gmat quant topic 2 statistics solutions

8/9/2019 GMAT Quant Topic 2 Statistics Solutions

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MEAN

1. 2002 total = 60, mean = 15, in 2003, total = 72, mean = 18, from 15 to 18,increase = 20%

2. A 200% increase over 2,000 products per month ould !e 6,000 products permonth. "#ecall that 100% = 2,000, 200% = $,000, and 200% over means

$,000 & 2,000 = 6,000.' (n order to avera)e 6,000 products per month overthe $ *ear period from 2005 throu)h 2008, the compan* ould need toproduce 6,000 products per month + 12 months + $ *ears = 288,000 totalproducts durin) that period. e are told that durin) 2005 the compan*avera)ed 2,000 products per month. -hus, it produced 2,000 + 12 = 2$,000products durin) 2005. -his means that from 2006 to 2008, the compan* illneed to produce an additional 26$,000 products "288,000 2$,000'. -hecorrect anser is /.

3. A$. 5. 6. /7. 8. . /10. 11. -his uestion deals ith ei)hted avera)es. A ei)hted avera)e is used to

com!ine the avera)es of to or more su!)roups and to compute the overallavera)e of a )roup. -he to su!)roups in this uestion are the men andomen. ach su!)roup has an avera)e ei)ht "the omen4s is )iven in theuestion the men4s is )iven in the rst statement'. -o calculate the overallavera)e ei)ht of the )roup, e ould need the avera)es of each su!)roupalon) ith the ratio of men to omen. -he ratio of men to omen oulddetermine the ei)ht to )ive to each su!)roup4s avera)e. oever, thisuestion is not asin) for the ei)hted avera)e, !ut is simpl* asin) for theratio of omen to men "i.e. hat percenta)e of the competitors ere omen'."1' (9:;3' of the competitors must have !een omen.onsider the folloin) rule and its proof.#;? -he ratio that determines ho to ei)ht the avera)es of to or moresu!)roups in a ei)hted avera)e A?:@ #


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9otice that the ei)hted avera)e is tice as close to the men4s avera)e as itis to the omen4s avera)e, and notice that this reCects the fact that thereere tice as man* men as omen. (n )eneral, the ratio of these distancesill ala*s reCect the relative ratio of the su!)roups. -he correct anser is "D', :tatement "2' A?@9 is suEcient to anser theuestion, !ut statement "1' alone is not.

12. e can simplif* this pro!lem !* usin) varia!les instead of num!ers. x =5$,820,

x & 2 = 5$,822. -he avera)e of "5$,820'2 and "5$,822'2 =

9o, factor x 2

& 2 x &2. -his euals x 2

& 2 x &1 & 1, hich euals " x & 1'2

& 1.

:u!stitute our ori)inal num!er !ac in for x as follos

" x & 1'2 & 1 = "5$,820 & 1'2 & 1 = "5$,821'2 & 1.

-he correct anser is /.

13.


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"12'"20' & $0n = 25n & "12'"25'15n = "12'"25' "12'"20' 15n = "12'"25 20'15n = "12'"5'

15n = 60n = $:ince it ould tae $ fort* oJ. !ottles alon) ith 12 tent* oJ. !ottles to *ield anavera)e volume of 25 oJ, 36 $ = 32 fort* oJ. !ottles must !e sold. -he correctanser is /.

15. -he avera)e num!er of vacation da*s taen this *ear can !e calculated !* dividin)the total num!er of vacation da*s !* the num!er of emplo*ees. :ince e no thetotal num!er of emplo*ees, e can rephrase the uestion as o man* totalvacation da*s did the emplo*ees of ompan* K tae this *earL"1' (9:;emplo*ee' = 80 da*s80 vacation days were taken last year. ence, the total number of vacationdays taken this year was 100 days.9ote (t is not necessar* to mae the a!ove calculations OO it is simpl* enou)h tono that *ou have enou)h information in order to do so "i.e., the information )ivenis suEcient'P -he correct anser is D.

16. -he uestion is asin) us for the weighted avera)e of the set of men and the set of omen. -o nd the ei)hted avera)e of to or more sets, *ou need to no theavera)e of each set and the ratio of the num!er of mem!ers in each set. :ince eare told the avera)e of each set, this uestion is reall* asin) for the ratio of thenum!er of mem!ers in each set. "1' :;


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have no idea ho man* men and omen. -his )ives us no indication of ho toei)ht the avera)es. -he correct anser is A.

