Time:

Instructor: Tosha Lamar Assignment: chp 8 examplesCourse: Fundamentals of Statistics (l)Book Sullivan: Fundamentals of Statistics,4e

-)7-7

t"'- =wyricriilii.l

Complete parts (a) through (d) for the sampling distribution of the sample mean shown in tlle

accompanying graph.

'*:-; Click the icon to view the graph.

(a) What is the value of p;?

The value of p- is .-VCC

(b) What is the value of o-?

rit

deviation of the population from which the sample was drawn

6ot\

/x I t+* f{srThe value of o-is 4fl

'/"li-''7tt''1tr7Llc, =dg

(c) If the sample size is n: 16, what is likely true about the shape of the population'?

, - r\ The shape of the population is skewed right.

.- !l The shape of the population is skewed left.,F-"

Qfn" shape of the population is normal.

' The shape of the population cannot be determined.

(d) If the sample size is n: 16, what is the standard deviation of the populafion from which the sample

was drawn?

The standard

fcv- 2 g'

Sampling Distribution L-,i. '.f LJt = L,i l@

2. Determine F; and o- from the given parameters of the population and sample size. =LlL: 43, o: 8, n: 30

I'r = /t$;: +3

f.i.t3.

'r:'C_o j,11

{ i1<r,;itii tr,i llrr',:,,i iir:iii,'iiri pi;i;*n :is

(--/{-

\11

q

--\34

'il

t & Lo'ititLct*-r\'-

L+4t)=1 Itro -: ff

Student:Date:

Instructor: Tosha Lamar Assignment: chp 8 examplesCourse: Fund"mentals of Statistics (l)Book Sullivan: Fundamentals of Statistics,4e

Time:

4. A simple random sample of size n: 64 is obtained from a populatidn with p: 86 and o :32.(a) Describe the sampling distribution of x.(b) What is P (x > 93.2)?(e) What is r (x < 77.8)?(d) What is P (81.4< *. gS)r

(a) Choose the correct description of the shape of the sampli distribution of x.

, -;A

The distribution is skewed left.

' .,il The distribution is uniform.

, . .r The distribution is skewed right.

---''.-9*" distribufion is approximately normal.

,-.,E The shape of the distribution is unknown.

ff1e 3zaLz rt+tzuL L) rWi +

*Tlurrern-

Find the mean and standard deviation of the sampling distribution of x.

t,t = l,l{ l"l; gA

u; : 86' r'o;: tr gr =# Ur= #.+&) p (; > %.2) :i

t*,,ig u,4uu,. crccinrar priucr ris ,rce dcrr. ,

{lo{wlal dll ( qz''f ic^ tt9' 8u' *)

(c) P (x < 77.8) : ,i tqut:,na to i,:u' ri*e ir:al placcs ;1:: rr,jittr;d. ,flt{ YlAlCdl (nnlftr7L'Klgb)4)-u,02LL

(d) P (81.4.*.q5): _ (R')linillr: !i,tir ri;:cir':uri i:lu;rs rs r:ctdu,i.i

t,\v1-1 lcrnta{cri+($ l,*t%t{b,4)

Page 3

Student:Date:

Instructor: Tosha Lamar Assignment: chp 8 examplesCourse: Fundamentals of Statistics (l)Book Sullivan: Fundamentals of Statistics,4e

The probability that the mean of a a random sample of 17 pregnancies is less thm 279 days is

The e mean of a a random sample of 33 pregnanqies is less than 279 days is

titii tll lbitt'iir..;r:;.r! itlliftr iis :r:":Ji,.i. i

{1"0r iqt.{Llcilf (-hoq*l ) 2:7':1 ) Ul l, 5, 2 2 z fi)

Time:

5. Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed withmean [r: 291 days and standard deviation o:30 days.

(a) What is the probability that a randomly selected pregnancy lastsJggsjhsl 279 days?- c,y+bThe probabllity that a randomly selected pregnancy lasts less than 279 days is approximately r ..

