) = F (55) = −0.00125  552 + 0.175  55 − 5.125 = 0.719
4.2.11
0, x  0

f ( x) = 0.5 exp(−0.5 x)  F ( x) =  x
 0.5 exp(−0.5 x)dx = 1 − exp(−0.5 x), x  0
0
P(40  X  60) = F (60) − F (40) = 2.06 10 −9
4.2.12 For 100  x  500 ,
x
F ( x) =  (−5.56  10− 4 + 5.56  10− 6 x)dx = (−5.56  10− 4 x + 2.78  10− 6 x 2 )
x
100
100
= (−5.56  10− 4 x + 2.78  10− 6 x 2 ) − (−5.56  10− 4  100 + 2.78  10− 6  1002 )
For 500  x  1000 ,
x
F ( x) = F (500) +  (4.44  10− 3 − 4.44  10− 6 x)dx
500
= 0.445 + (4.44  10− 3 x − 2.22  10− 6 x 2 )
−3
−6
x
500
= 0.445 + (4.44  10 x − 2.22  10 x ) − (4.44  10− 3  500 − 2.22  10− 6  5002 )
2
= 0.445 + (4.44  10− 3 x − 2.22  10− 6 x 2 ) − 1.665
Applied Statistics and Probability for Engineers, 7th edition
0, x  100

−6 2
−4
−2
2.78  10 x − 5.56  10 x + 2.78  10 ,100  x  500
F ( x) = 
−6 2
−3
− 2.22  10 x + 4.44  10 x − 1.22,500  x  1000
1, x  1000

Section 4-3
1
4.3.1
x4
E ( X ) =  1.5 x dx = 1.5
4
−1
1
=0
3
−1
1
1
V ( X ) =  1.5 x 3 ( x − 0) 2 dx = 1.5  x 4 dx
−1
= 1.5
−1
5
x
5
1
−1
= 0.6
4
4
4.3.2
x3
E ( X ) =  0.125x dx = 0.125
= 2.6667
3 0
0
2
4
4
V ( X ) =  0.125x( x − ) dx = 0.125 ( x 3 − 163 x 2 + 649 x)dx
8 2
3
0
0
= 0.125( x4 − 163
4
3
x
3
4
+ 649  12 x 2 ) = 0.88889
0

4.3.3
E ( X ) =  xe − x dx . Use integration by parts to obtain
1

E ( X ) =  xe − x dx = − e − x ( x + 1)

0
= 0 − (−1) = 1
0

V ( X ) =  ( x − 1) 2 e − x dx Use double integration by parts to obtain
0

V ( X ) =  ( x − 1) 2 e − x dx = −( x − 1) 2 e − x − 2( x − 1)e − x − 2e − x )

0
=1
0
50
4.3.4
x3
x2
E ( X ) =  x(0.0025x − 0.075)dx = 0.0025 − 0.075
3
2
30
70
+  x(−0.0025x + 0.175)dx = − 0.0025
50
3
2 70
x
x
+ 0.175
3
2
50
50
30
2
1
== 21 + 28 = 50
3
3
Applied Statistics and Probability for Engineers, 7th edition
50
x4
x3
E ( X ) =  x (0.0025x − 0.075)dx = 0.0025 − 0.075
4
3
30
2
50
2
70
+  x 2 (−0.0025x + 0.175)dx = − 0.0025
50
3 70
4
x
x
+ 0.175
4
3
50
30
2
2
= 950 + 1616 = 2566
3
3
2
1
V ( X ) = E ( X ) − [ E ( X )] = 2566 − 2500 = 166
3
3
2
4.3.5
2
Probability distribution from exercise 4.1.8 used.
Here, a = 5.56E-6, b = -5.56E-4, c = -4.44E-6, d = 4.44E-3
500
500
x3
x2
E ( X ) =  x(ax + b)dx = a + b
3
2 100
100
1000
x3
x2
+  x(cx + d )dx = c + d
3
2
500
1000
= 163.0933 + 370 = 533.0933
500
500
500
x4
x3
E ( X ) =  x (ax + b)dx = a + b
4
3 100
100
2
2
1000
x4
x3
+  x (cx + d )dx = d + d
4
3
500
70
= 63754.67 + 254375 = 318129.7
2
50
V ( X ) = E( X ) − [ E( X )] = 318129.7 − 533.09332 = 33941.16
2
4.3.6
2
a)
1210
E( X ) =
 x0.1dx = 0.05x
2 1210
1200
= 1205
1200
( x − 1205) 3
V ( X ) =  ( x − 1205) 0.1dx = 0.1
3
1200
1210
1210
= 8.333
2
1200
 x = V ( X ) = 2.887
b) Clearly, centering the process at the center of the specifications results in the greatest
proportion of cables within specifications.
1205
P(1195  X  1205) = P(1200  X  1205) =
 0.1dx = 0.1x
1200
120
4.3.7
a) E ( X ) =
x
100
600
120
dx = 600 ln x 100 = 109.39
2
x
1205
1200
= 0.5
Applied Statistics and Probability for Engineers, 7th edition
120
120
600
V ( X ) =  ( x − 109.39)
dx = 600  1 −
x2
100
100
2
2 (109.39 )
x
= 600( x − 218.78 ln x − 109.39 2 x −1 )
120
100
+
(109.39 ) 2
x2
dx
= 33.19
b) Average cost per part = $0.50*109.39 = $54.70
Section 4-4
4.4.1
a) E(X) = (-1+1)/2 = 0
(1 − (−1)) 2
V (X ) =
= 1 / 3, and  x = 0.577
12
x
b) P(− x  X  x) =

