Question

the time between visits to a u.s. emergency room for a member of the general population follows an exponential distribution with a mean of 2.5 years. what proportion of the population:

Answer 1

Using the exponential distribution, it is found that the proportion of the population that will visit an emergency room in the next six months is of 0.1813 = 18.13%.

Exponential distribution

The exponential probability distribution, with mean m, is described by the following mass function:

$f(x) = \mu e^{-\mu x}$

In this function, $\mu = \frac{1}{m}$ represents the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

The solution of the definite integral gives the cumulative mass function as follows:

$P(X \leq x) = 1 - e^{-\mu x}$

In this problem, the mean is of 2.5 years, hence the decay parameter is calculated as follows:

$\mu = \frac{1}{2.5} = 0.4$

The proportion of the population that will visit the emergency room in the next six months is P(X < 0.5), as 6 months = 0.5 years, hence:

$P(X \leq 0.5) = 1 - e^{-0.4 \times 0.5} = 0.1813.$

Missing Information

The problem asks for the proportion of the population that will visit an emergency room in the next six months.

More can be learned about the exponential distribution at https://brainly.com/question/14634921

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