{Rr:urrrl iii rirr,ri: iice iri:*i ririi:s,rs ncc,.lcri. ttltifl1lAl Cd( kiCAQQl '2741 /til, lq(b)Whatistheprobabilitythatafan@gnancieshasameangestationperiodof279days or less? pL = fi_l' r'^ bi{ u{'# -7,27607

*]:l*"i; ,iii$']" "''iir

iii;i'ra' ''":*:..;;?;i dlutl f tcn'rl,xQ,Lqt)1,2teg7)

(c)Whatistheprobabilitythatarandom@cieshasameangdstationperiodof279daysorless?

-(,_=-+r$,ZZZZ-{}

Page 4

Student:Date:

Instructor: Tosha Lamar Assignment: chp 8 examplesCourse: Fundamentals of Statistics (l)Book Sullivan: Fundamentals of Statistics,4e

(a) What is the probability that a randomly selected time interval between eruptions isJonger than 94minutes?

Time:

6. Suppose a geyser t ur@me between eruptions of E*irut"q. If the interval of time between the

eruptions is normally ifrsqffgkalwith standard deviation 22 minutes, answer the following questions.

is longer thm 94 minutes is approximately

c,*ilL(b) What is the probability that a randor$ sample of 15 time intervals between eruptions has a meanlonger thang4minutes?

n =i5 f;; = # _ 5u A\C 47hThe probability that the mean of a random sample of 15 time irt.#, is.mBlXtfran q+ minutes is

approximately.1Hi.i'ilti,ritii*i,l-i.r-ii:iir,-illii:i;.i,:':i.:::']iijijiii.-i.,;,y,;.W,K1,5,$e4./dD, C5ab t

(c) What is the probability that a random sample of 28 time intervals between eruptions has a meanIonger thang4minutes?

-

n=F (7= #:4,tS'/((:C'lr. -LXThe probability that the mean of a random sample of 28 time intervaldis inore than 94 minutes is

approximat.rv-.i,l}.r:lt;.lliiir1l,iirl.il.,:il.:iiiii1li:'l,:i":ill:,:lii:':iiii'i.j;;;;;iWl,ir;rr)c,tl5L

(d) What effect does increasing the sample size have on the probability'? Provide an explanation forthis result. Choose the correct answer below.

,* The probability decreases because the variability in the sample mean increases as the sample

1)*,size increases.

( )he proUrbility decreases because the variability in the sample mean decreases as the sample\--,/ size increases. -frc,-rrrl ,11U

, . The probability increases becaLse thJ'variabilify in the sample mean increases as the samplesize increases.

, ;-r The probability increases because the variabilify in the sample mean decreases as the samplesize increases.

(e) What might you conclude if a random sample of 28 time intervals between eruptions has a meanlonger than 94 minutes? Choose the best answer below.

The population mean is 85 minutes, and this is an example of a typical sampling.

1-,The population mean may be greater than 85.

The population mean must be more than 85, since the probability is so low.

The population mean must be less than 85, since the probabilify is so low.

l'ltrntil t't|{(r1, rcr(fi,85 )},a)The probability that a randomly selected time interval

iFiillillii Ti.l iirifii dlc!;::i,i i:iiti.": :i:,, i:eril;o!. )

Page -i

Studeut:Date:

Instructor: Tosha lamar Assignment: chp 8 examplesCourse: Fundamentals of Statistics (l)Book: Sullivan: Fundamentals of Statistics,4e

The shape of the distribution of the time required to get an oil change at a 20-minute oil-change facilityis unknown. However, records indicate that the mean time is2l.2-mi utes, and the standard deviation,.:!ggryu. Completeparts (a) through (c).

(a) To compute probabilities regarding the sample mean using the normal model, what size samplewould be required'?

Time:

or below? This will be the goal established by the manager.

There is a p9fchancee(Qerng at or below a mean oil-change |imeillr:ll;iii ft L;ri; iic"_r.:j:.dJilg&rb nc,id*ii. i

r]=+c

/N( irrr, ilcr fi (c,, D)'= - l,l\

D The normal model cannot be used if the shape o tion is unknown.

(b) What is the probability that a random sample of n:45 oil changes results in a sample mean time

flt"(c) Suppose the manager agrees to pay each unployee a $50 bonus if they meet a certain goal. On afypical Saturday, the oil-change facility will perform 40 oil changes between l0 A.M. and 12 P.M.Treatingthisasarandomsample,whatmeanoil-ch@al0Yochanceofbeingat

brtio, c( ?

