(a) The pmf ofthe random variable x is
2 52
CE I j-,x=o,1,2.
The value of C is calculated is as follows.
2
Z p|:x) :1, we have
:Ir-IJ
:>p|:0)+p[:l)+p|:2:|=l
250 9C C
:>+=
7" 7" 7"
1
The value of C is calculated is as follows.
2
z p|:x) :1, w
(as) The sample size is as: 25.
The sample mean isf = 1221:
as
1-1
= 2;[26+2?+.+21)
_ ass
as
E: 22.4
The known standard desiati an is J: 3.
Te eenstruet a 93% eenfidenee interval estimate efthe mean breaking strength efthe wire.
_ J C"
XEsngx+3
a at?
2241
The simple rand-am sample pf size as = 125 SAT see-res.
The sample mean isp.r = 1522.
The standard deviatien is J: 333.
rCensti'net a 95% eenfidenee interval estimate of the mean SAT seere is
20EE+ET
=15221.96[ 333 gig in _1522+1. 96[ 333]
.1125
=15221.
(g) Inee-me:
The level of signicance is 1 cfw_1: = 5.99
The sample size is as = 91
The sample mean is I? = 43
The standard desiatinn is 3 =123
The formula fer ealeulating the 99% ennfi denee interval is
_ .S'
1: 53+
4T .55 EraINF;
243%? 12s 12s
_1 E in E
Instructions to obtain Chi-s quare test for Asscoicati on output using Minitab:
* Store the Sea: in the column Cl and Polytical party in the rest ofthe columns of the Minitab
worksheet.
* Choose Stat Tables 4,: Chis quare test for Assoicaition.
* Choose S
Instructions to obtain Chi-s quare test for Asscoicati on output using Minitab:
* Store the Media in the column Cl and Age ICrroup in the rest ofthe columns of the h-linitab
worksheet.
* Choose Stat Tables 4,: Chi-square test for Assoicaition.
* Choose Su
(:1) Determine the value ofthe confidence coefficients for 1ct= 0.33.
(an:
51: 0.1?
cfw_1 0.1?
=:~=
2 2
290.035
We want the 2 values which have 33% centered on the mean.
That is the same as the z values which have a left tail of 0.035% and a right tail of
(a) The eonfidenee level is l cfw_3:090
The level of signifieanee is es 21090
= [3.10
mm
ICED) The eonfidenee level is 1513: 0.99.
The level of signifieanee is es 21099
= [3.01
mm
ICC) [n)The margin of error can be determined if vou know onlyr the eonfide
(a) The given p.-:1.f. is given hv,
fix) =cfw_Cxagm .2:- U
U ethenvise
The value of C is obtained hv integrating the denaitv funetian with reapeet to U to 0: and equate it to
1.
jf[x)ax=1
u
ICxede=1
e
CIxade=1
n
:Ltu =3: anddv=d:
E5
E E
Letnzjthen
Explanation:
The central limit theorem maintains that the sampling distribution of statistics will follow a
normal distribution.
Honors, tho answor is that tho distribution basod on samplos tond to bo normal.
The average salary for a eertain prdfessi-sn is p! = $37, 000.
assume that the standard deviation of such salaries is 0': $36,000.
The rand-am sample of 30 people in this professidn and let
3 represent the mean salary for the sample.
(a) The mean is #3. =
The sample size is as = 24.
The mean noise level vvas E = 41.6 deeihels.
The standard deviation is J: T.5deeihels.
The 95% eonfidenee interval is
J
Xzgj $5,515 X+zmj
7".5 TS
=4l.dl.96 5n541.6+l.96[]
[#24] #24
=4l.6l.96 i Err541o+l96 i
4.3989T9486 4.89897
(a) The manufacturer elaims that the life ofthese tires follows
a normal distribution with a mean ofpr = 60, 000
The standard deviation is J: 5, 000 miles.
The proportion of the tires should last less than 50,000 miles is
PU: cfw_50000) = P[X# (W)
J J
[
[2] Estimate the differenee in a way that provides information about the preei si on and reliability.
2
32
The 95% eonfiolenee interal for 1511 1.112: [ijizm 31 +
222 22
1._3:*Jr
=.(1o2 6 123 51+125
122 +122
1+69 4
129 +129
= 16i1.96JU.U131UUB+U.031UUB
=
[njlnstruetions to obtain Chi squsre test for i'ltsseoiestion output using Mnitsh:
* Store the Wins in the eolumn Cl and Sport is in the rest of the eolumns of the Mnitsh
worksheet.
