Patients arrive at the emergency room of Costa Valley Hosipital at an average of 5 per day. The
demand for emergency room treatment at Costa Valley follows a Poisson distribution.
(a) Using a Poisson appendix, compute the probability of exactly 0,1,2,3,4 and 5 arrivals per day.
(b) What is the sum of these probabilities, and why is the number less than 1?
Please show work. Thank you..
Solution
Mean value for the poisson\'s distribution = 5
So, the probability for various numbers is as follows:
P(X= 0) = 50e-5 / 0! = e-5 = 0.0067
P(X = 1) = 51e-5/ 1! = 5e-5 = 0.0336
P(X = 2) = 52e-5/2! = 25e-5/2 = 0.0842
P(X = 3) = 53e-5/3! = 125e-5/6 = 0.1403
P(X=4) = 54e-5/4! = 625e-5/24 = 0.1754
P(X=5) = 55e-5/5! = 3125e-5/120 =0.1754
Sum of al these probabilities = 0.6157 which is less than one because X can take even more
values like 6,7,8,9, etc. and so on each having a significant probability value.
Hope this helps. Please Rate it ASAP. Thanks..
Patients arrive at the emergency room of Costa Valley Hosipital at a.pdf
1. Patients arrive at the emergency room of Costa Valley Hosipital at an average of 5 per day. The
demand for emergency room treatment at Costa Valley follows a Poisson distribution.
(a) Using a Poisson appendix, compute the probability of exactly 0,1,2,3,4 and 5 arrivals per day.
(b) What is the sum of these probabilities, and why is the number less than 1?
Please show work. Thank you..
Solution
Mean value for the poisson's distribution = 5
So, the probability for various numbers is as follows:
P(X= 0) = 50e-5 / 0! = e-5 = 0.0067
P(X = 1) = 51e-5/ 1! = 5e-5 = 0.0336
P(X = 2) = 52e-5/2! = 25e-5/2 = 0.0842
P(X = 3) = 53e-5/3! = 125e-5/6 = 0.1403
P(X=4) = 54e-5/4! = 625e-5/24 = 0.1754
P(X=5) = 55e-5/5! = 3125e-5/120 =0.1754
Sum of al these probabilities = 0.6157 which is less than one because X can take even more
values like 6,7,8,9, etc. and so on each having a significant probability value.
Hope this helps. Please Rate it ASAP. Thanks.