F,of 37+

{,t* - 7'7 =C,fl\7't't"'d' n+5

rl --{jl.j--"x lrt

St,*hts is flt" z--Yrra

x - di,A

0,t5 b+

11 Any sample size could be used.

.,. .,D The sample size needs to be less than or equal to 30.

--u

Iess than 20 minutes?

The probability is a

2.2'

i;-'----*t-'N vn

?'x;

rn1lLl cd f (*lCt'iq, k,,) l, 1, t, 34 t 7

3,q-14a

s Co 55ltpl,

a *l,At =

Page6 c,5{7',1 G l'ls) = f, -)t' l,-

"lc?,351 -- [ -at'e

*c,'{tlkt8 = {.

Student:Date:

Instructor: Tosha Lamer Assignment: chp 8 examplesCourse: Fundamentals of Statistics (l)Book Sullivan: Fundamentals of Statistics,4e

The acceptable level for insect filth in a certain food item is 3 insect fragments (larvae, eggs, bodyparts, and so on) per 10 grams. A simple ymdom samplegllfllgn-gram portions of the food item is

obtained and results in a sample mean of x : 3.5 insecffragments per ten-gram portion. Complete parts(a) through (c) below. /" ^?tl = bc@J(hv is the sampling distribution of x approximately normal'?

/,-)) ) i. a. .,, a, . x>y( ,' lhe sampling distribution is approximately normal because the sample size is large enough.\-/ - -;ytr.'e.

F.: The sampling distribution is approximately normal because the population is normally I C

distributed and the sample size ls large enough. r -r ---- - - - - -J

3'A(-' The sampling distribution is approximately normal because the popluation is normally

distributed.

. .i ' The sampling distribution is assumed to be approximately normal.

(b) What is the mean and standard deviation of the sampling distribution of x assumirrfirrdo:{: r Y-

lf lt, W-*p : ? iF,.'irrtri I:l ii:t,'f l!;;ri;i,iiiii:r,;ts iri ri,'..'iit,,i. , / /'x J r*

6;: ii{*ui:j i,r iiri'c*,lurirr.ir *i.i;". ri,t i!"=ricii.i [i. = g - Ao A+f i+E'1713't'rA+\ :- "x n6i

(c) What is the probability a simple random sample of 50 ten-gram portions of the food item results ina mean rf d mlA_fj3gect fragments?

: = 5t)n,;= rffi;1r3, !,, ii)i{i ii,, ;rj,:ri i;:itiri:i ii:; *iriii:ii.i

ttirnlttj tll( S,s, ICA{,Ftr?r/ Le+)

=(nLlL(11'1t*+

Time:

a

Is this

This result is unusual because its probability is large.

This result is not unusual because its probabilify is small.

This result is not unusual because its probability is large.

What might we conclude?

is result is unusual because its probabil iry {ffi) { 1 , ( 5

5.

(coni. )

Since this result is not unusual,

i Since this result is not unusual,is higher than 3.

it is reasonable to conclude tnut tlre population mean is

it is reasonable to conclude that the population mean

reasonable to conclude that the population mean is 3.

reasonable to conclude that the population mean

3.

. Since this result is unusual,/-\-( y'ince rhis result is unusual,

is higher than 3.

it is

it is

- ; { i(l'i

CACbc

9. The amount of time adults spend watching television is closely monitored by firms because this helpsto determine advertising pricing for commercials. Complete parts (a) through (d).

(a) Do you think the variable "weekly time spent watching television" would be normally distributed?

l, No

. yes

If not, what shape would you expect the variable to have?

The variable "weekly time spent watching television" is likely qrmmetric.

The variable "weekly time spent watching television" is likely skewed left.

The variable "weekly time spent watching television" is likely uniform.

The variable "weekly time spent watching television" is likely skervg

(b) According to a certain suryey, adults spend 2.25 hours per day watching television on a weekday.Assume that the standard deviation for "time spent watching television on a weekday" is 1.93 hours. Ifa random sample of 50 adults is obtained, describe the sampling distribution of x, the mean amount oftime spent watching television on a weekday.

/l= 5O

C.2721+3 (7<,//

a'= fr-\

ubt

has a uniform distribution 5i and o;: , -. ',.

