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Y Dosraniad Poisson

Y Dosraniad Poisson. The Poisson Distribution.

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Y Dosraniad Poisson

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  1. Y Dosraniad Poisson The Poisson Distribution Mae’r Dosraniad Poisson yn cael ei ddefnyddio pan mae gennym nifer cyfartalog y troeon mae digwyddiad yn digwydd. The Poisson Distribution is used when we have the average number of times an event occurs. e.e. nifer y galwadau ffôn a dderbynnir mewn swyddfa mewn 1 diwrnod, the number of phone calls received in an office in 1 day, nifer y diffygion mewn hyd penodol o ddefnydd, the number of flaws along a specified length of material, nifer cymedrig y damweiniau a geir ar ddarn penodol o’r A470 mewn mis. the mean number of accidents along a specific stretch of the A470 in a month.

  2. Os oes gan X ddosraniad Poisson, rydym yn ysgrifennu If X has a Poisson distribution, we write X ~ Po ( μ ) μ = y cymedr μ = the mean P(X = x) = e-µ µ x x! Mae’n bosibl hefyd defnyddio tablau i ddarganfod tebygolrwydd ar gyfer dosraniad Poisson. It is also possible to use tables to find the probability for the Poisson distribution.

  3. Cymedr ac Amrywiant y Dosraniad Poisson The Mean and Variance of the Poisson Distribution Mae’r cymedr yn cael ei roi i ni fel µ mewn dosraniad Poisson, ac mae’r amrywiant yn hafal i’r cymedr bob amser. The mean is given to us as µ and the variance is always equal to the mean in a Poisson distribution. E(X) = Var(X) = µ Cofiwch mai ail-isradd yr amrywiant yw’r gwyriad safonol. Remember that the standard deviation is the square root of the variance. Gwyriad Safonol = √Var(X) = √µ Standard Deviation = √Var(X) = √µ

  4. Enghraifft - Example • Mae gan X ddosraniad Poisson, cymedr 5. Darganfyddwch • X has the Poisson distribution with mean 5. Find • P(X = 4) • P(X ≥ 6) • P(2 ≤ X ≤ 4) • P(X < 2) • gymedr ac amrywiant Y pan mae Y = 4X - 2

  5. X ~ Po ( 5 ) = 0.175 = e-µ µ x x! = e-5 54 4! a) P(X = 4) = 0.384 b) P(X ≥ 6) ( yn defnyddio’r tablau) (using tables) = P(X ≥ 2) - P(X ≥ 5) = 0.9596 – 0.5595 c) P(2 ≤ X ≤ 4) = 0.4001 d) P(X < 2) = P(X = 0) + P(X = 1) = e-5 50 + e-5 51 0!1! = 0.00674 + 0.0337 = 0.0404 = 4E(X) - 2 = 18 e) E(Y) = E(4X – 2) = 4 x 5 - 2 Var(Y) = Var(4X – 2) = 42 Var(X) =16 x 5 = 80

  6. Ymarfer/Exercise 4.6a 4.6b 4.6c Mathemateg - Ystadegaeth Uned S1 – CBAC Mathematics Statistics Unit S1 - WJEC

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