1
2
dt = 0.5t
x
= 0.5(2 x) = x
−x
−x
Therefore, x should equal 0.90.
0,
x  −1


c) F ( x) = 0.5 x + 0.5, − 1  x  1

1,
1 x

4.4.2
a) The distribution of X is f(x) = 10 for 0.95 < x < 1.05. Now,
0,



FX ( x) = 10 x − 9.5,


1,


x  0.95
0.95  x  1.05
1.05  x
b) P( X  1.02) = 1 − P( X  1.02) = 1 − FX (1.02) = 0.3
c) If P(X > x) = 0.90, then 1 − F(X) = 0.90 and F(X) = 0.10. Therefore, 10x - 9.5 = 0.10 and x =
0.96.
(1.05 − 0.95) 2
= 0.00083
d) E(X) = (1.05 + 0.95)/2 = 1.00 and V(X) =
12
4.4.3
a) The distribution of X is f(x) = 100 for 0.2050 < x < 0.2150. Therefore,
0,


F ( x) = 100 x − 20.50,


1,
x  0.2050
0.2050  x  0.2150
0.2150  x
b) P( X  0.2125) = 1 − F (0.2125) = 1 − [100(0.2125) − 20.50] = 0.25
c) If P(X > x) = 0.10, then 1 − F(X) = 0.10 and F(X) = 0.90.
Applied Statistics and Probability for Engineers, 7th edition
Therefore, 100x - 20.50 = 0.90 and x = 0.2140.
d) E(X) = (0.2050 + 0.2150)/2 = 0.2100 m and
(0.2150 − 0.2050) 2
= 8.33  10 −6 m 2
V(X) =
12
4.4.4
Let X denote the changed weight.
Var(X) = 42/12 and Stdev(X) = 1.1547
4.4.5 a) Let X be the time (in minutes) between arrival and 8:30 am.
1
f ( x) =
, for 0  x  90
90
x
Therefore, F ( x) =
, for 0  x  90
90
b) E ( X ) = 45 , Var(X) = 902/12 = 675
c) The event is an arrival in the intervals 8:50-9:00 am or 9:20-9:30 am or 9:50-10:00 am so that
the probability = 30/90 = 1/3
d) Similarly, the event is an arrival in the intervals 8:30-8:40 am or 9:00-9:10 am or 9:30-9:40 am
so that the probability = 30/90 = 1/3
4.4.6
a) E(X) = (380 + 374)/2 = 377
(380 − 374) 2
V(X ) =
= 3, and  x = 1.7321
12
b) Let X be the volume of a shampoo (milliliters)
375
375
1
1
1
P( X  375) =  dx = x
= (1) = 0.1667
6
6 374 6
374
c) The distribution of X is f(x) = 1/6 for 374  x  380 .
Now,
0, x  374


