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The number of people entering the intensive care unit at a
Chapter 3, Problem 207SE(choose chapter or problem)
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.