17. -he mean or avera)e of a set of consecutive inte)ers can !e found !* tain) theavera)e of the rst and last mem!ers of the set. Fean = "O5' & "O1' > 2 = O3. -he

correct anser is D.

18. -he formula for calculatin) the avera)e "arithmetic mean' home sale price isas follos

A suita!le rephrase of this uestion is Shat as thesum of the homes sale prices, and ho man* homesere soldLT

"1' :;


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must !e euidistant from the eGtreme values of that set, hich are x and y . :o thedistance from x to 75 must !e the same as the distance from 75 to y . e can eGpressthis al)e!raicall* 75 x = y 75 150 x = y 150 = y & x e are ased to ndthe value of 3 x & 3 y . -his is euivalent to 3" x & y '. :ince x & y = 150, e no that3" x & y ' = 3"150' = $50. Alternativel*, the median of a set of consecutive inte)ersis eual to the avera)e of the eGtreme values of the set. 7 = 530$ q"s m' ="7'"530$'.

-o consider possi!le value for the diNerence !eteen the sale price and the costper unit, s m, let4s loo at the prime factoriJation of "7'"530$'"7'"530$' = 7 + 2 + 2 + 2 + 3 + 13 + 17

:ince q and "s m' must !e multiplied to)ether to )et this num!er and q is aninte)er "i.e. Y of units', s m must !e a multiple of the prime factors listeda!ove.A )rid.

"1' (9:;


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needed here is a ei)hted avera)e of the avera)e A;F per customer for the 10!ans. ach !ranch4s avera)e A;F per customer needs to !e ei)hted accordingto the number of customers at that branch hen computin) the overall avera)eA;F per customer for the hole !an.

?et4s loo at a simple eGample to illustrate

(f e tae a simpleavera)e of theavera)e num!er ofapples per personfrom the to rooms,

e ill come up ith "2 & 3' > 2 = 2.5 apples>person. -his value has norelationship to the actual total avera)e of the to rooms, hich in this case is 2.6apples. (f e too the simple avera)e "2.5' and multiplied it !* the num!er ofpeople in the room "10' e ould 9@- come up ith the num!er of apples in theto rooms. -he onl* a* to calculate the actual total avera)e "short of noin)the total num!er of apples and people' is to ei)ht the to avera)es in thefolloin) manner $"2' & 6"3' > 10.

:;$ = 2 apples>person

#oomD

18 6 18>6 = 3 apples>person

-otal 26 1026>10 = 2.6apples>person


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17 5 85

25 3 75

-he S, or total num!er of runs, is 60 & 85 & 75 = 220. -he , or num!er ofpla*ers, is 2 & 5 & 3 = 10. A = 220>10 = 22. -he collective, or ei)hted,avera)e is 22, so e can denitivel* anser the uestion 9o. "#emem!er that

no is a suEcient anser. @nl* ma*!e is insuEcient.'(9:; num!er of items'. hen an additional eGam of score z is addedin, the ne sum ill !e xy & z . -he ne avera)e can !e eGpressed as the ne sum divided !* x & 1, since thereis no one more eGam in the lot. 9e avera)e = " xy & z '>" x & 1'. -he uestion ass us if the ne avera)e represents an increase in 50% over theold avera)e, y . e can rerite this uestion as /oes " xy & z '>" x & 1' = 1.5 y L(f e multipl* !oth sides of the euation !* 2" x & 1', !oth to )et rid of thedenominator eGpression " x & 1' and the decimal "1.5', e )et 2 xy & 2 z = 3 y " x &1'


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mem!er receives ill !e eual to the net chan)e in the overall avera)e. (n thiscase the eGtra 12 ill increase the avera)e !* 12>6 = 2.[ut mathematicall*, chan)e in avera)e = "the ne term eGistin) avera)e' > "thene Y of terms'e could have used this formula to rephrase the uestion a!ove " z y ' > " x & 1'= 0.5 y

A)ain if e multipl* !oth sides of the eGpression !* 2" x & 1', e )et 2 z 2 y = xy & y @# 2 z = xy & 3 y . :ometimes this method of dealin) ith avera)echan)es is more useful than dealin) ith sums, especiall* hen the sum isdiEcult or cum!ersome to nd.

