978-1285867045 Chapter 6 Solution Manual Part 2

subject Type Homework Help
subject Pages 9
subject Words 1496
subject Authors David R. Anderson, Dennis J. Sweeney, Thomas A. Williams

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page-pf1
25.
= 204 and
= 55
a.
z=x-
m
s
=225-204
55 =.38
b.
z=x-
m
s
=140-204
55 = -1.16
c. For x = 300,
For x = 200,
z=x-
m
s
=200 -204
55 = -.07
P(200
£
x
£
300) = P
(z£1.75)
-
P
(z£-.07)
= .9599
-
..4721 = .4878
d. The upper 20%, or area =
1-.20 =.80
occurs for
z=.84
xz

=+
= 204 + .84(55) = $250
The 20% most expensive hotel rooms will cost $250 or more per night.
Using Excel: EXPON.DIST(6,1/8,TRUE) = .5276
b. P(x 4) = 1 - e-4/8 = 1 - .6065 = .3935
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Using Excel: EXPON.DIST(6,1/8,TRUE)-EXPON.DIST(4,1/8,TRUE) = .1342
27. a.
0/3
0
( ) 1 x
P x x e
=
b. P(x 2) = 1 - e-2/3 = 1 - .5134 = .4866
Using Excel: EXPON.DIST(2,1/3,TRUE) = .4866
c. P(x 3) = 1 - P(x 3) = 1 - (1 -
e3 3/
) = e-1 = .3679
28. a. With
= 
/20
1
() 20
x
f x e
=
b. P(x ≤ 15) =
/ 15/20
11
x
ee
−−
=
= .5276
c. P( x > 20) = 1 - P(x ≤ 20)
= 1 - (1 -
20/20
e
) =
1.3679e=
page-pf3
b. P(x 12) = 1 - e-12/12 = 1 - .3679 = .6321
Using Excel: EXPON.DIST(12,1/12,TRUE) = .6321
30. a.
/ /2
11
() 2
xx
f x e e
−−
==
for x > 0
P(x < x0) =
0/
1x
e
P(x < 1) =
1/2
1e
=
.5
1e
−=
1 - .6065 = .3935
2/2
1.0
(1 2) ( 2) ( 1)P x P x P x =
= .6321 - .3935 = .2386
c. For this customer, the cable service repair would have to take longer than 4 hours.
( 4) 1 ( 4)P x P x =
= 1
4/ 2 2.0
(1 ) .1353ee
−−
= =
31. a.
/ /25
11
() 25
xx
f x e e
−−
==
for x > 0
P(x < x0) =
0/
1x
e
P(x < 20) =
20/25
1e
−=
.80
1e
−=
1 - .4493 = .5507
page-pf4
b. P(x 30) = 1 - P(x 30) = 1 - (1 - e-30/25) = e-1.2 = .3012
5:45 p.m. Since the order was placed at 5:20 p.m., the order must to be ready within 25 minutes.
P(x
25) = 1 -
25/25
e
= 1 -
1
e
= 1 - .3679 = .6321
for a mean of 1.6 calls per hour, the mean time between calls is
m
=60 minutes/hour
1.6 calls/hour =37.5 minutes
per call
37.5
è
çö
ø
÷e-x/37.5
x³0
where x is the minutes between 911 calls.
c. Using time in minutes,
P x <60
( )
=1-e-60/37.5 =1-.2019 =.7981
d.
P(x³30)=1-P x £30
( )
=1-(1-e-30/37.5)=1-.5507 =.4493
1 for 200 225
() 25
0 elsewhere
x
fx

