214k views
2 votes
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)=

User Meika
by
8.1k points

1 Answer

0 votes

Answer:

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

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) = (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) = (e^(-\mu)*\mu^(x))/((x)!)


P(X = 0) = (e^(-4)*4^(0))/((0)!) = 0.0183


P(X = 1) = (e^(-4)*4^(1))/((2)!) = 0.0733


P(X = 2) = (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.

User Chizh
by
8.4k points