Question 1081996
Given Distribution
<table border=1><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td></tr><tr><td>P(X)</td><td>0.6</td><td>0.1</td><td>0.04</td><td>0.02</td><td>c</td></tr></table>


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
<table border=1><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td></tr><tr><td>P(X)</td><td>0.6</td><td>0.1</td><td>0.04</td><td>0.02</td><td>0.24</td></tr></table>


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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 <font color=red>2.2 days</font>