Question 1081996: Please help me solve the following probability question
The average length of stay in a hospital is useful for planning purposes. Suppose that the following is the distribution of the length of stay in a hospital after a minor operation.
x = 1, 2, 3, 4, 5
P(x) = 0.6, 0.1, 0.04, 0.02, c
What is the average length of stay in the hospital?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Given Distribution
| X | 1 | 2 | 3 | 4 | 5 | | P(X) | 0.6 | 0.1 | 0.04 | 0.02 | c |
First we need to know the value of c. The probabilities must add to 1 for this to be a true probability distribution
0.6+0.1+0.04+0.02+c = 1
0.76+c = 1
0.76+c-0.76 = 1-0.76
c = 0.24
So the probability distribution updates to
| X | 1 | 2 | 3 | 4 | 5 | | P(X) | 0.6 | 0.1 | 0.04 | 0.02 | 0.24 |
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To find the average, multiply the probabilities with their corresponding x values. Then add up those products.
1*0.6+2*0.1+3*0.04+4*0.02+5*0.24 = 2.2
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Final Answer: The average length of stay in the hospital is 2.2 days
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