CH06 Notes Lecture

8/18/2019 CH06 Notes Lecture

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Chapter 6

The Normal Distribution

1Copyright © 2013 The McGraw-Hill Companies, Inc. ermission re!"ire# $or

repro#"ction or #isplay.


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Outline

6The Normal Distribution

6-1 Normal Distributions

6-2 Applications of the Normal Distribution

6-3 The Central Limit Theorem

6-4 The Normal Approximation to the BinomialDistribution


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Objectives


1 I#enti$y #istri%"tions as symmetric or s&ewe#.

2 I#enti$y the properties o$ a normal #istri%"tion.

3 'in# the area "n#er the stan#ar# normal #istri%"tion,gi(en (ario"s ) (al"es.

4 'in# pro%a%ilities $or a normally #istri%"te# (aria%le

%y trans$orming it into a stan#ar# normal (aria%le. 'in# speci$ic #ata (al"es $or gi(en percentages, "sing

the stan#ar# normal #istri%"tion.


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Objectives


6 *se the central limit theorem to sol(e pro%lemsin(ol(ing sample means $or large samples.

! *se the normal appro+imation to comp"te pro%a%ilities $or a %inomial (aria%le.


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.1 ormal istri%"tions

/ Many contin"o"s (aria%les ha(e #istri%"tions thatare %ell-shape# an# are calle# approximatel"approximatel"

normall" #istribute# $ariablesnormall" #istribute# $ariables.

/ The theoretical c"r(e, calle# the bell cur$ebell cur$e or the

%aussian #istribution%aussian #istribution, can %e "se# to st"#y many

(aria%les that are not normally #istri%"te# %"t are

appro+imately normal.

Bluman, Chapter 6 5


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ormal istri%"tions

2 2 2

2

− −

=

X e

y

µ σ

σ π

The mathematical equation for the normaldistribution is:

2.1

3.14

pop"lation mean

pop"lation stan#ar# #e(iation

where

e

π

µ

σ

≈

≈

=

=

Bluman, Chapter 6 6


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ormal istri%"tions

/ The shape an# position o$ the normal #istri%"tionc"r(e #epen# on two parameters, the meanmean an# the

stan#ar# #e$iationstan#ar# #e$iation.

/ 5ach normally #istri%"te# (aria%le has its own

normal #istri%"tion c"r(e, which #epen#s on the

(al"es o$ the (aria%le6s mean an# stan#ar# #e(iation.

Bluman, Chapter 6 7


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ormal istri%"tion roperties

/ The normal #istri%"tion c"r(e is %ell-shape#./ The mean, me#ian, an# mo#e are e!"al an# locate#

at the center o$ the #istri%"tion.

/The normal #istri%"tion c"r(e is unimo#alunimo#al i.e., ithas only one mo#e.

/ The c"r(e is symmetrical a%o"t the mean, which is

e!"i(alent to saying that its shape is the same on

%oth si#es o$ a (ertical line passing thro"gh thecenter.

Bluman, Chapter 6 9


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/ The c"r(e is contin"o"s7i.e., there are no gaps orholes. 'or each (al"e o$ X , there is a correspon#ing

(al"e o$ Y .

/ The c"r(e ne(er to"ches the x-a+is. Theoretically, no

matter how $ar in either #irection the c"r(e e+ten#s,

it ne(er meets the x-a+is7%"t it gets increasingly

closer.

Bluman, Chapter 6 10


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/ The total area "n#er the normal #istri%"tion c"r(e ise!"al to 1.00 or 1008.

/ The area "n#er the normal c"r(e that lies within

9 one stan#ar# #e(iation o$ the mean isappro+imately 0. 8.

9 two stan#ar# #e(iations o$ the mean is

appro+imately 0.:; :;8.

9 three stan#ar# #e(iations o$ the mean isappro+imately 0.:: ::.8.



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z (al"e


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Area under the Standard Normal

Distribution Curve

1& To the left of an" z $alue'

?oo& "p the z (al"e in the ta%le an# "se the area

gi(en.



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Distribution Curve

2& To the ri(ht of an" z $alue'

?oo& "p the z (al"e an# s"%tract the area $rom 1.



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Distribution Curve

3& Bet)een t)o z $alues'

?oo& "p %oth z (al"es an# s"%tract the

correspon#ing areas.



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Chapter

ormal istri%"tions


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5+ample -1> Area "n#er the C"r(e

'in# the area to the le$t o$ z B 2.0.

The value in the 2.0 row and the 0.06 column of Table E

is 0.9803. The area is 0.9803.



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Chapter

ormal istri%"tions


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'in# the area to the right o$ z B 91.1:.

The value in the –. row and the .09 column of Table E

is 0.!0. The area is – 0.!0 " 0.8830.