FX ( x ) = ( x − 374) / 6, 374  x  380

1, 380  x

P(X > x) = 0.95, then 1 – F(X) = 0.95 and F(X) = 0.05.
Therefore, (x - 374)/6 = 0.05 and x = 374.3
d) Since E(X) = 377, then the mean extra cost = (377-375) x $0.002 = $0.004 per container.
4.4.7
(a) Let X be the arrival time (in minutes) after 9:00 A.M.
V (X ) =
(120 − 0) 2
= 1200 and  x = 34.64
12
b) We want to determine the probability the message arrives in any of the following intervals:
9:05-9:15 A.M. or 9:35-9:45 A.M. or 10:05-10:15 A.M. or 10:35-10:45 A.M.. The probability of
this event is 40/120 = 1/3.
Applied Statistics and Probability for Engineers, 7th edition
c) We want to determine the probability the message arrives in any of the following intervals:
9:15-9:30 A.M. or 9:45-10:00 A.M. or 10:15-10:30 A.M. or 10:45-11:00 A.M. The probability of
this event is 60/120 = 1/2.
4.4.8
Let X denote the kinetic energy of electron beams.
3+ 7
a)  = E ( X ) =
=5
2
2
b)  2 = V ( X ) = (7 − 3)  1.33
12
c) P( X = 3.2) = 0 because X has a continuous probability distribution.
3+b
d)  = E ( X ) =
= 8 . Therefore, b = 13
2
2
e)  2 = V ( X ) = (b − 3)  0.75 . Therefore, b = 6
12
Section 4-5
4.5.1
a) P(Z<1.32) = 0.90658
b) P(Z<3.0) = 0.99865
c) P(Z>1.45) = 1 − 0.92647 = 0.07353
d) P(Z > −2.15) = p(Z < 2.15) = 0.98422
e) P(−2.34 < Z < 1.76) = P(Z<1.76) − P(Z > 2.34) = 0.95116
4.5.2
a) Because of the symmetry of the normal distribution, the area in each tail of the distribution
must equal 0.025. Therefore the value in Table III that corresponds to 0.975 is 1.96. Thus, z =
1.96.
b) Find the value in Table III corresponding to 0.995. z = 2.58
c) Find the value in Table III corresponding to 0.84. z = 1.0
d) Find the value in Table III corresponding to 0.99865. z = 3.0
4.5.3
a) P(X < 13) = P(Z < (13−10)/2) = P(Z < 1.5) = 0.93319
b) P(X > 9) = 1 − P(X < 9) = 1 − P(Z < (9−10)/2) = 1 − P(Z < −0.5) = 0.69146
14 − 10 
 6 − 10
Z
 = P(−2 < Z < 2)
 2
2 
c) P(6 < X < 14) = P
= P(Z < 2) −P(Z < − 2)]= 0.9545
4 − 10 
 2 − 10
Z

 2
2 
d) P(2 < X < 4) = P
= P(−4 < Z < −3) = P(Z < −3) − P(Z < −4) = 0.00132
e) P(−2 < X < 8) = P(X < 8) − P(X < −2)


= P Z 
4.5.4
8 − 10 
−2 − 10 

 − P Z 
 = P(Z < −1) − P(Z < −6) = 0.15866

2 
2 


a) P(X > x) = P Z 
x −10
x − 10 
 = 0.5. Therefore, 2 = 0 and x = 10.
2 
Applied Statistics and Probability for Engineers, 7th edition


b) P(X > x) = P Z 


Therefore, P Z 
x − 10 
x − 10 

 = 0.95
 = 1 − P Z 
2 
2 

x − 10 
x −10
 = 0.05 and 2 = −1.64. Consequently, x = 6.72.