* Choose Stat Tshles 4,: Chisquare test for i'l'rssoiesition.
* Choose Su
(or) The sample size is 22: 12.
The sample mean is f = 32502252
The sample standard deviation is 0': 31622282252
The level of significance is or = 0.05.
The critical values of the normal distribution is
3.222 2322522
The formula for calculating the 95% co
Explanation:
One interesting property ofs seores is the fast that the sample 3 seores will alwavs
have amean of U and standard deviation of 1. This means that vou ean eomputezseores
for several datasets and and eompare them direetlv as to the relative var
Instructions to obtain Chis quare test for Asscoicati on output using Minitalo:
* Store the Success rates in the column Cl and Therapies in the rest of the columns of the Mnitah
worksheet.
* Choose Stat Tahles 4; Chisquare test for Assoicaition.
* Choose
The proportions are:
n1: s13; = 0050,
91 =1P1
=1 0.?50
as: = s13; = 0.015
9: =1 F's
= 1 0.615
55?; = 0.ss5
The ee-nfi denee intervals for difference of props-rti ens is
_H"-\.
p1_g,;izg_ _ 40.0100, 0.2701)
PI F-z
At first, we need to ealeulate the value o
Stepl:We set up the null hvpethesis that the number efhen1e team
and visiting team wins is independent of the sport.
He : The number at henie team and visiting team wins is dependent efthe speit.
H1The nuinher efhenie team and visiting team wins is indepe
(a) The probability of success is p = 0.25.
The probability of failure is
9 =1- .9
= l 0.25
The number of trails is 29 = 10
Let)? be the random variable follows a binomial distribution.
X~Bn=l0p=025j
The pmf of binomial distribution is
me 0 =90)
22
[ Jpqu
The sample size is as =12.
The sample mean is f = 4.56
The sample standard desiatien is J: [131]
The eenfi denee level is 1 CE = [3.93
21 [3.93
Ct: [3.132
The eritieal values efthe nermal distnhutipn is
3m: = 3seats
The formula far calculating the 93% epn
(a) Step1:We set up the null hvpothesis that the qualitv of the flash drive produetion
and the produeti on shift are independent variables.
H0 :The qualitv of the flash drive produetion and the produetion
shift are dependent variables .
H1 :The qualitv of
Instiuetions to obtain Chisquare test for Asseoieation output using Minitah:
* Store the |Ereneler in the eolumn Cl and Health problems in the rest of the eolumns of the Mnitab
worksheet.
* Choose Stat rTables r Chisquare test for Assoieaition.
* Choose S
(e) Age:
The level of signifieanee is 1 or: 0.95
The sample size is as = 51
The sample mean isf=45
The standard deviation is s = 6.2.
The formula for ealeulating the 95% eonfi denee interval is
Xr_1i5ngf+r
i
a "109
as as
=45r _ 52545+: _
511 E51 511 [51
(:1) Instructions to obtain Chisquare test for i'l'isseoiestion output using Minitsh:
* Store the Computer i'l'is sistedordering in the eolumn C1 and Delivery Time
in the rest of the eolumns of the h-Iinitsh worksheet.
* Choose Stet Tshles 4,: Chis qusre
(is) The eonfidenee level fot the interval
Eamisnsfwm
a ' E
z ,2 = 2.14
To determine the eonfi denee level, vve use eveel funetion to get the requrieol
eonfi denee level.
=nor1nsdi shire
=nor1nsdist(2.l4)
=9933323
Therefore, the eonfiolenee level is appro
The sample size is as = 40.
The sample mean isf=44 eunees
The sample standard deviatien is J: 5.0 eunees
The ee-nfidenee level islcfw_l= 0.90
= l 0.99
.51: 0.01
The eritieal values at the normal distribution is
zmE =3asirs
= 2.5?5
3is JZIIIIS
The fermula
(a) In the binomial distribution, The probability of a sueeess,
denoted by p, remains eonstant from trial to trial and repeated trials are independent.
Henee the answer is B.
(is) The sum ofthe probabilities of all possibleouteomes must be 1.
Henee the an
Major difference:
Heroin:
The population ean be divided into groups ealled the strata.
The strata look different from eaeh other, but their individual members are all the same
in some sense.
Disasters:
The population ean be divided into groups ealled the