CHAPTER 4
THE NORMAL DISTRIBUTION
NOTE: A linear interpolation was used in the solution of most problems using the normal table. Thus
small discrepancies may be found when comparing the answers with solutions obtained without an
interpolation.
4-1.
P(–1 < Z < 1) = 2(.3413) = 0.6826
P(–1.96 < Z < 1.96) = 0.95
P(–2.33 < Z < 2.33) = 0.9802
Template (Normal Distribution, Mean=0, stddev=1)
x1
-1.00
-1.96
-2.33
P(x1<X<x2)
0.6827
0.9500
0.9802
x2
1.00
1.96
2.33
4-2.
P(–2 < Z < 1) = .4772 + .3413 = 0.8185
4-3.
P(–2.5 < Z < –0.89) = .4938 – .3133 = 0.1805
4-4.
P(Z > 3.02) = .5 – .4987 = 0.0013
4-5.
P(2 < Z < 3) = .4987 – .4772 = 0.0215
(Normal Distribution, Mean=0, stddev=1)
x1
2.00
P(x1<X<x2)
0.0214
x2
3.00
4-6.
P(Z < –2.5) = .5 – .4938 = 0.0062
4-7.
P(Z > –2.33) = .5 +.4901 = 0.9901
(Normal Distribution, Mean=0, stddev=1)
x
-2.33
P(X>x)
0.9901
4-8.
P(–2 < Z < 300) = a number very close to .5 + .4772 = 0.9772
4-9.
P(Z < –10) = a very small number, close to zero.
4-10.
P(–.01 < Z < .05) = .0199 + .0040 = 0.0239
(Normal Distribution, Mean=0, stddev=1)
x1
-0.01
4-11.
P(x1<X<x2)
0.0239
x2
0.05
P(–2 < Z < 2) = 2(.4772) = 0.9544
85
4-12.
z =  1.645
4-13.
Not likely. P(Z < –4) = .5 – .49997 = 0.00003
4-14.
P(Z > z) = 0.85
z = –1.036 (approximately).
(Normal Distribution, Mean=0, stddev=1)
x
-1.04
P(>x)
0.85
4-15.
P(Z < z) = 0.685
z = 0.48
4-16.
P(Z > z) = 0.5
z=0
Do not need a normal table to find this: the Z variable has mean = 0 by definition.
4-17.
P(Z > z) = 0.12
z = 1.175
4-18.
Look for z > 0 such that P(0 < Z < z) = 0.40/2 = 0.20:
(Normal Distribution, Mean=0, stddev=1)
 P(0 < Z < z) = 0.5 – 0.12 = 0.38
z =  0.524
Symmetric Intervals
P(x1<X<x2)
x1
x2
-0.524401
0.4
0.524401
z =  1.96
4-19.
Look for z > 0 such that P(0 < Z < z) = 0.95/2 = 0.475:
4-20.
Look for z > 0 such that 0.5 – P(0 < Z < z) = 0.01/2 = 0.005, which means
P(0 < Z < z) = 0.495: z =  2.576
(  2.575 is the value obtained by a linear interpolation,  2.576 is more accurate.)
4-21.
P(|Z| > 2.4) = P(Z > 2.4 or Z < –2.4) = 2P(Z > 2.4) = 2(.5 – .4918) = 0.0164
(Normal Distribution, Mean=0, stddev=1)
P(X<x)
0.0082
4-22.
x
-2.4
x
2.4
P(X>x)
0.0082
X ~ N(650, 402)


P(X < 650) = P  Z 
600  650 
40
 = P(Z < –1.25) = .5 – .3944 = 0.1056

86
4-23.
X ~ N(410, 22)
415  410 
 407  410
<Z<
 = P(–1.5 < Z < 2.5) =
2
2


P(407 < X < 415) = P 
.4332 + .4938 = 0.927
Normal Distribution
Mean
410
4-24.


