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8/3/2019 Asignment Sum Solutionn q.A
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3.3 MedianArrange terms in ascending order:073, 167, 199, 213, 243, 345, 444, 524, 609, 682There are 10 terms.Since there are an even number of terms, the median is the average of the
two middle terms:
Median = 243 + 345 = 588 = 294
Using the formula, the median is located at the n + 1 th term.2
n = 10 therefore 10 + 1 = 11 = 5.5th term.2 2
The median is located halfway between the 5th and 6th terms.
5
th
term = 243 6
th
term = 345Halfway between 243 and 345 is the median = 2942 2
3.6 Rearranging the data into ascending order:
11, 13, 16, 17, 18, 19, 20, 25, 27, 28, 29, 30, 32, 33, 34
25.5)15(100
35==i
P35 is located at the 5 + 1 = 6th term
P35 = 19
25.8)15(100
55==i
P55 is located at the 8 + 1 = 9th term
P55 = 27
Q1 = P25
75.3)15(100
25
==i
Q1 = P25 is located at the 3 + 1 = 4th term
Q1 = 17
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Q2 = Median
The median is located at the termthth
82
115=
+
Q2 = 25
Q3 = P75
25.11)15(100
75==i
Q3 = P75 is located at the 11 + 1 = 12
th term
Q3 = 30
3.11 x x - (x-)26 6-4.2857 = 1.7143 2.93882 2.2857 5.22444 0.2857 .08169 4.7143 22.22461 3.2857 10.79583 1.2857 1.65305 0.7143 .5102
x = 30 x-= 14.2857 (x -)2 = 43.4284
2857.4730 ===
Nx
a.) Range = 9 - 1 = 8
b.) M.A.D. = ==
7
2857.14
N
x 2.041
c.) 2 =7
4284.43)( 2 =N
x = 6.204
d.) = 204.6)(2
=
N
x = 2.491
e.) 1, 2, 3, 4, 5, 6, 9
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Q1 = P25
i = )7(100
25= 1.75
Q1 is located at the 1 + 1 = 2th
term, Q1 = 2
Q3 = P75:
i = )7(100
75= 5.25
Q3 is located at the 5 + 1 = 6th term, Q3 = 6
IQR = Q3 - Q1 = 6 - 2 = 4
f.) z =491.22857.46
= 0.69
z =491.2
2857.42= -0.92
z =491.2
2857.44= -0.11
z =491.2
2857.49= 1.89
z =491.2
2857.41= -1.32
z =491.2
2857.43= -0.52
z =491.2
2857.45= 0.29
3.1614, 15, 18, 19, 23, 24, 25, 27, 35, 37, 38, 39, 39, 40, 44,46, 58, 59, 59, 70, 71, 73, 82, 84, 90
Q1 = P25
i = )25(100
25= 6.25
P25 is located at the 6 + 1 = 7th term
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Q1 = 25
Q3 = P75
i = )25(
100
75= 18.75
P75 is located at the 18 + 1 = 19
th term
Q3 = 59
IQR = Q3 - Q1 = 59 - 25 = 34
3.19
xxx
2)( xx
7 1.833 3.3615 3.833 14.694
10 1.167 1.36112 3.167 10.0289 0.167 0.0288 0.833 0.694
14 5.167 26.6943 5.833 34.028
11 2.167 4.69413 4.167 17.3618 0.833 0.694
6 2.833 8.028106 32.000 121.665
12
106=
=
n
xx = 8.833
a) MAD =12
32=
n
xx= 2.667
b) s2 =11
665.121
1
)(2
=
n
xx= 11.06
c) s = 06.112 =s = 3.326
d) Rearranging terms in order:
3 5 6 7 8 8 9 10 11 12 13 14
Q1 = P25: i = (.25)(12) = 3
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Q1 = the average of the 3rd and 4th terms:
Q1 = (6 + 7)/2 = 6.5
Q3 = P75: i = (.75)(12) = 9
Q3 = the average of the 9th and 10th terms:
Q3 = (11 + 12)/2 = 11.5
