This document provides study material for the course Statistics for Managerial Decisions. It includes topics such as quarterly opening price for stocks, computation of mean and median, probability calculations, and normality probability plots.
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STATISTICS FOR MANAGERIAL DECISIONS STUDENT ID: [Pick the date]
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Question 1 (a)Quarterly opening price for the two stocks Stem and leaf plotfor the two stocks (b)Relative frequency histogram and frequency polygon graph 2
(c)For six selected publicly listed companies from the healthcare industry, the market capitalisation on December 31, 2018 is captured through the bar graph below. (d)The comparable information for the given two stocks is shown below which is useful for decision making. Based on the above data, it would make sense to invest in RMD because of the following reasons. 1)Lower risk as reflected from beta and hence lower volatility 2)Higher dividend yield which augers well for stakeholders 3)Lower P/E ratio which would imply lower valuations Question 2 (a)Computation of mean, median, first quartile and third quartile of the weekly rents 3
(b)Computation standard deviation, mean absolute deviation and range of the weekly rents (c)The box and whisker display for the weekly rents (d)The comparison of the weekly rent thought the box plot clearly highlight that lowest rents on average are observed for Perth while the highest rents are observed for Sydney. The variation is rent is also the lowest for Perth while the highest for Sydney, The comparable quotes for rent as obtained from Airbnb are comparatively on the lower side for a comparable property. This is especially the case for Melbourne and Sydney listings. Question 3 (a)Probability that a tourist would pick canola field in Australia P (Canola field in Australia) =2538678/ ((4623527+2538678+371339+955321+11720277)) P (Canola field in Australia)=0.1256or 12.56% 4
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(b)Probability that a tourist would pick Wheat field in NSW P (Wheat field in NSW) =9556517/ ((2755310+1201045+403121+483081+9556517)) P (Wheat field in NSW)=0.6637 or 66.37% (c)YieldofBarley(Aust.)=Production Area=12920461 4623527=2.795 YieldofBarley(NSW)=Production Area=2755310 1008269=2.733 YieldofBarley(VIC)=Production Area=3100466 954176=3.249 YieldofBarley(QID)=Production Area=413023 141960=2.909 YieldofBarley(SA)=Production Area=2744507 922787=2.974 Requisite probability = (2.974)/(2.974+2.909+3.249+2.733+2.795) = 0.2029 or 20.29% (d)Referring to the ABS data from the link provided, it is apparent for the grain sorghum figures corresponding to South Australia, the figures would be considered unreliable owing to standard error being in excess of 50%. Question 4 (a)Given distribution of rainfall “Poisson distribution” (i)Probability that there is zero rainfall in week Days on which rainfall was experienced in year = 170 Meanλ=170 52=3.3077and zero rainfall (X = 0) 5
P(X=0|λ=3.3077)=3.30770×e−3.30779 0!=0.0366 (ii)Probability that there would be 3 orhigher than 3 days of rainfall in a week P(X≥3) P(X≥3|λ=3.3077)=1−3.30773×e−3.30779 3!=1−0.5786=0.4214 (b)Given distribution of rainfall Normal distribution Mean rainfall amount per week = 83.75 mm and standard deviation = 152.74 mm Standarderror=Standarddeviation ඥሺ52ሻ=21.18 (i)Probability that rainfall will be between 10 and 50 mm Pሺ10<X<50ሻ=P൬ 10−83.75 21.18<Z50−83.75 21.18൰=PሺZ2<−1.59ሻ−PሺZ1<−3.48ሻ Pሺ10<X<50ሻ=0.055526−0.000249(Normaldistributiontable) Pሺ10<X<50ሻ=0.055277 (ii)Rainfall amount for the case when 12% of weeks have amount or higher than the amount PሺRainfallamount>xሻ=125 z=NORMSINVሺ0.12ሻ=−1.17499 Let therainfallamountbe X mm z=X−mean Standarderror;−1.17499=X−83.75 21.18 X=58.86mm 6
Question 5 (a)The normality probability plot for the given variables is as illustrated below. 7
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All the variables for which normal probability plot have been drawn tend to show a linear pattern in their respective graph especially if some outliers are not considered. Hence, it would be fair to conclude that all the given variables can be taken to be approximately normal in their respective distribution. (b)90% confidence interval for the variables for the given two cases (With heart disease and without heart disease) 8
With heart diseaseWithout heart disease With heart diseaseWithout heart disease With heart diseaseWithout heart disease 9
With heart diseaseWithout heart disease For a significant difference between the patients in terms of the given four variables, it is imperative that the confidence interval estimated for patients with heart disease and for those who do not have a heart disease must not overlap. This is visible for three variables namely, oldpeak, maximum heart rate achieved and resting. Hence, these variables can be used for differentiating between the patients with and without heart disease. 10