How Do You Calculate Standard Deviation for Linear Combinations?

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The discussion centers on calculating the standard deviation for the linear combination D = A - B - C, given the expected values and standard deviations for A, B, and C. The user calculated the variance V(D) as the sum of the squares of the individual standard deviations, resulting in V(D) = 0.0225 and thus sigma(D) = 0.15. They noted that the book's answer is 0.1225, which does not match their calculation. Other participants in the discussion agree with the user's calculation, suggesting that the book may contain an error. The conversation highlights the importance of verifying calculations in statistical problems.
doubled
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I have attached the problem of interest.
I think I am having trouble calculating the standard deviation for part a.

D=A-B-C
E(A)=10 E(B)=2 E(C)=2 sigma(A)=0.1 sigma(B)=0.05 sigma(C)=0.1
where sigma=standard deviation

so to find SD...:
V(D)=V(A-B-C)=sigma(A)2+sigma(B)2+sigma(C)2=(0.1^2)+(0.05^2)+(0.1^2)=0.0225
sigma(D)=(V(D))^0.5=0.0225^0.5=0.15

The book's answer is 0.1225. I do not see any glaring mistakes. Could someone please take a look?
The book may be wrong. I just want to make sure that it is not me that's wrong.
 

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doubled said:
I have attached the problem of interest.
I think I am having trouble calculating the standard deviation for part a.

D=A-B-C
E(A)=10 E(B)=2 E(C)=2 sigma(A)=0.1 sigma(B)=0.05 sigma(C)=0.1
where sigma=standard deviation

so to find SD...:
V(D)=V(A-B-C)=sigma(A)2+sigma(B)2+sigma(C)2=(0.1^2)+(0.05^2)+(0.1^2)=0.0225
sigma(D)=(V(D))^0.5=0.0225^0.5=0.15

The book's answer is 0.1225. I do not see any glaring mistakes. Could someone please take a look?
The book may be wrong. I just want to make sure that it is not me that's wrong.
I agree with your answer.
 
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