Question 319346: suppose that the time waiting in line has an expected value of 4 minutes, with a standard deviation of 1.2 minutes, and the time it takes to be served by the teller has an expected value of 5.5 minutes, with a standard deviation of 1.5 minutes. Compute the
a. expected value of the total time it takes to be served at the bank
b. standard deviation of the total time it takes to be served at the bank
I think a. is just the total of the expected values. 4+5.5=9.5 but I cannot figure out b.
any help would be appreciated.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! suppose that the time waiting in line has an expected value of 4 minutes, with a standard deviation of 1.2 minutes, and the time it takes to be served by the teller has an expected value of 5.5 minutes, with a standard deviation of 1.5 minutes. Compute the
a. expected value of the total time it takes to be served at the bank
The mean of the sum = the sum of the means
u(wait + serve) = u(wait) + u(serve) = 4+5.5 = 9.5
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b. standard deviation of the total time it takes to be served at the bank
The variance of a sum = the sum of the variances
V(wait + serve) = v(wait) + v(serve
= 1.2^2 + 1.5^2
= 3.69
Therefore std(sum) = sqrt(3.69) = 1.921
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Cheers,
Stan H.
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I think a. is just the total of the expected values. 4+5.5=9.5 but I cannot figure out b.
any help would be appreciated.
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