26. -o solve this pro!lem, use hat *ou no a!out avera)es. (f e are to compare HodieMs avera)e monthl* usa)e to DrandonMs, e can simplif* the pro!lem !*dealin) ith each personMs total usa)e for the *ear. :ince DrandonMs avera)emonthl* usa)e in 2001 as q minutes, his total usa)e in 2001 as 12q minutes. -herefore, e can rephrase the pro!lem as follosas HodieMs total usa)e for the *ear less than, )reater than, or eual to 12qL:tatement "1' is insuEcient. (f HodieMs avera)e monthl* usa)e from Hanuar* toAu)ust as 1.5q minutes, her total *earl* usa)e must have !een at least 12q.oever, it certainl* could have !een more. -herefore, e cannot determinehether HodieMs total *earl* use as eual to or more than DrandonMs.:tatement "2' is suEcient. (f HodieMs avera)e monthl* usa)e from April to/ecem!er as 1.5q minutes, her total *earl* usa)e must have !een at least13.5q. -herefore, her total *earl* usa)e as )reater than DrandonMs. -he correct anser is D :tatement "2' alone is suEcient, !ut statement "1'alone is not suEcient.

27.Defore she made the pa*ment, the avera)e dail* !alance as U600, from theda*, !alance as U300. hen e nd in hich da* she made the pa*ment, ecan )et it.:tatement 1 is suEcient.


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and z eual. oever, the uestion stem also tells us that x , y and z are consecutiveinte)ers, ith x as the smallest of the three, y as the middle value, and z as thelar)est of the three. :o, if e can determine the value of x , y , or z , e ill no thevalue of all three. -hus a suita!le rephrase of this uestion is Shat is the value of x ,

y , or z LT

"1' :;3' y . Anser choice / can!e eliminated. As for anser choice , there is no possi!le a* tocreate :et " ith a median of "2>7' y . h*L e no that y is either 1,2, 3, $, 5, or 6. -hus, "2>7' y ill *ield a value that is some fraction ithdenominator of 7.

and z must !e 11, 12 and 13, respectivel*. -hus the avera)e of x , y ,and z is

11 & 12 &13

3

=

"2' :;


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-he possi!le values of "2>7' y are asfollos

2

7,$

7,6

7

,1

1

7

,1

3

7

,1

5

7oever, the median of a set of inte)ers must ala*s !e either an inte)er ora fraction ith a denominator of 2 "e.). 2.5, or 5>2'. :o "2>7' y cannot !e themedian of :et " . -he correct anser is .

6. :ince S contains onl* consecutive inte)ers, its median is the avera)e of the

eGtreme values a and b. e also no that the median of S is . e canset up and simplif* the folloin) euation

:ince set # contains onl* consecutive inte)ers, its median is also the avera)eof the eGtreme values, in this case b and c. e also no that the median of

# is . e can set up and simplif* the folloin) euation

e can nd the ratio of a to c as follos -ain) the rst euation,

and the second euation, and settin) them eual to each other, *ieldsthe folloin)

. :ince set R contains onl* consecutive inte)ers, its median is

the avera)e of the eGtreme values a and c . e can use the ratio

to su!stitute for a


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-hus the median of set R is . -he correct anser is .

7. :ince a re)ular *ear consists of 52 ees and Him taes eGactl* to ees ofunpaid vacation, he ors for a total of 50 ees per *ear. is Cat salar* fora 50Oee period euals 50 + U200 = U10,000 per *ear. Decause the num!erof *ears in a 5O*ear period is odd, Him4s median income ill coincide ith hisannual income in one of the 5 *ears. :ince in each of the past 5 *ears thenum!er of uestions Him rote as an odd num!er )reater than 20, his

commission compensation a!ove the Cat salar* must !e an odd multiple of .:u!tractin) the U10,000 Cat salar* from each of the anser choices, illresult in the amount of commission. -he onl* odd values are U15,673,U18,$23 and U21,227 for anser choices D, /, and , respectivel*. :ince thetotal amount of commission must !e divisi!le !* , e can anal*Je each ofthese commission amounts for divisi!ilit* !* . @ne eas* a* to determinehether a num!er is divisi!le !* is to sum the di)its of the num!er and seeif this sum is divisi!le !* . -his anal*sis *ields that onl* U18,$23 "sum of thedi)its = 18' is divisi!le !* and can !e Him4s commission. ence, U28,$23could !e Him4s median annual income. -he correct anser is choice /.

