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 3.1. A binomial distribution with mean 6.2 and standard deviation 1.76. A binomial distribution with mean 6.2 and standard deviation 3.8. A Poisson distribution with mean 6.2 and standard deviation 6.2. (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:

1)  Poisson distribution with mean 6.2 and standard deviation 2.49.

2) 0.0536

Step-by-step explanation:

We are given the following information in the question:

Mean of daily surgeries = 6.2

a) The count of daily surgeries can be treated as a Poisson distribution.

• The Poisson distribution is the discrete probability distribution of the number of events occurring in a given time period, given the average number of times the event occurs over that time period.
• The variance of Poisson distribution is equal to the mean of Poisson distribution.

Variance = 6.2

Standard deviation = $\sqrt{6.2} = 2.49$

a) Poisson distribution with mean 6.2 and standard deviation 2.49.

2) P( 2 or less surgeries)

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

$P( x \leq 2) = P(x =0) + P(x=1) + P(x=2)\\\\= \displaystyle\frac{6.2^0 e^{-6.2}}{0!} + \displaystyle\frac{6.2^1 e^{-6.2}}{1!} + \displaystyle\frac{6.2^2 e^{-6.2}}{2!}\\\\ = 0.00203+ 0.01259 + 0.03900\\\\= 0.0536 = 5.36\%$