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Chapter

ormal istri%"tions


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'in# the area %etween z = +1. an# z = 1.3.

The values for # " $.68 is 0.9%3% and for

# " –.3! is 0.08%3. The area is 0.9%3% – 0.08%3 " 0.8682.



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Chapter

ormal istri%"tions


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5+ample -4> ro%a%ility

a. 'in# the pro%a%ility> ! 0 z 2.32

The values for # " 2.32 is 0.9898 and for # " 0 is 0.%000.

The &robabilit' is 0.9898 – 0.%000 " 0.(898.



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0.750

Find the area under the normal distribution curve.

Between z = 0 and z = 0.75

area = 0.2734

Exercise 7


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0 0.7! ".2#

0.3!!7 $ 0.2#52 = 0.""45

Find the area under the normal distribution curve.between # = 0.7! and # = ".2#. %he area is

&ound b' loo(in) u* the values 0.7! and".2# in table E and subtractin) the areasas shown in Bloc( 3 o& the+rocedure %able.

Exercise "5


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0 2.#3

Find *robabilities, usin) the standard normal distribution+-z 2.#3/.%he area is &ound b' loo(in) u* # = 2.#3 in

%able E then subtractin) the area &rom 0.5 as shown in Bloc( 2 o& the+rocedure %able.

0.5 $ 0.4!77 = 0.0023

Exercise 3"


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Find the # value that corres*onds to the )iven area.

0.#!2 $ 0.5 = 0.3!2

0 #

0.#!2

Exercise 45


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Find the # value that corres*onds to the )iven area.

1sin) %able E, &ind the area 0.3!2 and

read the correct # value corres*ondin)

to this area to )et ".2. Finall',because the # value lies to the le&t

o& 0, z = $ ".2.

0 #

0.#!2


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Chapter

ormal istri%"tions


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5+ample -;> ro%a%ility'in# the z (al"e s"ch that the area "n#er the stan#ar#

normal #istri%"tion c"r(e %etween 0 an# the z (al"e is

0.2123.

)dd 0.%000 to 0.223 to *et the cumulative area of

0.!23. Then loo+ for that value inside Table E.



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5+ample -;> ro%a%ility

The # value is 0.%6.

)dd .%000 to .223 to *et the cumulative area of .!23.Then loo+ for that value inside Table E.



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.2 Applications o$ the ormal

istri%"tions/ The stan#ar# normal #istri%"tion c"r(e can %e "se#

to sol(e a wi#e (ariety o$ practical pro%lems. The

only re!"irement is that the (aria%le %e normally or

appro+imately normally #istri%"te#.

/ 'or all the pro%lems presente# in this chapter, yo"

can ass"me that the (aria%le is normally or

appro+imately normally #istri%"te#.



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Applications o$ the ormal

istri%"tions/ To sol(e pro%lems %y "sing the stan#ar# normal

#istri%"tion, trans$orm the original (aria%le to a

stan#ar# normal #istri%"tion (aria%le %y "sing the z

(al"e $orm"la.

/ This $orm"la trans$orms the (al"es o$ the (aria%le

into stan#ar# "nits or z (al"es. Dnce the (aria%le is

trans$orme#, then the roce#"re Ta%le


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Chapter

ormal istri%"tions


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5+ample ->


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5+ample ->


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5+ample ->


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Chapter

ormal istri%"tions


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5ach month, an American ho"sehol# generates an a(erage o$ 2

po"n#s o$ newspaper $or gar%age or recycling. Ass"me thestan#ar# #e(iation is 2 po"n#s. I$ a ho"sehol# is selecte# at

ran#om, $in# the pro%a%ility o$ its generating %etween 2 an# 31

po"n#s per month. Ass"me the (aria%le is appro+imately

normally #istri%"te#.

raw the normal #istri%"tion c"r(e.

5+ample -a> ewspaper Fecycling



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5+ample -a> ewspaper Fecycling 'in# z (al"es correspon#ing to 2 an# 31.

Table E *ives us an area of 0.9332 – 0.308% " 0.62(!. The

&robabilit' is 62,.

2 20.;

2

−= = − z

te& 3/ ind the area between # " 10.% and # " .%.

31 21.;

2

−= = z



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Chapter

ormal istri%"tions


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Americans cons"me an a(erage o$ 1.4 c"ps o$ co$$ee per #ay.

Ass"me the (aria%le is appro+imately normally #istri%"te# witha stan#ar# #e(iation o$ 0.24 c"p.

I$ ;00 in#i(i#"als are selecte#, appro+imately how many will

#rin& less than 1 c"p o$ co$$ee per #ay

5+ample -> Co$$ee Cons"mption



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5+ample -> Co$$ee Cons"mption



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te& (/ To find how man' &eo&le dran+ less than cu& of

coffee multi&l' the sam&le si#e %00 b' 0.0038 to *et

.9.

ince we are as+in* about &eo&le round the answer to 2 &eo&le. ence a&&ro4imatel' 2 &eo&le will drin+ less

than cu& of coffee a da'.