=
b. P(x 215) = (1 / 25) (225 - 215) = 0.40
c. P(x < 210) = (1 / 25)(210 - 200) = 0.40
value approach to decision making would suggest she should wait.
34. a. Find the z value that cuts off an area of .10 in the lower tail.
page-pf5
19,000
1.28 2100
x
z
= =
x = 19,000 1.28(2100) = 16,312
10% of athletic scholarships are valued at $16,312 or less.
Using Excel: NORM.INV(.90,19000,2100) = 16,308
22,000 19,000 1.43
x
−−
7.64% of athletic scholarships are valued at $22,000 or more.
Using Excel: 1-NORM.DIST(22000,19000,2100) = .0766
c. Find the z value that cuts off an area of .03 in the upper tail: z = 1.88. Solve for x,
19,000
x
x = 19,000 + 1.88(2100) = 22,948
3% of athletic scholarships are valued at $22,948 or more.
Using Excel: NORM.INV(.97,19000,2100) = 22,949.6666
b. P(defect) = 1 - P(9.85 x 10.15)
= 1 - P(-3 z 3) = 1 - .9974 = .0026
Expected number of defects = 1000(.0026) = 2.6
So,
610 658
1.88z
= =
page-pf6
610 658 25.5319
1.88
==
700 658 1.65
25.5319
x
−−
At 600,
600 659 2.31
25.5319
x
z
−−
= = =
P(600 < x < 700) = P(-2.31 < z < 1.65) = .9505 - .0104 = .9401
c. z = 1.88 cuts off approximately .03 in the upper tail
x = 658 + 1.88(25.5319) = 706.
P
(x<5)
= P
(z<-.61)
= .7291
Using Excel: NORM.DIST(5,4.5,.82,TRUE) = .7290
b.
z=x-
m
s
=3-4.5
.82 = -1.83
Using Excel: 1-NORM.DIST(3,4.5,.82,TRUE) = .9663
c. For x = 4,
z=x-
m
s
=4-4.5
.82 = -.61
m
Using Excel: NORM.DIST(4,4.5,.82,TRUE) NORM.DIST(3,4.5,.82,TRUE) = .2373
d. 85% of the weekly cargo volumes can be handled without requiring the port to extend
operating hours; 15% of the time the tons of cargo is so large it requires the port to extend operating
page-pf7
b. Expected Profit = Expected Revenue - Expected Cost
= 200 - 150 = $50
39. a. Mean monthly revenue = ($368)(330)/12 = $10,120
b.
= 10,120 and
= 2200
s
2200 =.85
P
(x>12,000)
= P
(z>.85)
= 1
-
P
(z£.85)
= 1 - .8023 = .1977
Using Excel: 1-NORM.DIST(12000,10120,2200,TRUE) = .1964
c.
z=x-
m
s
=7,500 -10,120
2200 = -1.19
P
(x<7,500)
= P
(z<-1.19)
= .1170
P
(x>12,000)
= P
(z>.18)
= 1
-
P
(z£.18)
= 1 - .5714 = .4286
z=x-
m
s
=7,500-11,550
2500 = -1.62
P
(x<7,500)
= P
(z<-1.62)
= .0526
page-pf8
500 450 .500
100
z
= = +
Area to left is .6915
P(400 x 500) = .6915 - .3085 = .3830
480 450 .30
100
z
==
Area to left is .6179
38.21% are acceptable.
41. a. At 100,000
100,000 88,592 .57
19,900
z
=
b. At 100,000
100,000 97,417 .12
21,800
z
=
P(x > 100,000) = P(z > .12) = 1 - P(z .12) = 1 - .5478 = .4522
The probability of a Los Angeles brand manager having a base salary in excess of $100,000 is .4522
page-pf9
The probability of a Los Angeles brand manager receiving a base salary below $75,000 is small:
.1515
d. The answer to this is the Houston brand manager base salary that cuts off an area of .01 in the upper
tail of the distribution for Houston brand managers.
A Los Angeles brand manager who makes $134,959 or more will earn more than 99% of the
Houston brand managers.
page-pfa
5.8
EXPON.DIST(5,1/5.8,TRUE) =.5777
6.2
EXPON.DIST(5,1/6.2,TRUE) =.5536
6.6
EXPON.DIST(5,1/6.6,TRUE) =.5312
7.0
EXPON.DIST(5,1/7,TRUE) =.5105
µ = 6.6 provides the closest probability to
( 5) .53Px=
b.
10/6.6
( 10) 1 ( 10) 1 (1 ) 1 .7802 .2198P x P x e
= = = =
The probability that a worker uses the office computer between four and eight hours is
.7024 .4545 .2479−=
Using Excel: EXPON.DIST(8,1/6.6,TRUE) - EXPON.DIST(4,1/6.6,TRUE) = .2479
44. a. Mean time between arrivals = 1/7 minutes
P(x > .2) = 1 - P(x < .2) = 1- [1- e-7(.2)] = e-1.4 = .2466
45. a.
/36.5 .0274
1.0274
36.5
xx
ee
−−
c. From part (b), P(x < 40) = .6658
P(x > 40) = P(x 40) = 1 - P(x < 40) = 1 - .6658 = .3342
105
b. Note: 30 seconds = .5 minutes
P(x .5) = 1 - e-.5/2 = 1 - .7788 = .2212
c. P(x 1) = 1 - e-1/2 = 1 - .6065 = .3935
page-pfb
d. P(x 5) = 1 - P(x < 5) = 1 - (1 - e-5/2) = .0821

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