8.


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[ossi!ilit* 2 -he median of :et A is )reater than the median of :et !ut less thanthe median of :et D.

[ossi!ilit* 3 -he median of :et A is eual to the median of :et D or the median of:et . :tatement "2' tells us that the median of :et A is )reater than the median of

:et . -his eliminates [ossi!ilit* 1, !ut e are still left ith [ossi!ilit* 2 and[ossi!ilit* 3. -he median of :et D ma* !e )reater than @# eual to the median of:et A. -hus, usin) :tatement "2' e cannot determine hether the median of :et Dis )reater than the median of :et A. om!inin) :tatements "1' and "2' still does not*ield an anser to the uestion, since :tatement "1' )ives no relevant informationthat compares the to medians and :tatement "2' leaves open more than onepossi!ilit*. -herefore, the correct anser is hoice /E $tatements /1 and /2,*E,3E+ are N, su4cient.

. -o nd the mean of the set W6, 7, 1, 5, x , y X, use the avera)e formulahere A = the avera)e, S = the sum of the terms, and n = the num!er ofterms in the set. ;sin) the information )iven in statement "1' that x & y = 7,

e can nd the mean . #e)ardless of

the values of x and y , the mean of the set is !ecause the sum of x and y does not chan)e. -o nd the median, list the possi!le values for x and y suchthat x & y = 7.


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mean of the set ill var*. Additionall*, since there are man* possi!le values for xand y , there are numerous possi!le medians. -he folloin) ta!le illustrates that ecan construct a data set for hich x $ y % 3 and the mean is )reater than themedian. -he ta!le A?:@ shos that e can construct a data set for hich x $ y % 3and the median is )reater than the mean.

x * /A-A :- F/(A9 FA9

22 1 1, 5, 6, 7, 1, 22 6.5 10

$ 1 1, 1, $, 5, 6, 7 $.5 $

-hus, statement "2' alone is not suEcient to determine hether the mean is)reater than the median. -he correct anser is "A' :tatement "1' alone issuEcient, !ut statement "2' alone is not suEcient.

10. Fedian ^ Fean11. 13

12.

:ince an* poer of 7 is odd, the product of this poer and 3 ill ala*s !e odd.Addin) this odd num!er to the dou!led a)e of the student "an even num!er, since itis the product of 2 and some inte)er' ill ala*s *ield an odd inte)er. -herefore, allluc* num!ers in the class ill !e odd.

-he results of the eGperiment ill *ield a set of 28 odd inte)ers, hose median ill!e the avera)e of the 1$th and 15th )reatest inte)ers in the set. :ince !oth of theseinte)ers ill !e odd, their sum ill ala*s !e even and their avera)e ill ala*s !ean inte)er. -herefore, the pro!a!ilit* that the median luc* num!er ill !e a nonOinte)er is 0%.

13. :ince the set Wa& b& c& d& e& f X has an even num!er of terms, there is no one middle

term, and thus the median is the avera)e of the to middle terms, c and d. -hereforethe uestion can !e rephrased in the folloin) manner(s "c & d'>2 ^ "a & b & c & d & e & f '>6 L(s 3"c & d' ^ a & b & c & d & e & f L(s 3c & 3d ^ a & b & c & d & e & f L(s 2c & 2d ^ a & b & e & f L"1' (9:;$on the left side of the uestion', e cannot anser the uestion. e can mae theanser to the uestion \es !* relativel* picin) small b and f "compared to c and d'

OO for instance, b = 2, c = 7, d = and f = 12 "still leavin) room for a and e, hich inthis case ould eual 1 and 11, respectivel*'. @n the other hand, e can mae theanser 9o !* chan)in) f to a ver* lar)e num!er, such as 1000."2' (9:;3'"c & d'. -he reasone no this is that the set of inte)ers is ascendin), so a ] ! and e ] f. -herefore a


