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.


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 $\mu_{walls} = n \mu_{row} = 4 \cdot 8 = 32$.


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.


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:

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

  • $\begingroup$ So the short answer is to multiply the standard deviation by $\sqrt n$ in order to solve the story problem. Thanks! $\endgroup$ – Code-Guru Nov 4 '12 at 22:21
  • $\begingroup$ Exactly! That is known as the Bienamyé formula and only works if your random variables are uncorrelated. en.wikipedia.org/wiki/… Also, consider accepting the answer, if it helped you (click the check mark). $\endgroup$ – pedrosorio Nov 4 '12 at 22:51
  • $\begingroup$ 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. ;-) $\endgroup$ – Code-Guru Nov 4 '12 at 23:06
  • $\begingroup$ The above answer is wrong! See here en.wikipedia.org/wiki/Variance $\endgroup$ – user78973 May 22 '13 at 15:57
  • 1
    $\begingroup$ @guangbo: If the variables are independent, then the answer is correct. If there is some dependence, then covariance needs to be involved. $\endgroup$ – robjohn May 22 '13 at 16:18

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