Question

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):

Answer 1

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$\sigma^2 = mean = 6.2$

hence, standard deviation is given as $\sqrt{6.2} = 2.48998$

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

$P(X \leq 2) = P(X =0) + P(X =1) + P(X =2)$

                 $= \frac{e^{-6.2} 6.2^0}{0!} +\frac{e^{-6.2} 6.2^1}{1!} +\frac{e^{-6.2} 6.2^2}{2!}$

                 = 0.002029 + 0.012582+ 0.039006

                 = 0.053617

                = 0.0536