IQR = Q3 - Q1 = 11.5 - 9 = 2.5
e.) z =
326.3
833.86= - 0.85
f.) CV =833.8
)100)(326.3(= 37.65%
3.27 Mean
Class f M fM0 - 2 39 1 392 - 4 27 3 814 - 6 16 5 806 - 8 15 7 105
8 - 10 10 9 9010 - 12 8 11 8812 - 14 6 13 78
f=121 fM=561
= 121561
=
f
fM= 4.64
Mode: The modal class is 0 2.The midpoint of the modal class = the mode = 1
3.30 Class f M fM fM2
5 - 9 20 7 140 9809 - 13 18 11 198 2,178
13 - 17 8 15 120 1,80017 - 21 6 19 114 2,16621 - 25 2 23 46 1,058
f=54 fM= 618 fm2= 8,182
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s2 =
53
7.70718182
53
54
)618(8182
1
)(22
2
=
=
n
n
fMfM
= 20.9
s = 9.202 =S = 4.57
3.31 Class f M fM fM218 - 24 17 21 357 7,49724 - 30 22 27 594 16,03830 - 36 26 33 858 28,31436 - 42 35 39 1,365 53,23542 - 48 33 45 1,485 66,82548 - 54 30 51 1,530 78,03054 - 60 32 57 1,824 103,96860 - 66 21 63 1,323 83,34966 - 72 15 69 1,035 71,415
f= 231 fM= 10,371 fM2= 508,671
a.) Mean:231
371,10=
=
=
f
fM
n
fMx = 44.9
b.) Mode. The Modal Class = 36-42. The mode is the class midpoint =39
c.) s2 =
230
5.053,43
230
231
)371,10(671,508
1
)(22
2
=
=
n
n
fMfM
=
187.2
d.) s = 2.187 = 13.7
3.32
a.) Mean
Class f M fM fM2
0 - 1 31 0.5 15.5 7.751 - 2 57 1.5 85.5 128.252 - 3 26 2.5 65.0 162.50
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3 - 4 14 3.5 49.0 171.504 - 5 6 4.5 27.0 121.505 - 6 3 5.5 16.5 90.75
f=137 fM=258.5 fM2=682.25
= 137 5.258=
ffM = 1.89
b.) Mode: Modal Class = 1-2. Mode = 1.5
c.) Variance:
2
=137
137
)5.258(25.682
)( 222
=
NN
fMfM
=
1.4197
d.) standard Deviation:
= 4197.12 = = 1.19155.10 n = 16 p = .40
P(x > 9): from Table A.2:
x Prob9 .084
10 .03911 .01412 .00413 .001
.142
P(3 < x < 6):
x Prob
3 .0474 .1015 .1626 .198
.508
n = 13 p = .88
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P(x = 10) = 13C10(.88)10(.12)3 = 286(.278500976)(.001728) = .1376
P(x = 13) = 13C13(.88)13(.12)0 = (1)(.1897906171)(1) = .1898
Expected Value = = n p = 13(.88) = 11.445.13n = 15 p = .20
a) P(x = 5) = 15C5(.20)5(.80)10 =
3003(.00032)(.1073742) = .1032
b) P(x > 9): Using Table A.2
P(x = 10) + P(x = 11) + . . . + P(x = 15) =
.000 + .000 + . . . + .000 = .000
c) P(x = 0) = 15C0(.20)0(.80)15 =
(1)(1)(.035184) = .0352
d) P(4 < x < 7): Using Table A.2
P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) =
.188 + .103 + .043 + .014 = .348
e)
5.20 = 5.6 days3 weeks
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a) Prob(x=0 = 5.6):
from Table A.3 = .0037
b) Prob(x=6 = 5.6):from Table A.3 = .1584
c) Prob(x > 15 = 5.6):
x Prob.15 .000516 .000217 .0001
x > 15 .0008
Because this probability is so low, if it actually occurred, theresearcher would
actually have to question the Lambda value as too low for this period.
5.24 n = 100,000 p = .00004
Prob(x > 7n = 100,000 p = .00004):
= = n p = 100,000(.00004) = 4.0
Since n > 20 and n p < 7, the Poisson approximation to this binomial
problem isclose enough.