555  500 
20
 = P(Z > 2.75) = .5 – .4970 = 0.003

X ~ N(–44, 162)


P(X > 0) = P  Z 
4-26.
P(x1<X<x2)
0.9270
X ~ N(500, 202)
P(X > 555) = P  Z 
4-25.
x1
407.00
Stdev
2
0  (44) 
 = P(Z > 2.75) = 0.003
16

X ~ N(0,42)


P(X > 2.5) = P  Z 
2.5  0 
 = P(Z > .625) = .5 – .234 = 0.266
4 
Normal Distribution
Mean
0
P(X>x)
x
2.5
Stdev
4
0.2660
4-27.
X ~ N(16, 32)
P(11< X < 20) = P(–1.667 < Z < 1.333) = .4521 + .4088 = .8609
P(17 < X < 19) = P(.33 < Z < 1) = .3413 – .1306 = 0.2107
P(X > 15) = P(Z > –.33) = .5 + .1306 = 0.6306
4-28.
X ~ N(45, 102)
a. P(X < 60) = P(Z < 1.5) = .5 + .4332 = 0.9332
b. P(X < 40) = P(Z < –.5) = .5 – .1915 = 0.3085
c. P(X > 70) = P(Z > 2.5) = .5 – .4938 = 0.0062
Normal Distribution
Mean
45
P(X<x)
0.9332
0.3085
Stdev
10
x
60
40
x
70
P(X>x)
0.0062
87
x2
415.00
4-29.
X ~ N(8,000, 1,0002)
P(X  9,322) = P(Z  1.322) = .5 – .4069 = 0.0931
4-30.
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
1.19
Stdev
0.03
a. P(X >1.25) = P(Z > 2.0) = 0.5 – 0.4772 = 0.0228
x
1.25
P(X>x)
0.0228
b. P(X<1.15) = P(Z < -1.333) = 0.5 – 0.4088 = 0.0912
P(X<x)
0.0912
x
1.15
c. P(1.16 <X <1.23) = P(-1.0 <Z< 1.333) = 0.3413 + 0.4088 = 0.7501
x1
1.16
4-31.
P(x1<X<x2)
0.7501
x2
1.23
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
50
Stdev
1.5
P(X > 55) = P(Z > 3.333) = 0.5 – 0.4996 = 0.0004
x
55
4-32.
P(X>x)
0.0004
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
10.42
Stdev
0.87
P(underweight) =0.05
P(<x)
0.05
x
8.99
A cat is underweight if it weighs less than 8.99 pounds.
P(overweight) = 0.10
x
11.53
P(>x)
0.1
A cat is overweight if it weighs more than 11.53 pounds.
A cat is normal weight if it weighs between 8.99 and 11.53 pounds.
4-33.
P(2520 < X < 2670) = P(–1.6 < Z < 1.4) = .4192 + .4452 = .8644
88
4-34.
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
12.5
Stdev
2.5
P(X> 17.5) = P(Z > 2.0) = 0.5 – 0.4772 = 0.0228
x
17.5
4-35.
P(X>x)
0.0228


a. P(X > 12) = P  Z 
12  10 .5 
 = P(Z > .316) = 0.50 – .1241 = .3759

4.75


0  10.5 


5  10.5 
b. P(X < 0) = P  Z 
a. P(X > 5) = P  Z 
 = P(Z < –2.211) = 0.50 – .4865 = .0135
4.75 
 = P(Z > –1.158) = 0.50 + .3766 = .8766
4.75 
Normal Distribution
Mean
10.5
Stdev
4.75
P(X<x)
x
0
0.0135
4-36.
x
12
5
P(X>x)
0.3761
0.8765
a. P(x>187.6) = P(z>0) = 0.5000
b. P(x>200) = P(z>1) = 0.1587
c. P(x>x*) = 0.05 x* =  + 1.645σ = 207.998 or 208 safety stock = 20.4
d. P(x>x*) = 0.01 x* =  + 2.326σ = 216.44 or 216 safety stock = 28.4


