Poisson distribution business statistics
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Poisson distribution business statistics

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Business Statistics Topic: Poisson Distribution

Business Statistics Topic: Poisson Distribution

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    Poisson distribution business statistics Poisson distribution business statistics Presentation Transcript

    • BUSINESS STATISTICS PRESENTATION – POISSON DISTRIBUTION Submitted By: Reshmi Raveendran 212027
    • POISSON DISTRIBUTION Poisson distribution is a limiting case of Binomial distribution in which : - The number of trials are indefinitely large i.e. n → ∞ - Constant probability of success for each trail is very small i.e. p →0 - If X has a poisson distribution, then mean of X = λ x = 0, 1,.. - Probability mass function (PMF) = P(x) = e-λ * λx x!
    • Question 1 Number of errors on a single page has poisson distribution with average number of errors of 1 per page. Calculate the probability that there is at least one error on a page? Sol: PMF = P(x) = e-λ * λx x = 0, 1, …… x! Where λ is called parameter of the distribution Here λ = 1 since average number of errors per page is 1 Now P(X>1) = 1 – P(X=0) = 1 – e-1 = .632
    • Question 2 Number of accidents on an express-way each day is a poisson random variable with average of 3 accidents per day. What is the probability that no accidents with occur today? Sol: PMF = P(x) = e-λ * λx x! x = 0, 1, ….. Where λ is called parameter of the distribution Here λ = 3 since average number of accidents per day is 3 Now P(X=0) = e-3 = 0.0498
    • Question 3 A car – hire firm has two cars, which it hires out day by day. The number of demands for a car on each day is distributed as a poisson distribution with mean 1.5. Calculate the proportion of days on which neither car is used and the proportion of days on which some demand is refused. (e-1.5) = .2231 Sol: PMF = P(x) = e-λ * λx x = 0, 1, …. x! Here λ = 1.5 Now, proportion of days on which neither car is used : P(X=0) = e -1.5(1.5)0 0! = 0.2231
    • Proportion of the days on which some demand is refused : =P(x>2) = 1- P(x<2) = 1- P( x = 0 or x = 1 or x =2 ) = 1 – P(x=0) + P(x=1) + P(x=2) = 1 – e-1.5(1.5)0 + e-1.5(1.5)1 0! 1! = 1 – e-1.5 (1 + 1.5 + 2.25/2) = 1 – (0.2231) (3.625) = 0.1913 + e-1.5(1.5)2 2!
    • Q. If X has a poisson distribution and P(X=0)=1/2. What is E(X)? Sol. Mean of X = E(X) = λ Since it a poisson distribution, its probability mass function (pmf) is given by: P(X) = e- λ λx X! 1/2 = e- λ λ0 0! 1/2 = e- λ loge (1/2)= - λ loge e (Taking log on both sides) loge 1 – loge 2 = - λ (loge e =1) loge 2 – loge 1 = λ Hence, λ = log 2 or 0.693
    • Q. If X has a poisson variate such that P(X=2)= 9 P(X=4)+ 90 P(X=6). Find P(X). Soln. Using poisson distribution, P(X) = e- λ λx X! e- λ λ2 = 9 e- λ λ4 + 90 e- λ λ6 2! 4! 6! λ4 + 3 λ2 – 4 = 0 (Solve using quadratic method) λ2=1 Or λ = 1 (λ cannot be negative)
    • THANK YOU