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):

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

hence, standard deviation is given as

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

                 = 0.002029 + 0.012582+ 0.039006

                 = 0.053617

                = 0.0536

50x5=$250

step-by-step explanation: