# According to a recent government report, the aging of the U.S. population is translating into many more visits to doctors’ offices and hospi

Question

According to a recent government report, the aging of the U.S. population is translating into many more visits to doctors’ offices and hospitals (USA Today, August 7, 2008). It is estimated that an average person makes four visits a year to doctors’ offices and hospitals. What is the probability that an average person makes at least one MONTHLY visit to doctors’ office and hospitals

in progress 0
1 year 2021-09-04T13:45:41+00:00 1 Answers 58 views 0

0.2835 = 28.35% probability that an average person makes at least one MONTHLY visit to doctors’ office and hospitals

Step-by-step explanation:

Mean during a time period means that we use the Poisson distribution.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$$\mu$$ is the mean in the given interval.

It is estimated that an average person makes four visits a year to doctors’ offices and hospitals.

A year has 12 months, which means that the monthly mean is $$\mu = \frac{4}{12} = \frac{1}{3}$$

What is the probability that an average person makes at least one MONTHLY visit to doctors’ office and hospitals?

This is:

$$P(X \geq 1) = 1 – P(X = 0)$$

In which

$$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$$

$$P(X = 0) = \frac{e^{-(\frac{1}{3})}*(\frac{1}{3})^{0}}{(0)!} = 0.7165$$

$$P(X \geq 1) = 1 – P(X = 0) = 1 – 0.7165 = 0.2835$$

0.2835 = 28.35% probability that an average person makes at least one MONTHLY visit to doctors’ office and hospitals