5+ample -> Co$$ee Cons"mption 'in# the z (al"e $or 1.

1 1.42.

0.24 z

−= = −

te& 3/ ind the area to the left of # " –2.6!. 5t is 0.0038.



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Chapter

ormal istri%"tions


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To !"ali$y $or a police aca#emy, can#i#ates m"st score in the

top 108 on a general a%ilities test. The test has a mean o$ 200an# a stan#ar# #e(iation o$ 20. 'in# the lowest possi%le score to

!"ali$y. Ass"me the test scores are normally #istri%"te#.


5+ample -:> olice Aca#emy



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The cutoff the lowest &ossible score to ualif' is 226.

5+ample -> olice Aca#emy


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Chapter

ormal istri%"tions


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'or a me#ical st"#y, a researcher wishes to select people in the

mi##le 08 o$ the pop"lation %ase# on %loo# press"re. I$ themean systolic %loo# press"re is 120 an# the stan#ar# #e(iation

is , $in# the "pper an# lower rea#ings that wo"l# !"ali$y

people to participate in the st"#y.


5+ample -10>


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Area to the le$t o$ the positi(e z > 0.;000 0.3000 B 0.000.

*sing Ta%le 5, ) ≈ 0.4.

Area to the le$t o$ the negati(e z

> 0.;000 9 0.3000 B 0.2000.*sing Ta%le 5, ) ≈ 90.4.

The mi##le 08 o$ rea#ings are %etween 113 an# 12.

5+ample -10>


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ormal istri%"tions

/ A normally shape# or %ell-shape# #istri%"tion isonly one o$ many shapes that a #istri%"tion can

ass"meJ howe(er, it is (ery important since many

statistical metho#s re!"ire that the #istri%"tion o$

(al"es shown in s"%se!"ent chapters %e normallyor appro+imately normally shape#.

/ There are a n"m%er o$ ways statisticians chec& $or

normality. Ke will $oc"s on three o$ them.



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Chec&ing $or ormality

/ Histogram/ earson6s In#e+ I o$


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Chapter

ormal istri%"tions


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A s"r(ey o$ 1 high-technology $irms showe# the n"m%er o$

#ays6 in(entory they ha# on han#. etermine i$ the #ata areappro+imately normally #istri%"te#.

; 2: 34 44 4; 3 4 4

1 :1 : : 113 11 1;1 1;

Metho# 1> Constr"ct a Histogram.

The histogram is appro+imately %ell-shape#.

5+ample -11> Technology In(entories



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Metho# 2> Chec& $or Chec& $or D"tliers.

'i(e-"m%er Technology In(entories

( )3 :.; .;3 I 0.14

40.;

−−= = =

X "D

s

:.;, .;, 40.;= = = X "D s



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A s"r(ey o$ 1 high-technology $irms showe# the n"m%er o$

#ays6 in(entory they ha# on han#. etermine i$ the #ata areappro+imately normally #istri%"te#.

; 2: 34 44 4; 3 4 4

1 :1 : : 113 11 1;1 1;

Concl"sion>

/The histogram is appro+imately %ell-shape#.

/The #ata are not signi$icantly s&ewe#.

/There are no o"tliers.

Th"s, it can %e concl"#e# that the #istri%"tion is appro+imately

normally #istri%"te#.

5+ample -11> Technology In(entories



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ection 4

6**lications o& the ormal 8istribution

9ha*ter

%he ormal 8istribution

ection 4 Exercise 3


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a. :reater than 700,000.

b. Between 500,000 and 00,000.

%he avera)e dail' ;ail *o*ulation in the 1nited tatesis "#,3"!.


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a. :reater than 700,000

7 ( # ".3) = 0.5 –0.44#4

= 0.05" or 5.">

# =

700,000$"#,3"!

50,200 = ".3

0 ".3

# = – µ

σ


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b. Between 500,000 and 00,000.

# =

500,000$"#,3"!

50,200 = $ 2.3

area = 0.4!0!

# =00,000$"#,3"!

50,200 = $ 0.3

area = 0."40

# = – µ

σ


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ection 4 Exercise ""


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%he avera)e credit card debt &or colle)e seniors is

@322.


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# = $

a. %hat the senior owes at least @"000

# ="000 $ 322

""00

= $ 2.0

area = 0.4#03

= 0.!#03 or !#.03> 7- # $ 2.0/ = 0.5 A 0.4#03

0.!#03 or !#.03>

$ 2.0 0


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# = $

b. %hat the senior owes more than @4000

# =4000 $ 322

""00 = 0.7

= 0.25"4 or 25."4> 7 - # 0.7/ = 0.5 $ 0.24#

area = 0.24#

0.25"4 or 25."4>

0.70


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# =

$

c. %hat the senior owes between @3000 and @4000.

