Question

Can you help me with the Poisson distribution? Please see instructions in the image. Thank you.

Distribution of Poisson Instructions: Please present the processes necessary to support the answer to the exercises and round the final results — not the intermediate values that appear during the calculations — to two decimal places, when necessary. 1. Explain when the Poisson probability distribution is used. 2. Consider a Poisson random variable with u = 3.5. Use the Poisson formula to calculate the following probabilities: a) P(0) b) P (2) c) P (1) 3. Using the Poisson distribution table, determine the probability when: a) u = 0.8, if x= 2 b) u = 5.0, if x= 5 c) u = 2.5, if x=3 4. The number x of people admitted to an intensive care unit in a private hospital, on any day, has a Poisson probability distribution with an average equal to five (5) people per day. a) What is the probability that the number of people admitted to an intensive care unit, on a particular day, is two (2)? Solve using the Poisson formula. 5. Calculate with a) the formula y b) the table, the Poisson probability when u = 4, if x= 8. Certify that with both methods you get the same result.

Answer 1

1. Poisson podbability distribution is used when the probability of the number of successful events occurring over a specified time or space needs to be calulated. It can be cred when the average rata at which the successful events over a period is known. The average rate is denoted by u and the poobability that there are during the specited time or space is given by the formula p (x)= M e M x Occurences 2 Given M= = 3-5. Hence P(x) = 3-5 3.5 0 @ P 0) -3-5 3-5 e O! X! -3-5 е. = 0.0302 ~ 0.03 2 (6) P2) 3.5 23.5 12.25 * 0-302 I X2 0.184975 V0-18 2. c) P(1) - 3.5 23.5 3.5*0.0302 - 0.1057 20:// = ca) from the poisson poobability table when p=0.8 probability of x = 2 ie P(2) is equal to 0.1438 v0.14 (1) When M=5; P (5) 0.1755 0.18 (e) When pe=2-5; P(3) = 0.2138 V0.21 Given that average number of people admitted to an intensive care unit is 5 people per day. Hence x Bllows porrson destofbutions with M-5. Hence Pcē5. I poobability that the number people admitted to an intensive care unit on a particular day is 2 P (2) -0.00674* 25 4. 21 -5 2 e 533 21 -0.08425 1x2 ~O.OS 5. O using the formula : P 4 8 8. 0.0183/56 *65536 0-02977 $0.02991 40320 Y 0.03 (b) prom table : When us 4 and x=8 probability is equal to 0.0298 V 0.03 Hence we get the same result using books the methods.