BUSINESS MATHEMATICS & STATISTICS

LECTURE 39 Patterns of probability: Binomial, Poisson and Normal Distributions Part 3

EXPECTED VALUE EXAMPLE A lottery has 100 Rs. Payout on average 20 turns Is it worthwhile to buy the lottery if the ticket price is 10 Rs. Expected win per turn = p(winning) x gain per win + p(losing) x loss if you loose = 1/20 x (100 – 10) + 19/20 x (- 10) Rs. = 90/20 –190/20 Rs. = = - 5 Rs. So on an average you stand to loose 5 Rs.

DECISION TABLES No.of Pies demanded % Occasions Price per pie = Rs. 15 Refund on return = Rs. 5 Sale price = Rs. 25 Profit per pie = Rs. 25 – 15 = Rs. 10 Loss on each return = Rs. 15 – 5 = Rs. 10 How many pies should be bought for best profit?

DECISION TABLES 25(0.1)30(0.2)35(0.25) 40(0.2) 45(0.15) 50(0.1) EMV Buy Expected profit 30 pies = 0.1 x x x x x x 300 = = 290 Rs. Best Profit 35 Pies = Rs. 310

DECISION TREE TOY MANUFACTURING CASE 1A Abandon 1B Go ahead>2A: Series appears (60%) >2B: No series (40%) 2A>3A: Rival markets (50%) 2A>3B: No Rival (50%) Production Series, no rival = units Series, rival = 8000 units No series = 2000 units Investment = Rs Profit per unit = Rs. 200 Loss if abandon = Rs What is the best course of action?

DECISION TREE Profit if rival markets, series appears = 8000 x 200 – = – = Rs. Profit if no rivals = x 200 – = – = Rs. Profit/Loss if no series = 2000 x 200 – = – = Rs. (No series) EMV = Rival markets and no rivals = 0.5 x x = (Series) EMV = 0.6 x x – = – = Rs. Conclusion Go ahead

THE POISSON DISTRIBUTION Either or situation No data on trials No data on successes Average or mean value of successes or failures Typical Poisson Situation Characteristics 1.Either/or situation 2.Mean number of successes per unit, m, known and fixed 3.p, chance, unknown but small, (event is unusual)

THE POISSON TABLES OF PROBABILITIES Give cumulative probability of r or more successes Knowledge of m required Table gives the probability of that r or more random events are contained in an interval when the average number of events per interval is m Example m = 7; r = 9; P(r or more successes) = Values given in 4 decimals

EXAMPLE Attendance in a factory shows 7 absences What is the probability that on a given day there will be more than 8 people absent? Solution m = 7 r = More than 8 = 9 or more P(9 or more successes) =

EXAMPLE An automatic production line breaks down every 2 hours Special production requires uninterrupted operation for 8 hours What is the probability that this can be achieved? Solution m = 8/2 = 4 r = 0 (no breakdown) p( 0 breakdown) = 1 – p(1 or more breakdowns) = 1 – = = 1.83%

EXAMPLE An automatic packing machine produces on an average one in 100 underweight bags What is the probability that 500 bags contain less than three underweight bags? Solution m = 1 x 500/100 = 5 p(r = less than three) = 1 – p(r= 3 or more) = 1 – = = 12.47%

EXAMPLE Faulty apple toffees in a production line average out at 6 per box The management is willing to replace one box in a hundred What is the number of faulty toffees that this probability corresponds to? Solution p = 1/100 = 0.1 m = 6 Look for value of p close to 0.1 p(r = 12) = p(r = 13) = Hence 13 or more faulty toffees correspond to this probability

BUSINESS MATHEMATICS & STATISTICS