AnsweronQuestion#60656 -Math-StatisticsandProbability Questiona)Thedistributionofmarksobtainedby500candidatesinaparticularexamisgivenbelow:Marksmorethan:01020304050Numberofcandidates50046040020010030Calculatethelowerquartilemarks.If70%ofthecandidatespassintheexam,findtheminimummarksobtainedbyapasscandidate.(5)Solutiona)

Thelowerquartilemarks:

𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑎𝑡𝑎 0𝟏𝟎20

− 𝑇ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑣𝑎𝑙𝑢𝑒

Median

𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑎𝑡𝑎 304050

𝑄! = 10

If70%ofthecandidatespassintheexam,findtheminimummarksobtainedbyapasscandidate:70% of the candidates = 500 ∙ 70% = 350candidates 400 (80%) people passed the examwith the assessment of 20 or above.That’swhythetheminimummarksof350candidatesis20.Theminimummarksobtainedbyapasscandidate:20.

Answer:Thelowerquartilemarks:𝑄! = 10Theminimummarksobtainedbyapasscandidate:20.

b)AnanalysisofmonthlywagespaidtotheworkersoftwofirmsAandBbelongingtothesameindustrygivesthefollowingresults:FirmAFirmBNumberofworkers500600Averagedailywages`186`175Varianceofdistributionofwages81100i)Whichfirm,AorB,hasalargewagebill?ii)Inwhichfirm,AorB,istheregreatervariabilityinindividualwages?iii)FindtheaveragedailywageandthevarianceofthedistributionofwagesofalltheworkersinthefirmsAandBtakentogether.

i) Whichfirm,AorB,hasalargewagebill?NumberofworkersonFirmA: 𝑁!! = 500 AveragedailywagesonFirmA: 𝐴!! = 186 WagebillonFirmA: 𝑁!! ∙ 𝐴!! = 500 ∙ 186 = 93000

NumberofworkersonFirmB: 𝑁!! = 600 AveragedailywagesonFirmB: 𝐴!! = 175 WagebillonFirmB: 𝑁!! ∙ 𝐴!! = 600 ∙ 175 = 105000

93000 < 105000 So,firmBhaslargerwagebill.

ii) Inwhichfirm,AorB,istheregreatervariabilityinindividualwages?For firm A VarianceofdistributionofwagesonFirmA: 𝑉!! = 81Standarddeviation:𝜎 = 𝑉!! = 9Coefficientofvariation= !

!!!∙ 100 = !

!"#∙ 100 = 4.8387

For firm B VarianceofdistributionofwagesonFirmB: 𝑉!! = 100Standarddeviation:𝜎 = 𝑉!! = 10Coefficientofvariation= !

!!!∙ 100 = !"

!"#∙ 100 = 5.7143

Variabilityofthefirmdependsuponcoefficientofvariation.Higherthecoefficientofvariation,higherthevariability.CoefficientofvariationoffirmBishigher.Hence,firmBshowsgreatervariabilityinindividualwages.

iii) FindtheaveragedailywageandthevarianceofthedistributionofwagesofalltheworkersinthefirmsAandBtakentogether.

Averagedailywage𝑥 =!!! ∙!!!!!!! ∙!!!

!!!!!!!= !"###!!"#"""

!""!!""= 180

Varianceofthedistributionofwages:Thestandarddeviationsoftwoseriescontaining𝑛!,and𝑛!,membersare𝜎!,and𝜎!,respectively,beingmeasurefortheirrespectivemeans𝑥!,and𝑥!. If the two series are grouped together as one series of 𝑛! + 𝑛! members, show that the standard deviation𝜎of this series, measuredfromitsmean𝑥,isgivenby:

𝜎! =𝑛! 𝜎!! + 𝑑!

! + 𝑛! 𝜎!! + 𝑑!!

𝑛! + 𝑛!

where𝑑! = 𝐴!! − 𝑥and𝑑! = 𝐴!! − 𝑥 This formula we obtain the following simplification of expression themeansquaredeviationoftwoseriestakentogether:

𝜎! =1

𝑛! + 𝑛!𝑓 𝑥 − 𝑎 !

!!!!!

!

𝑑! = 186− 180 = 6 ⇒ 𝑑!

! = 36𝑑! = 175− 180 = −5 ⇒ 𝑑!

! = 25

𝜎! =500 81+ 36 + 600 100+ 25

500+ 600 =58500+ 75000

1100 = 121.36

Answer:i) firmBhaslargerwagebill.ii) firmBshowsgreatervariabilityinindividualwages.iii) Average daily wage𝑥 = 180, variance of the distribution of wages

𝜎! = 121.36

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