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Answer :
To determine which distribution the count of daily surgeries at a hospital follows, we need to consider the characteristics of the distributions provided:
Binomial Distribution: This distribution models the number of successes in a fixed number of independent trials, each with the same probability of success. It is characterized by two parameters: [tex]n[/tex] (number of trials) and [tex]p[/tex] (probability of success). However, for a daily count of surgeries where we only know the average, a binomial distribution is not applicable without knowing the fixed number [tex]n[/tex] and the probability [tex]p[/tex].
Poisson Distribution: This distribution is used to model the number of occurrences of an event in a fixed interval of time or space. It is applicable here because surgeries can be considered independent, random events happening at a constant average rate. The Poisson distribution is characterized by its mean [tex]\lambda[/tex], and for a Poisson distribution, the mean is equal to its variance.
Given that the mean number of daily surgeries is 6.2, for a Poisson distribution, the mean ([tex]\lambda[/tex]) and variance are both 6.2.
Let's evaluate the options:
Option A: A binomial distribution with mean 6.2 and variance 3.8 is not appropriate because it assumes a fixed number of trials and probability, which are not known here.
Option B: A Poisson distribution with mean 6.2 and variance 6.2 matches the properties of the Poisson distribution which applies to this scenario.
Option C: A Poisson distribution with mean 6.2 and variance 2.49 does not match because, in a Poisson distribution, mean and variance must be the same.
Option D: A binomial distribution, with mean 6.2 and variance 1.95, also does not fit for the same reasons as Option A.
Therefore, Option B is correct, as surgeries at a hospital being independent events occurring at a constant rate fits a Poisson distribution with mean and variance both equal to 6.2.
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