2
x − 10 

 x − 10

 Z  0  = P( Z  0) − P Z 

2 
 2


x − 10 

 = 0.2.
= 0.5 − P Z 
2 

c) P(x < X < 10) = P


Therefore, P Z 
x − 10 
 = 0.3 and x − 10 = −0.52. Consequently, x = 8.96.
2 
2
d) P(10 − x < X < 10 + x) = P(−x/2 < Z < x/2)= 0.95. Therefore, x/2 = 1.96 and x = 3.92
e) P(10 − x < X < 10 + x) = P(−x/2 < Z < x/2) = 0.99. Therefore, x/2 = 2.58 and x = 5.16
4.5.5
a) 1 − (2) = 0.0228
b) Let X denote the time.
X ~ N(129, 142)
 100 − 129 
P( X  100) = 
 =  (−2.0714) = 0.01916
 14 
−1
c)  (0.95) 14 + 129 = 152.0280
Here 95% of the surgeries will be finished within 152.028 minutes.
d) 199 >> 152.028 so the volume of such surgeries is very small (less than 5%).
4.5.6
a) Let X denote the time.
X distributed N(260, 502)
P( X  240) = 1 − P( X  240) = 1 − (
240 − 260
) = 1 − (−0.4) = 1 − 0.3446=0.6554
50
−1 (0.25)  50 + 260 = 226.2755
−1 (0.75)  50 + 260 = 293.7245
−1
c)  (0.05)  50 + 260 = 177.7550
b)
4.5.7
0. 5 − 0. 4 
 = P(Z > 2) = 1 − 0.97725 = 0.02275
0.05 
0.5 − 0.4 
 0.4 − 0.4
Z
b) P(0.4 < X < 0.5) = P

0.05 
 0.05


a) P(X > 0.5) = P Z 
= P(0 < Z < 2) = P(Z < 2) − P(Z < 0) = 0.47725


c) P(X > x) = 0.90, then P Z 
x − 0 .4 
 = 0.90.
0.05 
x − 0 .4
Therefore, 0.05 = −1.28 and x = 0.336.
Applied Statistics and Probability for Engineers, 7th edition
4.5.8 Let X denote the cholesterol level.
X ~ N(159.2, σ2)
200 − 159.2
a) P( X  200) = (
) = 0.841
200 − 159.2


−1
=  (0.841)
=
200 − 159.2
= 40.8582
 −1 (0.841)
b) 
−1
(0.25)  40.8528 + 159.2 = 131.6452
−1 (0.75)  40.8528 + 159.2 = 186.7548
−1
c)  (0.9)  40.8528 + 159.2 = 211.5550
d) (2) − (1) = 0.1359
e) 1 − (2) = 0.0228
f) Φ(−1) = 0.1587
4.5.9 Let X denote the height.
X ~ N(1.41, 0.012)
a) P( X  1.42) = 1 − P( X  1.42) = 1 − (
b)
1.42 − 1.41
) = 1 − (1) = 0.1587
0.01
−1 (0.05)  0.01 + 1.41 = 1.3936
1.43 − 1.41
1.39 − 1.41
c) P(1.39  X  1.43) = (
) − (
) = (2) − (−2) = 0.9545
0.01
0.01
4.5.10 Let X denote the height.
X ~ N(64, 22)
a) P(58  X  70) = (
70 − 64
58 − 64
) − (
) = (3) − (−3) = 0.9973
2
2
−1 (0.25)  2 + 64 = 62.6510
−1 (0.75)  2 + 64 = 65.3490
−1
c)  (0.05)  2 + 64 = 60.7103
−1 (0.95)  2 + 64 = 67.2897
b)
d) [1 − (
68 − 64 5
)] = [1 − (2)]5 = 6.0942 10−9
2
5000 − 7000 
 = P(Z < −3.33) = 0.00043
600

x − 7000 

x − 7000
 = 0.95 and 600 = −1.64
b) P(X > x) = 0.95. Therefore, P Z 
600 



4.5.11 a) P(X < 5000) = P Z 
Consequently, x = 6016
Applied Statistics and Probability for Engineers, 7th edition


c) P(X > 7000) = P Z 
7000 − 7000 
 = P( Z  0) = 0.5
600

P(Three lasers operating after 7000 hours) = (1/2)3 =1/8
4.5.12 Let X denote the demand for water daily.
X ~ N(310, 452)
a) P( X  350) = 1 − P( X  350) = 1 − (
b)
350 − 310
40
) = 1 − ( ) = 0.1870
45
45
−1 (0.99)  45 + 310 = 414.6857
−1 (0.05)  45 + 310 = 235.9816
c)
d) 𝑋 ~𝑁(𝜇, 452 )
P( X  350) = 1 − P( X  350) = 1 − (
(
350 − 
) = 0.99
45
350 − 
) = 0.01
45
 = 350 − −1 (0.99)  45 = 245.3143 million gallons is the mean daily demand.
The mean daily demand per person = 245.3143/1.4 = 175.225 gallons.
13 − 12 
 = P(Z > 2) = 0.02275
0 .5 
13 − 12 