# = 3000 $ 322

""00

= $ 0.24

area = 0.0!4#

= 0.3434 or 34.34> 7- $ 0.24 ? # 0.7/ = 0.0!4# A 0.24#

0.3434 or 34.34>

0.7$ 0.24 0

ection 4 *xercise +2!


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6n advertisin) com*an' *lans to mar(et a *roduct tolowincome &amilies. 6 stud' states that &or a *

@2

ar

4,

ticu

5!

lar

area, the avera)e income *er &amil' is and thestandard devia @25tion is .


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%he bottom "#> means that 32> o& the area is between

# and 0. %he corres*ondin) # score will be .$ 0.!2

@"#,#40.4# @24,5!

0.320."#

= $ 0.!2-25/ A 24,5!

= @"#,#40.4#


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ection 5

%he 9entral imit %heorem

9ha*ter


ection 5 *xercise +13


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%he avera)e *rice o& a *ound o& sliced bacon is @2.02.6ssume the standard deviation is @0.0#.


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@2.00 @2.02

%he avera)e *rice o& a *ound o& sliced bacon is @2.02.6ssume the standard deviation is @0.0#.


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%he avera)e time it ta(es a )rou* o& adults to com*letea certain achievement test is 4.2 minutes. %he

standard deviation is # minutes. 6ssume the variableis normall' distributed.

6vera)e time = 4.2 minutes, tandard deviation = # minutes, variable is normall' distributed.

a. Find the *robabilit' that a randoml'selected adult will com*lete the test

in less than 43 minutes.

b. Find the *robabilit' that, i&

50 randoml' selected adults ta(ethe test, the mean time it ta(es the

)rou* to com*lete the test will be

less than 43 minutes.

6vera)e time = 4.2 minutes, tandard deviation = #


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c. 8oes it seem reasonable that an adult

would &inish the test in less than43 minutesC Ex*lain.

d. 8oes it seem reasonable that the

mean o& 50 adults could be less

than 43 minutesC Ex*lain.

e a)e t e utes, ta da d de at o #minutes, variable is normall' distributed.

a. Find the *robabilit' that a randoml'


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selected adult will com*lete the test inless than 43 minutes.

# = $

= 43 $ 4.2#

= $ 0.4

area = 0."554

7

- # 9

$0.4/ = 0.5 $0."554 = 0.344or 34.4>

43 4.2

0.344or 34.4>

b. Find the *robabilit' that, i& 50 randoml'selected adults ta(e the test the mean


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selected adults ta(e the test, the meantime it ta(es the )rou* to com*lete thetest will be less than 43 minutes.

# = 43 $ 4.2#

50

= $ 2.#3

area = 0.4!77

7- # 9 $ 2.#3/ = 0.5 $0.4!77 = 0.0023or 0.23>

43 4.2

0.0023or 0.23>

c. 8oes it seem reasonable that anadult would &inish the test in less


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adult would &inish the test in lessthan 43 minutesC Ex*lain.

Des, since it is within onestandard deviation o& the mean.

d. 8oes it seem reasonable that the

mean o& 50 adults could be lessthan 43 minutesC Ex*lain.


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%he avera)e cholesterol o& a certain brand o& e))s is2"5 milli)rams, and the standard deviation is "5

milli)rams. 6ssume the variable is normall'distributed.

a.


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# = $

*robabilit' that the cholesterol content will be )reater than milli)r220 ams.

= 220$

2"5"5

area = 0."2!3

= 0.33

7- # 0.33/ = 0.5 $ 0."2!3 = 0.3707or 37.07>

2202"5

0.3707or37.07>

b.


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# =

8 $

n

=

220$ 2"5

"5

25

= ".7

area = 0.4525

the mean o& the sam*le will be lar)er than milli)r220 ams.

2"5 220

7- # ".7/ = 0.5 $ 0.4525 = 0.0475 or 4.75>


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ection

%he ormal 6**roximation to

%he Binomial 8istribution

9ha*ter


ection Exercise 5


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%wo out o& &ive adult smo(ers acuired the habit b' a)e

"4.


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"!.5"0

7- "!.5/ = 0.5 $ 0.3340 = 0."0

ection Exercise 7


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%he *ercenta)e o& 6mericans 25 'ears or older whohave at least some colle)e education is 50.!>.


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"75.5"52.7

7- "75.5/ = 0.5 $ 0.4!57 = 0.0043

ection Exercise ""


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omen com*rise #3.3> o& all elementar' schoolteachers.


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50.550."

7- 50.5/ = 0.5 $ 0.023! = 0.47"

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