Question

In: Accounting

Patients arrive at the emergency room of Costa Val- ley Hospital at an average of 5...

Patients arrive at the emergency room of Costa Val- ley Hospital at an average of 5 per day. The demand for emergency room treatment at Costa Valley fol- lows a Poisson distribution. (a) Using Appendix C , What is the sum of these probabilities, and why is the number less than 1?

The Appendix C, compute the probabilty of exactly 0,1,2,3,4, and 5 arrivals per day. Put in excel.

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A B C D E F G H I J
2
3 As per Poisson distribution, the probability of observing k events over time is
4 P(X=k) = (?ke-?)/k!
5
6 Where ? is the average number of times the event occurs over the time period.
7
8 Since number of patient arriving the energency room is 5 per day,
9 therefore ? is 5 per day and period is day.
10
11 ?= 5 per day
12
13 As per Poisson distribution, the probability of observing k arrivals per day will be
14 P(X=k) = (?ke-?)/k!
15
16 P(X=0) = (?0e-?)/0!
17 =POISSON.DIST(0,D11,FALSE) =POISSON.DIST(0,D11,FALSE)
18
19 Hence probability of zero arrivals per day is =D17
20 Similarly probability of other number of arrivals can be calculated as below:
21
22 P(X=1) = (?1e-?)/1!
23 =POISSON.DIST(1,$D$11,FALSE) =POISSON.DIST(1,$D$11,FALSE)
24
25 P(X=2) = (?2e-?)/2!
26 =POISSON.DIST(2,$D$11,FALSE)
27
28 P(X=3) = (?3e-?)/3!
29 =POISSON.DIST(3,$D$11,FALSE)
30
31 P(X=4) = (?4e-?)/4!
32 =POISSON.DIST(4,$D$11,FALSE)
33
34 P(X=5) = (?5e-?)/5!
35 =POISSON.DIST(5,$D$11,FALSE)
36

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