Prob(x > 7 = 4):
Using Table A.3 x Prob.7 .05958 .0298
8/3/2019 Asignment Sum Solutionn q.A
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9 .013210 .005311 .001912 .000613 .0002
14 .0001x > 7 .1106
Prob(x>10 = 4):
Using Table A.3 x Prob.11 .001912 .000613 .0002
14 .0001x > 10 .0028
Since getting more than 10 is a rare occurrence, thisparticulargeographicregionappears to have a higher average rate than other regions. Aninvestigationofparticular characteristics of this region might be warranted.
5.29 N = 17 A = 8 n = 4
a) P(x = 0) =417
4908
C
CC =
2380
)126)(1(= .0529
b) P(x = 4) =417
0948
C
CC =
2380
)1)(70(= .0294
c) P(x = 2 non computer) =417
2829
C
CC =
2380
)28)(36(= .4235
5.30 N = 20 A = 16 white N - A = 4 red n = 5
a) Prob(x = 4 white) =520
14416
C
CC =
15504
)4)(1820(= .4696
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b) Prob(x = 4 red) =520
11644
C
CC =
15504
)16)(1(= .0010
c) Prob(x = 5 red) = 520
01654
C
CC
= .0000 because 4C5 is impossible
determine.The participant cannot draw 5 red beads if there are only 4 to draw from
5.32 N = 16 A = 4 defective n = 3
a) Prob(x = 0) =560
)220)(1(
316
31204=
C
CC= .3929
b) Prob(x = 3) =560
)1)(4(
316
01234=
C
CC= .0071
c) Prob(x > 2) = Prob(x=2) + Prob(x=3) =316
11224
C
CC
+ .0071 (from
part b.) =
560
)12)(6(
+ .0071 = .1286 + .0071 = .1357
d) Prob(x < 1) = Prob(x=1) + Prob(x=0) =
8/3/2019 Asignment Sum Solutionn q.A
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316
21214
C
CC + .3929 (from part a.) =
560
)66)(4(+ .3929 = .4714 + .
3929 = .86436.3 a = 2.80 b = 3.14
=2
14.380.2
2=+ba = 2.97
=12
80.214.3
12
=
ab= 0.10
P(3.00 < x < 3.10) =80.214.3
00.310.3
= 0.2941
6.4 a = 11.97 b = 12.03
Height = 97.1103.12
11
=ab = 16.667
P(x > 12.01) =97.1103.12
01.1203.12
= .3333
P(11.98 < x < 12.01) =97.1103.12
98.1101.12
= .5000
6.5 = 2100 a = 400 b = 3800
=12
4003800
12
=
ab= 981.5
Height =12
40038001 =
ab= .000294
P(x > 3000) =3400
800
4003800
30003800=
= .2353
P(x > 4000) =.0000
P(700 < x < 1500) =3400
800
4003800
7001500=
= .2353
6.8 = 22 = 4
a) Prob(x > 17):
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z =4
2217 =
x= -1.25
area betweenx = 17 and = 22 from table A.5 is .3944
Prob(x > 17) = .3944 + .5000 = .8944
b) Prob(x < 13):
z =4
2213=
x= -2.25
from table A.5, area = .4878
Prob(x < 13) = .5000 - .4878 = .0122
c) P(25 $2000) = .5000 - .3212 = .1788
b) Prob(owes money) = Prob(x < 0):
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z =725
13320 =
x= -1.84
from Table A.5, the z = -1.84 yields: .4671
Prob(x < 0) = .5000 - .4671 = .0329
c) Prob($100 < x < $700):
z =725
1332100 =
x= -1.70
from Table A.5, the z = -1.70 yields: .4554
z =
725
1332700 =
x= -0.87
from Table A.5, the z = -0.87 yields: .3078
Prob($100 < x < $700) = .4554 - .3078 = .1476
6.11 = $30,000 = $9,000
a) Prob($15,000 < x < $45,000):
z =000,9
000,30000,45 =
x= 1.67
From Table A.5, z = 1.67 yields: .4525
z =000,9
000,30000,15 =
x= -1.67
From Table A.5, z = -1.67 yields: .4525
Prob($15,000 < x < $45,000) = .4525 + .4525 = .9050
b) Prob(x > $50,000):
z =000,9
000,30000,50 =
x= 2.22
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From Table A.5, z = 2.22 yields: 4868
Prob(x > $50,000) = .5000 - .4868 = .0132
c) Prob($5,000 < x < $20,000):
z =000,9
000,30000,5 =
x= -2.78
From Table A.5, z = -2.78 yields: .4973
z =000,9
000,30000,20 =
x= -1.11
From Table A.5, z = -1.11 yields .3665
Prob($5,000 < x < $20,000) = .4973 - .3665 = .1308
d) 90.82% of the values are greater than x = $7,000.
