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I am helping someone study for a statistics exam. I am quite good at most other math classes but it's been a while since I studied statistics. I am stuck on one of the exercise problems that we worked on.

Exercise:

This is a paraphrase of the exercise:

A mason lays a row of cement blocks with a layer of motar. Each row (bricks and mortar) has a mean height of 8 inches with standard deviation 0.1 inches. What is the probability that a wall built from 4 rows of cement blocks differs from 32 inches by more than a half-inch? Assume that the height of each row is independent and distributed normally.

What I can do:

I know that I need to find the mean $\mu_{wall}$ and standard deviation $\sigma_{wall}$ for the wall made from four rows of cement blocks. I'm given that the mean $\mu_{row}$ and standard deviation $\sigma_{row}$ of one row of blocks are $\mu_{row} = 8$ and $\sigma_{row} = 0.1$. So for $n = 4$ rows, we have $\mu_{walls} = n \mu_{row} = 4 \cdot 8 = 32$.

Question:

How do I calculate $\sigma_{wall}$ knowing that $\sigma_{row} = 0.1$? My first guess would be to simply multiply by 4, but I'm not sure that is correct. Once I have the standard deviation, doing a Z test is easy.

1 Answers 1

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No. Variances are additive, so:

$$var\left(\sum_{i=1}^n x_i\right) = \sum_{i=1}^n var(x_i) = n \cdot var(x)$$

Which means:

$$std \left(\sum_{i=1}^n x_i\right) = \sqrt{n} \cdot std(x)$$

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    So the short answer is to multiply the standard deviation by $\sqrt n$ in order to solve the story problem. Thanks!2012-11-04
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    Exactly! That is known as the Bienamyé formula and only works if your random variables are uncorrelated. http://en.wikipedia.org/wiki/Variance#Sum_of_uncorrelated_variables_.28Bienaym.C3.A9_formula.29 Also, consider accepting the answer, if it helped you (click the check mark).2012-11-04
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    I was waiting for some more views before accepting your answer. I didn't want to scare anyone off if they had something else to add. ;-)2012-11-04
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    The above answer is wrong! See here http://en.wikipedia.org/wiki/Variance2013-05-22
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    @guangbo: If the variables are independent, then the answer is correct. If there is some dependence, then covariance needs to be involved.2013-05-22