This document provides solutions for Applied Quantitative Methods Assignment II. It covers topics such as number of passengers at train station, frequency histogram, central tendency measures, correlation coefficient, regression line calculation, probability calculations, Bayesian probability, binomial distribution, Poisson distribution, and more.
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APPLIED QUANTITATIVE METHODS ASSIGNMENT II Student Name [Pick the date]
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Question 1 (a)Number of passengers at train station (Melbourne) (b)Frequency histogram (c)Central tendency measures 1
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Question 2 (a)The dataset provided is labelled as a sample as it corresponds to selected values which have been chosen from the population of all weeks in a year. If the given data corresponded to the population data, then all the 52 weeks contained in a year would have been represented. Instead only selected values have been captured which denote a sample. (b)standard deviation (c)Inter Quartile Range 3
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Sorted In cases where the underlying data is skewed, standard deviation does not provide an accurate picture and as a result, it is better to rely on IQR to capture the extent of dispersioninthedata,IQRisnotimpactedbyextremevalues(Erikssonand Kovalainen, 2015). (d)Correlation coefficient Correlationcoefficient(r)=0.968 4
The correlation coefficient indicated above highlights that the underlying relationship between the given variables is positive. Also, the strength of the relationship between thevariablesisquitestrongwhichisapparentfromthehighmagnitudeofthe correlation coefficient (Hair et al., 2015). Question 3 Calculation for regression line Slope=a=แ แบ47618แปโแบ1676576แปโแบ3392แปโแบ23425707แป {7โ(1676576)2โแบ3392แป2}แ=1628.689 Intercept=b=แ 7โ23425707โแบ3392แปโแบ47618แป {7โ(1676576)2โแบ3392แป2}แ=10.677 แบy=a+bxแป;y=1628.689+10.677x Chocolatebarssold=1628.689+(10.677โStudentWeeklyAttendance) The two examples to refers the regression line ๏ทHolmes is closed in holidays which implies that the weekly attendance is 0. Numberofchocolatebarssold=1628.689+(10.677โweeklyattendance) Numberofchocolatebarssold=1628.689+(10.677โ0.00)=1628.689or1629 ๏ทIf there is an increase in the student attendance by 10, then the chocolates bars sold would be increased by 1068 units approximately which can also be derived by 5
multiplying the slope of regression equation by the change in weekly attendance of students. (b) Calculation of R2 R2= (Corelation coefficient)2= (0.968)2= 0.937 The above value highlights that 93.7% of the variation in chocolate bars sale can be accounted for by corresponding changes in the weekly attendance of students. This clearly implies that the model is a good fit with high prediction capability (Flick, 2015). Question 4 (a)Probability (grassroots training or Holmes University) (b)Probability (scientific training and external). (c)Probability (scientific training and Holmes University) 6
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(d)The two events i.e. training (P) and recruitment (Q)will be independent when. PแบPแปรPแบQแป=P(PandQ) P(P)=P(External)=(54+12)/(35+92+54+12)=0.3412 P(Q)=P(Scientifictraining)=(35+54)/(35+92+54+12)=0.4611 PแบPแปรPแบQแป=0.1573 PแบPandQแป=PแบExternalandScientifictrainingแป=0.2576 PแบPแปรPแบQแปโ P(PandQ) Tr aining (P) and recruitment (Q) will be independent because the condition has not satisfied. Question 5 (a)ProbabilitySegmentA(preferX Y,Z) 7
PแบSegmentAแป=0.55,P(SegmentB)=0.30 PแบSegmentCแป=0.10,P(SegmentD)=0.05 P(ProductX|SegmentA)=PแบXa=Aแป=0.20 P(ProductX|SegmentB)=PแบXa=Bแป=0.35 P(ProductX|SegmentC)=PแบXa=Cแป=0.60 P(ProductX|SegmentD)=PแบXa=Dแป=0.90 P(X)=(0.55โ0.2)+(0.35โ0.3)+(0.6โ0.1)+(0.9โ0.05)=0.32 Bayesian Probability PแบSegmentAa=ProductX)=PแบAa=Xแป=PแบAแปโP(X|A) P(X)=0.55โ0.2 0.32=0.35375 (b)Preference of customer be product X P(X)=(0.55โ0.2)+(0.35โ0.3)+(0.6โ0.1)+(0.9โ0.05)=0.32 Question 6 (a)Probability 2 or less customers will buy Binomial Distribution Pแบxโค2แป=Pแบx=0แป+Pแบx=1แป+Pแบx=2แป Number of trials = 8 Probability of success = 0.1 Pแบxโค2แป={แ8 0แ(0.1)0แบ0.9แป8}+{แ8 1แแบ0.1)1แบ0.9แป7แฝ+{แ8 2แแบ0.1)2แบ0.9แป6แฝ=0.96 (b)Probability that 9 customers will come for buying in 2 min 8
Poissondistribution ฮณ=8 x=9 Pแบx=9แป=eโฮณฮณx x!=๐โ889 9!=0.124 Question 7 (a)Probability (Apartment will sell > $2 Mn) ฮผ=$1.1million=$1100000,ฯ=$385,000,x=$2million=$2000000 z=แxโฮผ ฯแ=เตฌ 2000000โ1100000 385000เตฐ z=2.34 From Standard Normal Table Pแบz>2.34แป=0.0097 Pแบx>$2millionแป=0.0097 (b)Probability (Apartment will sell b/w $1Mnand $1.1 Mn) 9
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ฮผ=$1.1million=$1100000,ฯ=$385,000 x1=$1million=$1000000,x2=$1.1million=$1100000 Pแบx1<x<x2แป=Pแบ1000000<x<1100000แป =Pแเตฌ 1000000โ1100000 385000เตฐ<แxโฮผ ฯแ<เตฌ 1100000โ1100000 385000เตฐแ Pแบx1<x<x2แป=Pแบ1000000<x<1100000แป=Pแบโ0.26<z<0แป From Standard Normal Table Pแบz<0แป=0.5000 Pแบz<โ0.26แป=0.3975 Pแบx1<x<x2แป=Pแบ1000000<x<1100000แป=0.5000โ0.3975 Pแบ1000000<x<1100000แป=0.1025 Question 8 (a) As per the Central Limit Theorem, it is fair to assume that the underlying sample distribution is normal provided the sample size is sufficiently large. This sample size has been defined as 30. It is known that the sample size provided in the given case is 50 which is significantly greater than 30 which implies that despite the original population not being normal, the given sample data can be assumed to be normal (Hillier, 2016). (b) Sample proportionp=11 45=0.244 10
References Eriksson, P. and Kovalainen, A. (2015)Quantitative methods in business research. 3rd ed. London: Sage Publications, pp. 115 Flick,U.(2015)Introducingresearchmethodology:Abeginner'sguidetodoinga research project.4th ed. New York: Sage Publications, pp. 206 Hair,J.F.,Wolfinbarger,M.,Money,A.H.,Samouel,P.,andPage,M.J. (2015)Essentials of business research methods.2nd ed. New York: Routledge, pp. 165 Hillier, F. (2016)Introduction to Operations Research.6th ed.New York: McGraw Hill Publications, pp. 189 12