Then x = $7,000 is in the lower half of the distribution and .9082 - .5000 =
.4082 lie between x and .
From Table A.5, z = -1.33 is associated with an area of .4082.
Solving for :
z =
x
-1.33 =
000,30000,7
= 17,293.23
e) = $9,000. If 79.95% of the costs are less than $33,000, x =$33,000 is inthe upper half of the distribution and .7995 - .5000 = .2995 of the
values liebetween $33,000 and the mean.
From Table A.5, an area of .2995 is associated with z = 0.84
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Solving for :
z =
x
0.84 = 000,9000,33
= $25,440
6.12 = 200, = 47 Determine x
a) 60% of the values are greater than x:
Since 50% of the values are greater than the mean, = 200, 10% or .1000 lie
between x and the mean. From Table A.5, the z value associated withan area
of .1000 is z = -0.25. The z value is negative since x is below the mean.Substituting z = -0.25, = 200, and = 47 into the formula and
solving for x:
z =
x
-0.25 =47
200x
x = 188.25
b) x is less than 17% of the values.
Since x is only less than 17% of the values, 33% (.5000- .1700) or .3300lie
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between x and the mean. Table A.5 yields a z value of 0.95 for an areaof
.3300. Using this z = 0.95, = 200, and = 47, x can be solved for:
z =
x
0.95 =47
200X
x = 244.65
c) 22% of the values are less than x.
Since 22% of the values lie below x, 28% lie between x and the mean
(.5000 - .2200). Table A.5 yields a z of -0.77 for an area of .2800.Using the z
value of -0.77, = 200, and = 47, x can be solved for:
z =
x
-0.77 =47
200x
x = 163.81
d) x is greater than 55% of the values.
Since x is greater than 55% of the values, 5% (.0500) lie between x andthe
mean. From Table A.5, a z value of 0.13 is associated with an area of .05.
Using z = 0.13, = 200, and = 47, x can be solved for:
z =
x
0.13 =47
200x
x = 206.11
6.22 n = 300 p = .53
= 300(.53) = 159
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= )47)(.53(.300= qpn = 8.645
Test: + 3 = 159 + 3(8.645) = 133.065 to 184.935
which lies between 0 and 300. It is okay to use the normal distribution asan approimation on parts a) and b).
a) Prob(x > 175 transmission)
correcting for continuity: x = 175.5
z =645.8
1595.175 = 1.91
from A.5, the area for z = 1.91 is .4719
Prob(x > 175) = .5000 - .4719 = .0281
b) Prob(165 < x < 170)
correcting for continuity: x = 164.5; x = 170.5
z =645.8
1595.170 = 1.33
z =645.8
1595.164 = 0.64
from A.5, the area for z = 1.33 is .4082the area for z = 0.64 is .2389
Prob(165 < x < 170) = .4082 - .2389 = .1693
For parts c and d: n = 300 p = .60
= 300(.60) = 180
= )40)(.60(.300= qpn = 8.485
Test: + 3 = 180 + 3(8.485) = 180 + 25.455
154.545 to 205.455 lies between 0 and 300
It is okay to use the normal distribution toapproimate c) and d)
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c) Prob(155 < x < 170 personnel):
correcting for continuity: x = 154.5; x = 170.5
z =485.8
1805.170
= -1.12
z =485.8
1805.154 = -3.01
from A.5, the area for z = -1.12 is .3686the area for z = -3.01 is .4987
Prob(155 < x < 170) = .4987 - .3686 = .1301
d) Prob(x < 200 personnel):
correcting for continuity: x = 199.5
z =485.8
1805.199 = 2.30
from A.5, the area for z = 2.30 is .4893