80  77.2 
4-37.
P(X < 80) = P  Z 
4-38.
X ~ N(120, 442)
P(X < x) = 0.56
P(Z < z) = 0.56, thus:
z = 0.15
x =   z = 120 + (.15)(44) = 126.6
3.1
 = P(Z < .903) = 0.50 + .3167 = .8167

Normal Distribution
Mean
120
Stdev
44
P(<x)
x
126.64
0.56
89
4-39.
X ~ N(16.5, 0.82) P(X > x) = 0.85
P(Z > z) = 0.85, thus:
z = –1.04
x =   z = 16.5 – 1.04(.8) = 15.67
4-40.
X ~ N(19,500, 4002)
P(X > x) = 0.02
P(Z > z) = 0.02, thus:
TA = 0.48 and z = 2.054
x =   z = 19,500 + 2.054(400) = 20,321.6
4-41.
X ~ N(88, 52)
P( x1 < X < x2 ) = 0.98
P(–z < Z < z) = 0.98
TA = .98/2 = .49
x1 = 88 – 2.33(5) = 76.35
x2 = 88 + 2.33(5) = 99.65
Normal Distribution
Mean
88
4-42.
X ~ N(32, 72)
Symmetric Intervals
P(x1<X<x2)
x1
x2
76.36829
0.98
99.63171
Stdev
5
Find x1 , x2 so that
P( x1 < X < x2 ) = .99
TA = .99/2 = .495
z =  2.576
x1, 2 = 32  2.576 (7)
= 13.97 and 50.03
4-43.
X ~ N(–61, 222)
P(X > x) = 0.25
TA = .25
z = 0.675 x = –61 + 0.675(22) = –46.15
4-44.
X ~ N(97, 102)
P(102 < X < x) = 0.05
x  97 
 102  97
Z
 = 0.05
10 
 10
P
P(0.5 < Z < z) = 0.05
TA(0.5) = 0.1915. We need z such that TA(z) = 0.1915 + 0.05 = 0.2415
z is approximately 0.65
x = 97 + (.65)(10) = 103.5
Check: P(102 < X < 103.5) = P(.5 < Z < .65)
= TA(.65) – TA(.5) = .2422 – .1915 = 0.0507
90
4-45.
X ~ N(600, 10,000)
 =
P(X > x1 ) = 0.01
z1 = 2.326
10,000 = 100
x1 = 600 + 2.326(100) = 832.6
P(X < x 2 ) = 0.05
z 2 = –1.645
x 2 = 600 – 1.645(100) = 435.5
4-46.
X ~ N(25,000, 5,0002)
P(X > x) = 0.15
TA = .35
z = 1.037
x =  + z = 25,000 + (1.037)5,000 = $30,185
Normal Distribution
Mean
25000
Stdev
5000
x
30182.16
P(>x)
0.15
4-47.
z =  1.96
X ~ N(27,009, 4,5302)
Interval: 27,009  1.96 (4,530) = [18,130.2, 35,887.8]
4-48.
X ~ N(11.2, 0.62)
0.99 probability implies TA = .99/2 = .495
z =  2.576
x1, 2 = 11.2  2.576 (0.6) = 9.654, 12.746
4-49.
0.95 probability of falling < z means we want TA = .9500 – .5 = .4500
z = 1.645
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
25000
Stdev
5000
P(X<x1) = 0.95 TA = 0.45
P(<x)
0.95
Z = 1.645 x1 = 25000 + 1.645(5000) = 33225 (approximately)
x
33224.27
91
4-50.
X ~ N(2.75, .152)
0.90 probability of being < z means we want TA = .90 – .5 = .400 so z = 1.282
Therefore X = 2.75 + 1.282(.15) = 2.9423 (approximately)
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
2.75
P(<x)
0.9
4-51.
Stdev
.15
x
2.94
X ~ N(2317, 2002)
0.80 probability of being > z means we want TA = .80 –.50 = .30, so z = .842
Therefore X = 2317 – .842(200) = 2148.60 (approximately)]
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
4-52.
Mean
2317
Stdev
200
x
2148.68
P(>x)
0.8
Assume independence of parties arriving for dinner.
P(success) = 0.7
P(X > 15) = 1 – F(15) = F(4) [using p = 0.3] = 0.2375
Using a normal approximation:
 = np = 20(.7) = 14
  npq = 2.049
P(X > 15) = P(X > 15.5) [using a continuity correction]