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& e ] ! & f, and ! & f = "$>3'"c & d' accordin) to this statement. oever, e donMtno hether a & e ] "2>3'"c & d'."1' A9/ "2' :;


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An eGample of Rust to sets are73, 80, 81 median = 80 61, 73, 100 median = 73 "2':;


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sum to 18, e can arran)e the values in increasin) order as follos

x , 2, 3, 8, 11, y

:ince 3 and 8 are the middle values, the median euals 5.5 as reuired. -heuestion ass for the maGimal value of x , so letMs increase x as far as possi!le

ithout chan)in) the median. As x increases to 3 "and y decreases to 15', themiddle values of 3 and 8 donMt chan)e, so the median remains at 5.5.oever, as x increases !e*ond 3, the median also increases, so the maGimalvalue of x that leaves the median at 5.5 is 3.

-he correct anser is /.

15.


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2.Am* as the 0th percentile of the 80 )rades for her class, therefore, 10% are hi)herthan Am*Ms, 10%Q80=8.1 of the other class as hi)her than Am*. -otall*, 8&1=27 -hen, the percentile is"180O27'>180=85>100

Anser is /27.AnnMs actual sale is $50OG, alMs 10&G, after corrected, Ann still hi)her than al, soAnn is the median.@r e can eGplain it in another a*$50OG=330, so G=120AnnMs actual sale is $50OG alMs 10&G,:uppose that either Ann or al can !e the median, if Ann is the median, than e )etthe previous anser hoever, if al is median "330', e ill have 10&G=330,G=1$0, then Ann"$50O1$0=310' ill less than al"330', that is incorrect. -his can eGplain h* al can not !e the median and Ann must

M"E

1. O8

2.

:tatement 1 tells us that the diNerence !eteen an* to inte)ers in the setis less than 3. -his information alone *ields a variet* of possi!le sets.


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1. 02. 3. $. A

5. /6. 7. 8. D. D10.

11. Defore anal*Jin) the statements, let4s consider diNerent scenarios for theran)e and the median of set A. :ince e have an even num!er of inte)ers inthe set, the median of the set ill !e eual to the avera)e of the to middlenum!ers. 2 ] "36 & x & y '>63" x & y ' ] 36 & " x & y ' 2" x & y ' ] 36 x & y ] 18. -herefore,statement "2' is :;


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12 months in a *ear, *ieldin) 12 distinct possi!ilities for the !irth month of astudent. month of !irth>)ender, some students

must share the same com!ination. -hat is, some students must have thesame )ender, !e !orn in the same month. and have the same VFA- score. -hus, statement "1' is suEcient to anser the uestion.:tatement "2' provides no information a!out the ran)e of student VFA-scores in the rstO*ear class. :ince there are 61 distinct VFA- scores!eteen 200 and 800, the total num!er of distinct com!inations of VFA-score>month of !irth>)ender on the !asis of statement "2' is 61 + 12 + 2 =1,$6$. :ince this num!er is )reater than the rstO*ear enrolment, there arepotentiall* enou)h uniue com!inations to cover all of the students, impl*in)that there ma* or ma* not !e some students sharin) the same 3parameters. :ince e cannot )ive a conclusive anser to the uestion,statement "2' is insuEcient.

-he correct anser is A.

1$.


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"1' :;


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= W$0, 50, 0, 70, 80X( = W50, 55, !!, 80, 0X oever, A3 could !e less than or eual to (3, main) theanser to the uestion SnoT

A = W$0, 50, 0, 70, 80X ( = W50, 60, 70, 80, 0X "2' :;


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-o mae the )reatest num!er as )reater as possi!le, these 7 num!ers should cost theran)e as little as possi!le. -he* ill !e, 2$, 23, 22, 21, 20, 1, 18.:o, the )reatest value that can fulll the ran)e is 18&25=$3

$,AN"A+" "E9#A,#N

1. 1.12 approG2. 3. /$. 5. 6.

7. (f ' ^ 0, then ' ^ and the median of A is )reater than the mean of set A. (f * $ , = 0, then * % , and the median of set ( is eual to the mean of set(.

(. 9@- 9::A#(?\ Accordin) to the ta!le, - ^ means that the standarddeviation of set A is )reater than that of set (. :tandard deviation is ameasure of ho close the terms of a )iven set are to the mean of the set. (f aset has a hi)h standard deviation, its terms are relativel* far from the mean. (f a set has a lo standard deviation, its terms are relativel* close to the mean.

