4. The hospital had 3000 deliveries each year, so if these happened randomly aro

Question

4. The hospital had 3000 deliveries each year, so if these happened randomly around the clock 1000 deliveries would be expected between the hours of midnight and 8.00 a.m. This is the time when many staff are off duty and it is important to ensure that there will be enough people to cope with the workload on any particular night. a) On how many days in the year would 3 or more deliveries be expected? b) Over the course of one year, what is the greatest number of deliveries expected in any night? c) Why might the pattern of deliveries not follow a Poisson distribution?

Explanation / Answer

The average number of deliveries per night = 1000/365 = 2.74

The number of deliveries can be modelled as following a Poisson distribution with = 2.74

(a) Probability of 0 baby per night, P(X=0) = 2.740 x e-2.74 / 0! = 0.0646

P(X=1) = 2.741 x e-2.74 / 1! = 0.1769

P(X=2) = 2.742 x e-2.74 / 2! = 0.2424

P(X3) = 1 - P(X2) = 1-0.0646 - 0.1769 - 0.2424 = 0.5161

Probable number of days on which there would be 3 or more babies = 365 x 0.5161 = 188.38

So the number of days is 188

(b) P(X=10) = 2.7410 x e-2.74 / 10! = 0.0004

Number of days on which 10 deliveries are expected = 365 x 0.0004 = 0.155. This is less than 1 day, so it is unlikely to have 10 babies delivered in a day

P(X=9) = 2.749 x e-2.74 / 9! = 0.0015. Number of days on which 9 deliveries are expected = 365x0.0015=0.5653

P(X=8) = 2.748 x e-2.74 / 8! = 0.0051. Number of days on which 8 deliveries are expected = 365x0.0051=1.857

So the greatest number of deliveries expected in any night = 8

(c) If deliveries were not random throughout the 24 hours; e.g. if a lot of women had elective caesareans done during the day, then the pattern of deliveries may not follow Poisson distribution.

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