Prob(x < 200) = .5000 + .4893 = .98936.23 p = .16 n = 130
Conversion to normal dist.: = n(p) = 130(.16) = 20.8
= )84)(.16(.130= qpn = 4.18
a) Prob(x > 25):
Correct for continuity: x = 25.5
z =18.4
8.205.25 = 1.12
from table A.5, area = .3686
Prob(x > 20) = .5000 - .3686 = .1314
b) Prob(15
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z =18.4
8.205.14 = -1.51
z =
18.4
8.205.23 = 0.65
from table A.5, area for z = -1.51 is .4345area for z = 0.65 is .2422
Prob(15 < x < 23) = .4345 + .2422 = .6767
c) Prob(x < 12):
correct for continuity: x = 11.5
z =18.4
8.205.11 = -2.22
from table A.5, area for z = -2.22 is .4868
Prob(x < 12) = .5000 - .4868 = .0132
d) Prob(x = 22):
correct for continuity: 21.5 to 22.5
z =18.4
8.205.21 = 0.17
z =18.4
8.205.22 = 0.41
from table A.5, area for 0.17 = .0675area for 0.41 = .1591
Prob(x = 22) = .1591 - .0675 = .0916
7.18 = 99.9 = 30 n = 38
a) Prob( x < 90):
z =38
30
9.9990
1
=
N
nN
n
x
= -2.03
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from table A.5, area = .4788
Prob( x < 90) = .5000 - .4788 = .0212
b) Prob(98 < x < 105):
z =38
30
9.99105
1
=
N
nN
n
x
= 1.05
from table A.5, area = .3531
z =
38
30
9.9998
1
=
N
nN
n
x
= -0.39
from table A.5, area = .1517
Prob(98 < x < 105) = .3531 + .1517 = .5048
c) Prob( x < 112):
z =38
30
9.99112
1
=
N
nN
n
x
= 2.49
from table A.5, area = .4936
Prob( x < 112) = .5000 - .4936 = .0064
d) Prob(93 < x < 96):
z =3830
9.9993
1
=
NnN
n
x
= -1.42
from table A.5, area = .4222
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z =38
30
9.9996
1
=
N
nN
n
x
= -0.80
from table A.5, area = .2881
Prob(93 < x < 96) = .4222 - .2881 = .1341
7.20 = $65.12 = $21.45 n = 45
Prob( x > 0x ) = .2300
Prob. x lies between 0x and = .5000 - .2300 = .2700
from Table A.5, z.2700 = 0.74
Solving for 0x :
z =
n
x
0
0.74 =
45
45.21
12.650 x
2.366 = 0x - 65.12
0x = 65.12 + 2.366 = 67.486
7.27 P = .48 n = 200
a) Prob(x < 90):
p =200
90= .45
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z =
200
)52)(.48(.
48.45. =
n
QP
Pp
= -0.85
from Table A.5, the area for z = -0.85 is .3023
Prob(x < 90) = .5000 - .3023 = .1977
b) Prob(x > 100):
p =200
100= .50
z =
200
)52)(.48(.
48.50. =
n
QP
Pp
= 0.57
from Table A.5, the area for z = 0.57 is .2157
Prob(x > 100) = .5000 - .2157 = .2843
c) Prob(x > 80):
p =200
80= .40
z =
200
)52)(.48(.
48.40.
=
n
QP
Pp
= -2.26
from Table A.5, the area for z = -2.26 is .4881
Prob(x > 80) = .5000 + .4881 = .98817.28 P = .19 n = 950
a) Prob( p > .25):
z =
950
)89)(.19(.19.25. =
n
QPPp
= 4.71
from Table A.5, area = .5000
Prob( p > .25) = .5000 - .5000 = .0000
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b) Prob(.15 < p < .20):
z =
950
)81)(.19(.
19.15. =
n
QP
Pp
= -3.14
z =
950
)89)(.19(.
19.20. =
n
QP
Pp
= 0.79
from Table A.5, area for z = -3.14 is .4992
from Table A.5, area for z = 0.79 is .2852
Prob(.15 < p < .20) = .4992 + .2852 = .7844
c) Prob(133 < x < 171):
1p =
950
133= .14 2p =
950
171= .18
Prob(.14 < p < .18):
z =
950
)81)(.19(.
19.14. =
n
QP
Pp
= -3.93
z =
950
)81)(.19(.
19.18. =
n
QP
Pp
= -0.79
from Table A.5, the area for z = -3.93 is .49997the area for z = -0.79 is .2852
P(133 < x < 171) = .49997 - .2852 = .21477
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