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4. At a local hospital, babies are born at an average rate of four births per night (consider night to be an 8 hour period
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Answer #1

We would be looking at the first 4 parts of the problem here as:

a) The average time between two consecutive births is computed here as:

= Total time for a night / Average number of births in a night

= 8/4 = 2

Therefore the average time required here is 2 hours.

b) Probability that after a birth occurs, next occurs in less than half an hour is computed here as:

= 1 - Probability of no birth in 30 minute period.

Average number of births in 30 minutes period = 4/16 = 0.25. Therefore the probability here is computed as:

= 1 - e^{-0.25} = 0.2212

Therefore 0.2212 is the required probability here.

c) Probability that exactly 5 births occur in a single night is computed here as:

P(X = 5) = \frac{4^5}{5!}e^{-4} = 0.1563

Therefore 0.1563 is the required probability here.

d) Probability that less than 5 births occur in a single night is computed here as:

P(X < 5) = P(X =0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X < 5) = e^{-4} + 4e^{-4} + \frac{4^2}{2}e^{-4} + \frac{4^3}{3!}e^{-4}+ \frac{4^4}{4!}e^{-4} = 0.4405

Therefore 0.4405 is the required probability here.

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