Chapter 8
16
Chapter 8 Theorem Limit Central the and Methods Sampling Problem Solved
description
Chapter 8. Problem 8-6. Problem 8-15. µ=60 σ =12. 24. 96. 72. 48. Problem 8-15(contd.). .2734. .2266. 63. Problem 8-15(contd.). m. =. 60. x. s. =. 4. x. -. m. -. x. 56. 60. =. =. =. x. z. 1. 0. s. 4. x. 56. Problems 8-16. - PowerPoint PPT Presentation
Transcript of Chapter 8
Chapter 8
TheoremLimit Central
theand Methods Sampling
Problems. Solved
Problem 8-6
x
6 12 4,8 10
6 12 4,8 9
4 8 4,4 8
5 10 2,8 7
3 6 2,4 6
3 6 2,4 5
5 10 2,8 4
3 6 2,4 3
3 6 2,4 2
2 4 2,2 1
MeanSumValuesSample a.
4 = 40/10 = 10 / ) 6 + 6 + 4 + 5 + 3 + 3 + 5 + 3 + 3 + (2 =
4 = 20/5 = 5 / 8) + 4 + 4 + 2 + (2=b.
6. to2 fromonly vary means sample the whereas8, to2 from variespopulation The
means. sample for thean that greater th is population for the dispersion The equal. areThey c.
Problem 8-15
0
0.05
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0.15
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0.25
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0.35
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0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
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4.5 5
0
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0.1
0.15
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0.35
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0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
µ=60σ=12
24 96
48 72
49
12
nx
60x
x
x
Problem 8-15(contd.)
0
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0.35
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0.45
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-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
2266.2734.5000.)63(. xPa
.2734
63
7504
6063
4
60
.x
zx
x
x
x
.2266
x
Problem 8-15(contd.)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
1587.3413.5000.)56(. xpb
6147.3413.2734.)6356(. xpc
56x
0.14
6056 =-
=-
=x
xx
zs
m
4
60
=
=
x
x
s
m
Problems 8-16
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0.35
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-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
Since n=40,the sampling distribution of means will approximatea normal distribution
40
5
5
75
40
n
n
x
74 76 77
40
5
75
x
x
.3962 .3962
.4943
x
Problem 8-16(Contd.)
0057.4943.5000.)77(.
53.2
405
7577
0981.3962.4943.)7776(.
7924.3962.3962.)7674(.
26.1
405
7574
1038.3962.5000.)74(.
=-=>
=-=-
=
=-=<<
=+=<<
-=-=-
=
=-=<
xpd
xz
xpc
xpb
xz
xpa
x
x
x
x
m
m
Problem 8-17
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0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
1950
50
250
2200
nx
x
x
07.7
50
25022001950
certaintyvirtualaor1)1950(
n
xz
xp
x
Problem 8-18
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
.2852
320
64912
330
.x
x
minutes649.1240
80. ===
na
x
79.
4080
330320
7852.5000.2852.)320(.
-=-=-
=
=+=>x
z
xpb
x
xm
Problem 8-18(Contd.)
.4429.2852
320 350
4080
330
x
x
58.1
40
80330350
7281.4429.2852.)350320(.
=-
=-
=
=+=<<
xz
xpc
xm
μx
x
Problem 8-18 contd..
.0571 .4429 - .5000 )350 x(P_
.d
.0571.4429
350 X
4080
330
x
x
=110,000
= 40,000/ 50
a. = / n = 40,000/ 50 = 5657
b. Normal distribution
x
x
x
Problem 8-34
c. .1368
.3632
112,000 X
=110,000
.35 Z
P(X> 112,000)=.5000-.1368=.3632
x
355000040
000110000112.
/,
,,z
n/
xz
Problem 8-34 cont’d
d.
.4616
100,000
-1.77
000110,x
7715000040
000110000100.
/,
,,z
n/
xz
Problem 8-34 cont’d
X
P( X> 100,000)=.4616 +.5000=.9616
Z
e.
.1368
.4616
100,000 112,000 X
-1.77 .35
000110,x
Problem 8-34 cont’d
P(100,000<X< 112,000)=.4616 + .1368 = .5984
Z
Problem 8-35
0
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0.1
0.15
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0.25
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0.35
0.4
0.45
-5
-4.5 -4
-3.5 -3
-2.5 -2
-1.5 -1
-0.5 0
0.5 1
1.5 2
2.5 3
3.5 4
4.5 5
xx
1.258.24
52
60
824
.
n
.
.3238
.93 Z
8238.3238.5000.)1.25(
93.
60
5.28.241.25
=+=<
=-
=-
=
xpn
xz x
m
