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StatisticsReview

# StatisticsReview - MGMT 36100 Operations Management(OM...

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Unformatted text preview: MGMT 36100 Operations Management (OM) Review of Probability Distributions We deal with some random variable, say X. X takes values between Xmin and Xmax. If X is discrete, we get Discrete distribution such as Poisson, binomial, uniform, empirical, ….. uniform, If X is continuous, we get Continuous distribution such as exponential, Weibull, “t”, “F”, Chi square, Weibull, uniform, normal… Review of Statistics StatisticsReview 1 StatisticsReview 2 MGMT 36100 MGMT 36100 Examples of discrete distributions (i) Toss a coin 4 times. X: number of heads. binomial distribution More on discrete distributions For a discrete distribution, quantity Σ{X * Pr(X)} is called expected 2 3 4/16 4 1/16 value of X and gives the average value (i.e. mean) of X. Toss a coin 4 times. X: number of heads. binomial distribution X 0 1 2 3 4 Average Let’s play a game Pr(X) X . Pr(X) Y 1/16 0 +5 4/16 4/16 -6 6/16 12/16 +5 -6 8/16 - 48/16 4/16 12/16 -6 1/16 4/16 +5 +5 8/16 + 40/16 - 0.5 Average 2.0 X 0 1 Pr(X) 1/16 4/16 6/16 4 5 6 (ii) Toss a die once. Y: number on the top Pr(Y) 1/6 1/6 1/6 1/6 1/6 1/6 face. uniform distribution (iii) A random sample of 200 engineering students contained 10 in category 1 (freshman), 20 in category 2, 40 in category 3, 56 in category 4 and 74 in category 5 (graduate). Y 1 2 3 We can build an X No. Pr(X) F(X) 1 10 2 20 3 40 4 56 5 74 Total 200 empirical probability distribution. Cumulative probabilities StatisticsReview Suppose you pay 6 dollars if 1 or 3 heads appear and I will Y pay 5 dollars otherwise. You Pr(Y) can define your winning as Y, Y . Pr(Y) create distribution of Y and calculate the average. 0.05 0.10 0.20 0.28 0.37 0.05 0.15 0.35 0.63 1.00 3 Average value of 2.0 for X means on an average 2 heads will appear. Average of - 0.5 for Y means on an average you will lose 50 cents. StatisticsReview 4 MGMT 36100 MGMT 36100 One more empirical distribution Demand (X) for a product is given by the following X Pr(X) X * Pr(X) 10 0.05 0.5 20 0.14 2.8 30 0.20 6.0 40 0.45 18.0 50 0.16 8.0 Average Demand 35.3 Continuous uniform distribution Given C, calculate Pr( X ≤ C) Pr(X ≤ C) = C–A B–A X varies between A and B Suppose you have 36 items in the stock. How many would you sell on an average? Define Y as the number sold 20 Y 10 30 Y * Pr(Y) 0.5 2.8 6.0 36 16.2 36 5.76 Average 31.26 A Suppose A = 100, B = 500. Find Pr (X ≤ 320) Pr (X 320) (320 100) (500 100) 0.55 Pr (X ≤ 320) = (320 – 100) / (500 – 100) = 0.55 C X→ B Newsvendor model Now find C such that Pr(X ≤ C) = 0.65. Use A = 100, B = 500 0.65. (C – A) / (B – A) = 0.65 C = A + 0.65 * (B – A) = 100 + 0.65 * (500-100) = 360 (500- Average demand is 35.3. Newsvendor model Average number of units sold is 31.26 StatisticsReview 5 StatisticsReview 6 MGMT 36100 MGMT 36100 StatisticsReview 1 Normal distribution Symmetrical, X varies from - ∝ to + ∝ mean = μx and standard deviation = σx When μx = 0 and σx = 1, we have standard 1, normal distribution. This variable is denoted by Z and tables for probability calculations are available. X∼ N[200, 25] : X is normally distributed, mean 200 and standard deviation 25. Normal distribution Given X ∼ N[100 , 15], Find Pr (X ≤ 130) = Pr [ Z ≤ (130 - 100)/15] = Pr [ Z ≤ 2] = 0.9772 Find Pr (85 ≤ X ≤ 122.5) 122.5) = Pr[ (85 - 100) (122.5 - 100) ≤Z≤ ] 15 15 = Pr ( - 1 ≤ Z ≤ 1.5) = [0.9332 – (1 – 0.8413)] 0.8413)] = 0.7745 μx We should be able to do the following two types of problems. Given X ∼ N[μx , σx], find 1. Pr(A ≤ X ≤ B) and 2. Find C such that Pr(X ≤ C) = specified value We need to convert these to standard normal. Pr(A ≤ X ≤ B) = Pr[ (A - μx) / σx ≤ Z ≤ (B - μx) / σx ] 100 130 85 100 122.5 8 StatisticsReview 7 StatisticsReview MGMT 36100 MGMT 36100 Normal distribution Sampling distributions Suppose we have a random variable X with mean = μx and standard and deviation = σx. Suppose we take random sample of size n and calculate the average value W and total Y. Then W and Y are also random variables. Y. W = ( X1 + X2 + … + Xn ) /n /n If X~N[μx , σx] then Y = ( X1 + X2 + … + Xn ) If X~N[μx , σx] then Y~N[μY= n*μx , σY = σx* (√n)] If X is not normal, Y becomes approximately normal if sample size n is large. Q / P systems Z0.9 = Area = α 1.29 Z0.95 = 1.65 Z0.98 = 2.06 Z1-α When we want to find Z value, we will not interpolate. We will select next higher value of Z. Q/P systems, Newsvendor model, Process capability studies StatisticsReview 9 W~N[μW = μx , σW = σx/ (√n)] If X is not normal, W becomes approximately normal if sample size n is large. Control charts StatisticsReview 10 MGMT 36100 MGMT 36100 StatisticsReview 2 ...
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