#ecall that a median separates the set into to as far as the num!er ofterms. -here is an eual num!er of terms !oth a!ove and !elo themedian. (f the median of a set is )reater than the mean, hoever, the terms!elo the median must collectivel* !e farther from the median than the termsa!ove the median. osite set) the meanof the com>osite set must either be between the means of theindividual sets or be e?ual to the mean of both of the individualsets. -o prove this, consider the simple eGample of one mem!er sets A =

3B, ( = 5B, A & ( = 3, 5B. (n this case the mean of A & ( is )reater than themean of A and less than the mean of (. e could easil* have reversed thisresult !* reversin) the mem!ers of sets A and (.

(((. 9@- 9::A#(?\ Accordin) to the ta!le, # . R implies that the median of the set A & (B is )reater than the mean of set A & (B. e can eGtend therule )iven in statement (( to medians as ell when two sets are combinedto form a com>osite set) the median of the com>osite set must eitherbe between the medians of the individual sets or be e?ual to the


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median of one or both of the individual sets. hile the median of set A is)reater than the mean of set A and the median of set ( is eual to the meanof set (, set A & (B mi)ht have a median that is )reater or less than themean of set A & (B. :ee the to ta!les for illustration

:et Fedian Fean #esult

A 1, 3, $ 3 2.67 Fedian ̂ Fean

( $, 5, 6 5 5 Fedian = Fean

A & ( 1, 3, $, $, 5, 6 $ 3.83 Fedian ̂ Fean

:et Fedian Fean #esult

A 1, 3, 3, $ 3 2.75 Fedian ̂ Fean

( 10, 11, 12 11 11 Fedian = Fean

A & ( 1, 3, 3, $, 10, 11, 12 $ 6.2 Fedian ] Fean

-herefore none of the statements are necessaril* true and the correct anser is .

8. 65, 85. 3010. a:11. a>! + ":'12. :

16.

9um!er of num!ers'. "0 & 1 & 1 & 1 & 2 & 2 & 2 & 3' b 8 = 12 b 8 = 3>2Fethod 2 Fultipl* each possi!le sum !* its pro!a!ilit* and add. "0 + 1>8' & "1 + 3>8' & "2 + 3>8' & "3 + 1>8' = 12>8 = 3>2Fethod 3 :ince the sums have a s*mmetrical form, spot immediatel* that the meanmust !e ri)ht in the middle. \ou have one 0, three 14s, three 24s and one 3 so themean must !e eGactl* in the middle = 1.5 or 3>2.


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-hen, to )et the standard deviation, do the folloin)"a' ompute the diNerence of each trial from the avera)e of 3>2 that as Rust determined. "-echnicall* it4s Savera)e minus trialT !ut the si)n does not mattersince the result ill !e suared in the neGt step.'"!' :uare each of those diNerences.

"c' 2 and 3>2', suare those and avera)e them, *ou

ill )et 5>$, and the standard deviation as . -his is incorrect !ecause it implies that the 3>2 and 3>2 diNerences are as commonas the and 1>2 diNerences. -his is not true since the and 1>2 diNerences occurthree times as freuentl* as the 3>2 and 3>2 diNerences.

1$. :tandard deviation is a measure of ho far the data points in a set fall from themean.


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follos hat is the diNerence !eteen each term in the set and the mean of thesetL "1' :;


26/26

@!viousl*, 70 and 75 can fulll the reuirements.Anser is D15.1 and 2 standard deviations !elo the mean=^num!er of the hours at most is 21O6=15, at least is 21O2Q6=.Anser is /

20.Fean 8.1:tandard deviation 0.3ithin 1.5 standard deviations of the mean=8.1O0.3Q1.5,8.1&0.3Q1.5B=7.65,8.55BAll the num!ers eGcept 7.51 fall ithin such intervalAnser is 1121. d2="a1Oa'2&"a2Oa'2&....&"anOa'2 B>nhen e added 6 and 6, the numerator remained unchan)ed !ut the denominatorincreased, so, the ne deviation is less than d.Anser is 22.

gmat quant topic 2 statistics solutions

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