 = 0.999
b) If P(X < 13) = 0.999, then P Z 
 



4.5.13 a) P(X > 13) = P Z 
Therefore, 1/

= 3.09 and

= 1/3.09 = 0.324


c) If P(X < 13) = 0.999, then P Z 
Therefore,
13− 
0.5
13 −  
 = 0.999
0 .5 
= 3.09 and  = 11.455
0.0026 − 0.002 
 = P(Z > 1.5) = 1-P(Z < 1.5) = 0.06681
0.0004

0.0026 − 0.002 
 0.0014 − 0.002
Z
b) P(0.0014 < X < 0.0026) = P

0.0004
0.0004




4.5.14 a) P(X > 0.0026) = P Z 
= P(−1.5 < Z < 1.5) = 0.86638
0.0026 − 0.002 
 0.0014 − 0.002
Z





0.0006 
 − 0.0006
Z

= P
 
 
c) P(0.0014 < X < 0.0026) = P
Applied Statistics and Probability for Engineers, 7th edition


Therefore, P Z 
0.0006 
 = 0.9975. Therefore,
 
0.0006

= 2.81 and

= 0.000214.
4.5.15
From the shape of the normal curve, the probability is maximized for an interval symmetric about
the mean. Therefore a = 23.5 with probability = 0.1974. The standard deviation does not affect the
choice of interval.


4.5.16 a) P(X > 10) = P  Z 
10 − 4.6 
 = P (Z > 1.8621) = 0.0313
2.9 
b) P(X > x) = 0.25, then


c) P(X < 0) = P  Z 
 −1 (0.75)  2.9 + 4.6 = 0.6745  2.9 + 4.6 = 6.5560
0 − 4.6 
 = P (Z < -1.5862) = 0.0563
2 .9 
The normal distribution is defined for all real numbers. In cases where the distribution is truncated
(because wait times cannot be negative), the normal distribution may not be a good fit to the data.
100 − 50.9 

 = P (Z > 1.964) = 0.0248
25


25 − 50.9 

b) P(X < 25) = P  Z 
 = P (Z < -1.036) = 0.1501
25 

4.5.17 a) P(X > 100) = P  Z 
c) P(X > x) = 0.05, then
 −1 (0.95)  25 + 50.9 = 1.6449  25 + 50.9 = 92.0213
4.5.18 Let X denote the value of the signal.
1.55 − 1.5 
 1.45 − 1.5
Z
 = P(−2.5  Z  2.5)  0.988
0.02 
 0.02
a) P (1.45  X  1.55) = P
b) P( X  a) = 0.95
a − 1.5 
a − 1 .5 


P Z 
 = 0.95 and P Z 
 = 1 − 0.95 = 0.05 .
0.02 
0.02 


a − 1.5
= −1.645 and a = 1.4671 .
Then,
0.02
c)
 + 2 −  

P( X   + 2 ) = P Z 
 = P( Z  2) = 1 − P( Z  2) = 1 − 0.97725 = 0.02275



4.5.19 P(V  1.5) = P( Z 
1.5 − 

) = 0.05 
1.5

= 1.645   = 0.912
4.5.20
Let X 1 and X 2 denote the right ventricle ejection fraction for PH subjects and control subjects,
respectively.
a) Find a such that P( X 1  a) = 0.05 .
Applied Statistics and Probability for Engineers, 7th edition
a − 36 
a − 36 


P ( X 1  a ) = P Z 
 = 1 − 0.05 = 0.95
 = 0.05 , then P Z 
12 
12 


a − 36
= 1.645 and a = 55.74
12
55.74 − 56 

b) P( X 2  55.74) = P Z 2 
 = P( Z 2  −0.0325) = 1 − P( Z 2  0.0325) = 0.487
8


From the standard normal table,
c) Higher EF values are more likely to belong to the control subjects.