The mean number of daily surgeries at a local hospital is 6.2. Assume that surgeries are random, independent events. (a) The count of daily surgeries follows approximately: A Poisson distribution with mean 6.2 and standard deviation 2.49. A binomial distribution with mean 6.2 and standard deviation 3.8. A binomial distribution with mean 6.2 and standard deviation 1.76. A Poisson distribution with mean 6.2 and standard deviation 6.2. A binomial distribution with mean 6.2 and standard deviation 3.1. (b) The probability that there would be only 2 or fewer surgeries in a given day is approximately (round to 4 decimal places):

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rejkjavik rejkjavik · Answer 1 · 2019-10-07T12:38:58+02:00

Answer:

a) correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

b) probability for only 2 or less than 2 surgeries in a given day is 0.0536

Step-by-step explanation:

Given data:

mean number is given as 6.2

correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

we know that variance is given as[tex] \sigma^2 = mean = 6.2[/tex]

hence, standard deviation is given as [tex]\sqrt{6.2} = 2.48998 [/tex]

thus standard deviation is 2.49

correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

b) probability for only 2 or less than 2 surgeries in a given day is

[tex]P(X \leq 2) = P(X =0) + P(X =1) + P(X =2)[/tex]

[tex]= \frac{e^{-6.2} 6.2^0}{0!} +\frac{e^{-6.2} 6.2^1}{1!} +\frac{e^{-6.2} 6.2^2}{2!}[/tex]

= 0.002029 + 0.012582+ 0.039006

= 0.053617

= 0.0536