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ACC544 - Decision Support Tools Report

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Charles Sturt University

   

Decision Support Tools (ACC544)

   

Added on  2020-03-04

ACC544 - Decision Support Tools Report

   

Charles Sturt University

   

Decision Support Tools (ACC544)

   Added on 2020-03-04

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Question 1a)Random variable is described as a variable from the random phenomenon whose possiblevalues are numerical results. The two types of random variables are discrete random variable and continuous random variable. Discrete random variables are variables that only assume finite /countable values or numbers while continuous random variables are the variables that assume infinite number of the possible values.b)Expected value is the value that is predicted in the variable, it is always engaged in the measurement or determination of the error residuals. Expected value is calculated by taking the sum of product of the variables (x1, x2, x3...) to their probabilities (p1, p2, p3...).E(x) =[x1p1+x2p2+x3p3+...]c)1)Sales units (x)Number of daysP(x)Exp valueMore thanLess than1201/151*1/15=1/151-1/15=14/151-1=02402/152*2/15=4/151-1/5=4/51-14/15=1/153201/53*1/5=3/51-2/5=3/51-4/5=1/54104/154*4/15=16/151-2/3=1/31-3/5=2/55101/35*1/3=5/31-1=01-1/3=2/3Total011/3 = 3.67 Probability space = 1+2+3+4+5 = 15 units2) P(1) or P(2) = 1/15+2/15 = 1/53) Average daily sales = expected number of sales which is given by 3.67 units4) P(3) or more = P(3)+P(4)+P(5)= 1/5 + 4/15 + 1/3 = 4/55) P(4) or less = 1-P(5)1-1/3 = 2/3d) Average = 5000standard deviation = 500 x=55001)Z=x−μσZ>5500−5000500>1.00P(Z>1.00) = 1-P(Z=1) = 1 - 0.8413 = 0.1587
ACC544 - Decision Support Tools Report_1
2)X=4900Z<4900−5000500<−0.2P(Z<-0.2) = 1 – P(Z<0.2) = 1 – 0.5793 = 0.42.73)X=4250 Z<4250−5000500<−1.5P(Z<-1.5) = 1 – P(Z<1.5) = 1 – 0.9332 = 0.0668Question 22) A table of age group and sexagemalefemaletotal0-1421054331997433410286615-2415289931451340298033325-5448625914691975955456655-6413477801369501271728165 and over168433919532693637608total115291361146351822992654Source (http://www.indexmundi.com/australia/age_structure.html)3)Probability space is given by = male + female = 115299136 + 11463518 = 22992654 Then P(male) = number of male/total population = 11529136/22992654 = 0.5014P(15 and 24) = total number of people in bracket (15 and 24)/total population = 2980333/22992654 = 0.1296P(female and aged 15 and 24) = P(female)*P(15 and 24)= 11463518/22992654*0.1296 = 0.4986 + 0.1296 = 0.6282P(25 and over|male) = P(male and 25 and over)/P(male)P(25 and over) = 1 – [P(0-14) + P(15-24)]=1 – [0.1784 + 0.1296] = 0.692P(25 and over|male) = (0.5014*0.692) = 0.347
ACC544 - Decision Support Tools Report_2

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