SOLUTION: 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

Question 1081995: 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
Find the variance and standard deviation
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
E(X)=0.6+0.2+0.12+0.08+1.2; the 1.2 comes from5*0.24, which is what c has to be to have the probability =1
E(X)=2.2
V(X)= sum (x-E(X))^2*p(x)
-(1.2)^2*0.6=0.864
-(0.2)^2*0.1=0.004
0.8^2*0.04=0.0256
1.8^2*0.02=0.0648
2.8^2*0.24=2.0256
Adds to 2.984
SD=1.73, sqrt of the variance
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