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

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

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
`√(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)`

`= (e^(-6.2) 6.2^0)/(0!) +(e^(-6.2) 6.2^1)/(1!) +(e^(-6.2) 6.2^2)/(2!)`

= 0.002029 + 0.012582+ 0.039006

= 0.053617

= 0.0536