The number of people entering the intensive care unit at a

Chapter 3, Problem 207SE

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QUESTION:

Problem 207SE

The number of people entering the intensive care unit at a hospital on any single day possesses a Poisson distribution with a mean equal to five persons per day.

a What is the probability that the number of people entering the intensive care unit on a particular day is equal to 2? Is less than or equal to 2?

b Is it likely that Y will exceed 10? Explain.

Questions & Answers

QUESTION:

Problem 207SE

The number of people entering the intensive care unit at a hospital on any single day possesses a Poisson distribution with a mean equal to five persons per day.

a What is the probability that the number of people entering the intensive care unit on a particular day is equal to 2? Is less than or equal to 2?

b Is it likely that Y will exceed 10? Explain.

ANSWER:

Solution :

Step 1 of 2:

Given the poisson distribution with a mean equal to five person.

Here =5.

Our goal is:

a). We need to find to find the probability that the number of people entering the intensive

     care unit on a particular is equal to 2.

b). We need to find is it likely that Y will exceed 10.Explain.

a).

Now we have to find the probability that the number of people entering the intensive care unit on a particular is equal to 2.

The formula of the poisson probability is

P(X=k)=

Where k is the actual number of successes that result from the experiment.

e is approximately equal to 2.71828.

Let k=2.

P(X=2)=

P(X=2)=

P(X=2)=

P(X=2)= 0.0842

Then the probability of less than or equal to 2 is

P(X2)=P(X=0)+P(X=1)+P(X=2)

Where, P(X=0)+P(X=1)+P(X=2) is obtained from Excel by using the function

“=poisson(x,,false)”

We know that =5.

Then the table is given below.

X

P(X 2)

0

0.006738

1

0.03369

2

0.084224

Total

0.124652

Now,

P(X2)= P(X=0)+P(X=1)+P(X=2)

P(X2)= 0.006738+0.03369+0.084224

P(X2)= 0.124652

Therefore the probability of less than or equal to 2 is 0.124652.


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