= PZ 
15 .5  14 
 = P(Z > .732) = 0.2321
2.049 
Normal Approximation of Binomial Distribution
n
20
p
0.7
x
15.5
P(X>x)
Mean
14
Stdev
2.04939
0.2321
92
4-53.
Random sampling from large population implies a binomial distribution.
n = 100
 = np = 40
p = 0.4
  npq = 4.899
P(X > 20) = P(X > 19.5) [continuity correction]


= PZ 
4-54.
19.5  40 
 = P(Z > –4.18) > 0.99997
4.899 
n = 45 p = 0.8
Assume independence of chips.
  npq = 2.683
 = 45(.8) = 36
P(X > 30) = P(X > 29.5) [cont. corr.] = P(Z > –2.42) = 0.9922
4-55.
P(x  200) = .3804
Normal Approximation of Binomial Distribution
4-56.
n
328
p
0.6
Mean
196.8
Stdev
8.87243
P(X<x)
x
x
199.5
P(X>x)
x1
P(x1<X<x2)
0.3804
Assume independence of students.
p = 0.25
n = 1,889
  npq = 18.82
 = np = 472.25


P(X > 500) = P(X > 499.5) [cont. corr.] = P  Z 
= P(Z > 1.448) = .5 – .4262 = 0.0738
93
499 .5  472 .25 
18.82


x2
4-57.
(Use template: Normal Distribution.xls, sheet: normal approximation)
Normal Approximation of Binomial Distribution
n
120
Mean
48
p
0.4
Stdev
5.36656
P(x > 50) = 0.3899
x
49.5
4-58.
P(X>x)
0.3899
(Use template: Normal Distribution.xls, sheet: normal approximation)
Normal Approximation of Binomial Distribution
n
1000
Mean
600
p
0.6
Stdev
15.4919
P(x < 400) = 0.00
P(X<x)
0.0000
4-59.
x
400.5
X ~ N(549, 682)


P(X > 500) = P  Z 
4-60.
500  549 
68
 = P(Z > –.72) = .5 + .2642 = 0.7642

X ~ N(785, 602)


P(X > 800) = P  Z 
800  785 
60
 = P(Z > .25) = .5 – .0987 = 0.4013

850  785 
 750  785
Z

60
60


P(750 < X < 850) = P 
= P(–.58333 < Z < 1.0833) = .2201 + .3606 = 0.5808
i.e., about 58% of all working days.
P(X < 665) = P(Z < –2) = .5 – .4772 = 0.0228
4-61.
X ~ N(650, 502)


700  650 


750  650 
P(X > 700) = P  Z 
P(X < 750) = P  Z 
50
50
 = P(Z > 1) = .5 – .3413 = 0.1587

 = P(Z < 2) = .5 + .4772 = 0.9772

94
4-62.
1
1
(Note that price is discrete: 48, 48 2 , 48 4 , etc.)
X ~ N(48, 62)


