11_Continuous Probability Distributions Part 2-1

# 3 what is the probability that the demand is between

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3. What is the probability that the demand is between 31 and 42? 25 45 1/20 x f(x) 31 42 What is the shaded area? P(31 < X < 42) = (42 - 31)/(45 - 25) = 11 / 20 = 0.55

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23 Uniform distribution example: iPad mini Daily demand for iPad mini at the Apple store on Walnut Street is uniformly distributed between 25 and 45. 3. What is the probability that the demand is between 31 and 42? 4. What are the mean and standard deviation? P(31 < X < 42) = (42 - 31)/(45 - 25) = 11 / 20 = 0.55 25 45 1/20 x f(x)
24 Uniform distribution example: iPad mini Daily demand for iPad mini at the Apple store on Walnut Street is uniformly distributed between 25 and 45. 3. What is the probability that the demand is between 31 and 42? 4. What are the mean and standard deviation? P(31 < X < 42) = (42 - 31)/(45 - 25) = 11 / 20 = 0.55 Mean = (a+b)/2 = (25+45)/2 = 35 25 45 1/20 x f(x)

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25 Uniform distribution example: iPad mini Daily demand for iPad mini at the Apple store on Walnut Street is uniformly distributed between 25 and 45. 3. What is the probability that the demand is between 31 and 42? 4. What are the mean and standard deviation? P(31 < X < 42) = (42 - 31)/(45 - 25) = 11 / 20 = 0.55 Mean = (a+b)/2 = (25+45)/2 = 35 Standard Deviation = 25 45 1/20 x f(x) ( b - a ) 2 12 = (45 - 25) 2 12 = 5.77
26 Exponential Distribution : Time/interval between successive arrivals/ “events of interest” is usually exponentially distributed Examples: Time to arrival of next customer Time to serve a given customer at bank/call-center/check-out Properties and Assumptions: Non-negative R.V. Can assume any value between 0 to +∞ Asymmetric: P(X > x) decreases as x increases, for any x ≥ 0 Related to Poisson distribution (more later) and many queueing systems Exponential distribution: X~Expon(λ)

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27 Exponential distribution formula: X~Expon() Characterized by λ : “arrival rate” (avg. rate of events per unit time) PDF CDF P(X≤x) Right tail prob. P(X>x) x Mean: Variance: Stdev: 0 ) ( = - x for e x f x λ λ 0 1 ) ( ) ( - = = - x for e x X P x F x λ 0 ) ( ) ( 1 = = - - x for e x X P x F x λ λ μ 1 ] [ = = X E 2 2 1 ] [ λ σ = = X Var λ σ 1 =
28 Exponential distribution formula: Example: X~Expon (= 5 cust. per hour) CDF P(X≤x) x 1. P(X≤1) = 2. P(X≤0.5) = 3. P(0.5<X≤1.5)= 4. P(X>2)= Mean: Variance: Stdev: unit? 1 – e-5 = 0.9933 1 – e-2.5 = 0.9179 (1-e-7.5) – (1-e-2.5) = 0.0815 e-10 = 0 (up to 4 digits) 0.2 hours 0.04 hr2 = 0.2 hr λ μ 1 ] [ = = X E 2 2 1 ] [ λ σ = = X Var λ σ 1 =

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29 Lab accidents at Drexel occur at an average rate of 3 accidents per month, and the number of accidents follows a Poisson distribution. 1. Describe the distribution and the mean time between accidents. λ =3 accidents per month E(X) = 1/ λ = 1/3 months per accident Exponential distribution example: Sci-fi insurance II
30 2. What is the probability that there is no accident in the next 2 months? 3. An NSF inspection has been scheduled for the first half of March. What is the probability that there will be no accident during the inspection period?

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