E n,_2.15 r- y

- L, )7 Z',l+\(c) Determine the probability that a random sample of 50 adults results in a mean time watchingterevisiononaweekdavorbetween2andrt#--

nr;;-liiiif (i ;,r'.,;;,"2-r, Yt,li)The probability is : iF:-i;ri,.;.i iri fr:.ir ii.--:i*:al placcs as reciled.)

-_ s\ e t 1 .)

---U,cllt_(d) One consequence of the popularity of the Internet is that it is thought to reduce television watching.Suppose that a randgrn sg.mp!go[45 individuals who consider themselves to be avid Internet usersresults in a mean time of 1.88 hours watching television on a u,eekday. Determine the likelihood ofobtaining a sample mean of^l.88 honrc or less from a population whose mean is presumed tobe 2.25hours.

The likelihood is ioiil li:i;::':! i.ii:.-s as lciricC. i

f\, +5 W/y)cr-rn&l adl fr / tt4

, | ,89 /':;" ,' A5, ( n as77)

: t;,-c33-L

,Lr'"{

=Litr l(=2,)s-/' ,/

-. ir13 - t\- _ _ L,Ag77w

---------*9.

{cont. }

is approximately normal

Based on the result obtained, do you think avid Internet users watch less television?

Student:Date:

Instructor: Tosha Lamar Assignment: chp 8 examplesCourse: Fundamentals of Statistics (l)Book Sullivan: Fundamentals of Statistics,4e

In the game of roulette, a wheel consists ofi8_Slpltnumbered 0, 00, 1, 2,..., 36. To play the game, ametal ball is spun around the wheel and is allowed to fall into one of the numbered slots. If the numberof the slot the ball falls into matches the numberyou selected, you win $35; otherwise you lose $1.Complete parts (a) through (g) below.

Time:

10.

tlt'utu+t-

(a) Construct a probability distribution for the random variable X, the-winnings of each spin.

enr slc I c*t o+ 3t slats_xP(x)jr5ffi1'3 Lb3 I / 38' oo ttab\

{\ - 1 , 6,Cn 37 (Wcautn I - C"tibl= C* q737)iI",rpc ini;gcrs ol ilecil"::als i'irirudcd io i*,,ri'ii*cinral piaces ns ncciicd,t

LL\LL ,I'emean;"P % L-Var3 S l L, tOrLU)> -J- u wfi fu,edi W4l*

4o,t 7= -oc5

a-hd- ux- 5"7b

:)otv /'"t- ,4. -_C"Cb)Z(c) Suppose that you play the gamEjq times so tha{n : 60. Describe the sampling distribution of x,rhemean**t*o"ww-' /

^ \rn e^it{<p t/r# c*Jr"Jc.*n

- l - i ---*_*-F -7 altfr-sWhat are the mean and standard deviation of the sampling distribution of x? Round your resi{ts to the

probability that the g$ils mean irgreal$.l!g;!j- n:60'.)

P(;> o) :'C,+1,31' iiii"t r;f (,; ) )a

^q?, -.cs, a d

(d) What is the probabilify of being ahead after playing the game 60 times? That is, what is the

LT'r'pc nn int*.1iti itr: de citnai riiunil"ij i.o i,;ul 11,:cir:'rai iliaci:s a.s fietileii.)

(e) What is the probability of being ahead after playing the game 180 times?

Page I I

Student:Date:

Instructor: Tosha Lamar Assignment: chp 8 examplesCourse: Fund"mentals of Statistics (l)Book Sullivan: Fundamentals of Statistics,Time:

r0.

(cont.)

4e

P(;> o) :'0bt51€ l)=lt} ff-= !ft 'S-ab- .-@

0.4Lqi trr

xt'Iypc an intcger rir fi;r+*ri rolllrdcd to f'our ticcimal plg.lJ;, ne cd-'il" )

(g) Compare the results of parts (d) through (f). What lesson does this teach you'?

ffiprobabilityofbeingaheaddecreasesasthenumberofgamesplayedincreases.V.,.:g The probability of being ahead increases as the number of games played increases.

-jc The probability of being ahead decreases as the number of games played decreases.

.:D The probability of being ahead remains the same as the number of games played varies.

n= bc\ x0,"352

nm. rrrrad{(c) I 0 n%r-,05) c':}?j50

5,'7b

Page 12