P(X > 60) = P  Z 
60  48 
6
 = P(Z > 2) = 0.0228

P(X < 60) = P(Z < 2) = 0.9772


P(X > 40) = P  Z 
40  48 
6
 = P(Z > –1.33) = .5 + .4088 = 0.9088

P(40 < X < 50) = P(–1.33 < Z < .33) = .4088 + .1306 = 0.5394
Limitation: stock prices are very discrete.
4-63.
X ~ N(800,000, 10,0002)
P(X > x) = 0.80
P(Z > z) = 0.80
TA = .30 z negative
z = –0.842
x =   z = 800,000 – (.842)10,000 = 791,580 barrels per day.
4-64.
X ~ N(8.95, 22)
P(X > x) = 0.90
z = –1.282
x = 8.95 – 1.28(2) = $6.39
Normal Distribution
Mean
8.95
x
6.39
4-65.
Stdev
2
P(>x)
0.9
X ~ N(8, 0.52)
find x1 x2 such that P x1  X  x2  = 0.95
z1, 2 =  1.96
x1, 2    z1, 2 = 8  1.96(.5) = [7.02, 8.98] percent.
4-66.
X ~ N(50,000, 3,0002)
Give 0.99 probability bounds:
z =  2.576
Bounds are 50,000  2.576(3,000) = [42,272, 57,728]
95
4-67.
X ~ N(2,348, 7622)
P(X > x) = 0.85
z = –1.04
x = 2,348 – 1.04(762) = 1,555.52
0.80 probability bounds: z =  1.28
x1 x2 = 2,348  1.28(762) = [1,372.64, 3,323.36], i.e., they are 80% sure that anywhere
from 1,373 to 3,323 people will sign up for the trip.
4-68.
X ~ N(12.1, 2.52)
P(X > x) = 0.75
z = –0.675
x = 12.1 – .675(2.5) = 10.4125%
P(X < x  ) = 0.75
z  = 0.675
x  = 12.1 + (.675)(2.5) = 13.7875%
4-69.
X ~ N(15.6, 4.12)
P(X > x) = 0.95
z = –1.645
x = 15.6 – 1.645(4.1) = 8.8555 KW.
4-70.
X ~ N(1.14, 0.32)
P(Z < z) = 0.95
z = 1.645 7A = .45
  z = 1.634 (approximately)
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
1.14
Stdev
0.3
P(<x)
x
1.63
0.95
4-71.
.5 – TA(z = 3.000) = .5 – .4987 = .0013, so the total area outside of z =  3.000 is 2(.0013) =
.0026, or 0.26%. Thus, define “too many” items as “more than 0.26%.” If noticeably more than
0.26% of the items are out of bounds, assuming the normal distribution for when the process is in
control, then a very low-probability event is being observed and the distribution is probably not
in fact a normal one, i.e., the process is probably not in control.
96
4-72.
X ~ N(  ,  2 )
P(X > 671,000) = 0.45
z1 = 0.126
0.126 =
P(X > 712,000) = 0.10
z 2 = 1.282
671,000  
1.282 =
712 ,000  

671,000 –  = 0.126 

712,000 –  = 1.282 
 = 35,467.128
41,000 = 1.156 
 = 671,000 – 0.126  = 666,531.1
4-73.
X ~ N(  ,  2 )
P(X > 65) = 0.45
z 2 = 1.0365
z1 = 0.126
65  

P(X > 70) = 0.15
= 0.126
70  

70 –  = 1.0365 
= 1.0365
65 –  = 0.126 
 = 5.49
5 = .9105 
 = 70 – 1.0365  = 64.31
4-74.
X ~ N(  ,  2 )
P(X > 1,000) = 0.1
P(X > 650) = 0.5
1,000  
= 1.28
z = 1.282
z=0
650  


1,000 – 650 = 1.28 
 = 650  = 273.01
4-75.
=0
X ~ N(4,500, 1,8002)
P(X < x) = 0.80
z = 0.842
x = 4,500 + (.842)1,800 = 6,015.6
Normal Distribution
Mean
4500
P(<x)
0.8
Stdev
1800
x
6014.92
97
4-76.
X ~ N(34,750, 3,5602)
P(X < x) = 0.75
z = 0.675
X = 34,750 + (.675)3,560 = 37,153 papers.
4-77.
P(X < 1.1) = P  Z 


