Question

Elizabeth is a busy pediatrician. On any given day, she diagnoses an average of four babies with middle-ear infections. Assume that the number of babies who come to her clinic with middle-ear infections is a Poisson random variable. Calculate the probability that fewer than three babies with middle-ear infections will come to her clinic tomorrow. Give your answer in decimal form precise to three decimal places. P(X<3)=

Answer 1

Answer:

There is a 23.81% probability that fewer than three babies with middle-ear infections will come to her clinic tomorrow.

Step-by-step explanation:

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.

In this problem

Elizabeth is a busy pediatrician. On any given day, she diagnoses an average of four babies with middle-ear infections. This means that $\mu = 4$.

Calculate the probability that fewer than three babies with middle-ear infections will come to her clinic tomorrow.

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X = 1) = \frac{e^{-4}*4^{1}}{(2)!} = 0.0733$

$P(X = 2) = \frac{e^{-4}*4^{2}}{(2)!} = 0.1465$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0183 + 0.0733 + 0.1465 = 0.2381$

There is a 23.81% probability that fewer than three babies with middle-ear infections will come to her clinic tomorrow.