1.1  1.0 
 = P(Z < 2.00) = 0.50 + .4772 = 0.9772
.05 
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
1.0
Stdev
0.05
P(X<x)
x
1.1
0.9772
4-78.
(Use template: Normal Distribution.xls, sheet: normal)
Normal Distribution
Mean
7.5
Stdev
0.5
P(X > 10) = P(Z > 5.0) = 0.000 (approximately)
x
10
P(X>x)
0.0000
4-79.
a) Since each phase of the project is normally distributed, we can add the individual phase
means and variances to get the overall project duration mean and variance:
E(Project) = E(Phase 1) + E(Phase 2) + E(Phase 3) = 84 + 102 + 62 = 248
V(Project) = V(Phase 1) + V(Phase 2) + V(Phase 3) = 9 + 16 + 4 = 29
Std dev (Project) = 5.3852
b) P(x  250) = 0.6448
c) P(x  240) = 0.0687
98
Normal Distribution
Mean
248
Stdev
5.3852
P(X<x)
x
250
240
0.6448
0.0687
P(X>x)
x
4-80.
Normal Distribution
Mean
487
Stdev
98
P(X<x)
x
x
500
P(X>x)
x
P(>x)
0.4472
x1
600.00
420.85
279.23
P(x1<X<x2)
0.1096
0.7500
0.7500
x2
700.00
841.70
558.45
Inverse Calculations
P(<x)
0.75
x
553.10
99
Symmetric Intervals
P(x1<X<x2)
x1
x2
374.2658
0.75
599.73424
a)
b)
c)
d)
e)
P(X >500) = 0.4472
P(600  X  700) = 0.1096
P(x ≥ X) = 0.75 z = -0.67 x = 487 – 0.67(98) = 421.34
narrowest interval containing 75% of the scores is: [374.27, 599.73]
the two intervals [x, 2x] are: [420.85, 841.70] [279.23, 558.45]
4-81.
Normal Distribution
Mean
11200
Stdev
8250
P(X<x)
x
0
-10000
0.0873
0.0051
x
P(X>x)
x
16764.55
P(>x)
x1
10000.00
P(x1<X<x2)
0.4148
x2
20000.00
Inverse Calculations
P(<x)
a)
b)
c)
d)
e)
x
0.25
Symmetric Intervals
P(x1<X<x2)
x1
x2
P(X < 0) = 0.0873
P( 10000  X  20000) = 0.4148
P(X > x) = 0.25, x = 16764.55
P(X < -10000) = 0.0051
There is a 91.27% probability of making a profit, and a 99.49% of not having to borrow
additional cash.
100
4-82
(Use template: Normal Distribution.xls, sheet: normal)
(i) Proportion of acceptable rods:
The manufacturing process is a normal distribution:
Normal Distribution
Mean
973.8
Stdev
0.32
The proportion within the specifications 974 ± 1.2 is 0.9991
x1
972.80
(ii) C p 
P(x1<X<x2)
0.9991
x2
975.20
Tolerance
1.2
1.2


 1.25
3* 
3 * 0.32 0.96
(iii)
A larger value of the process capability index, given the tolerance level, would ensure more units
would meet the customer’s specifications because the standard deviation would be decreasing as
Cp increases.
(iv) C pk  C p 
Offset
0.2

 0.208
3* 
0.96
(v)
Mean
974
Stdev
0.32
x1
972.80
P(x1<X<x2)
0.9998
x2
975.20
The proportion increases to 0.9998
(vi) Tolerance = 3*σ*Cp =3*1*1.2=3.6
Offset = (Cp - Cpk )*3 σ = (1.2-0.9)*3 = 0.9
specification: 1000 ± 3.6
Process: 999.1 ± 1.0
x1
996.40
P(x1<X<x2)
0.9965
x2
1003.60
101
4-83
(Use template: Normal Distribution.xls, sheet: normal)
(i) yes the total revenue will be normally distributed.
(ii) Mean Total Revenue = 5780 + 641 + 712 = 7133
Standard deviation = square root(1422 + 782 + 722) = 177.29
Mean
7133
Stdev
177.29
(iii)
P(X > 7000) = 0.7734
x
7000
P(X>x)
0.7734
102