A company that makes fleece clothing uses fleece produced from two farms, Northern Farm and Western Farm. Let the random variable X represent the weight of fleece produced by a sheep from Northern Farm. The distribution of X has mean 14.1 pounds and standard deviation 1.3 pounds. Let the random variable Y represent the weight of fleece produced by a sheep from Western Farm. The distribution of Y has mean 6.7 pounds and standard deviation 0.5 pound. Assume X and Y are independent. Let W equal the total weight of fleece from 10 randomly selected sheep from Northern Farm and 15 randomly selected sheep from Western Farm. How do you calculate the standard deviation of W?

Is this calculated as

A) sqrt(10 * 1.3^2 + 15 * 0.5^2), or

B) sqrt(10^2 * 1.3^2 + 15^2 * 0.5^2) ?

I am finding sources citing each of the above as correct. If it's for example A), is there a condition that would change it to B) or vice versa?

Those formulas are often confused but they cover two distinct situations. It depends on how the given mean and standard deviation were obtained. In general the variance of a sum is the sum of the variances of the summed values. Since variance is the square of the SD this allows you to calculate the SD of a sum.

The way your question was worded leads me to think formula A is the correct one. If the mean and SD for each farm is derived from recording the weight of wool from each individual sheep, then the variance for each individual sheep is as given. Multiply that variance by the number of sheep to be added and take the square root of the result to get the SD of the sun - hence formula A.

Where formula B would come in is in the case where it is not individual measurements that are recorded, but rather each recorded value is the mean of some number of replicate measurements. For the formula B you give to apply, the weights recorded at Northern farm would have to be not individual weights, but rather a mean of the weight recorded for 10 sheep. That is data point 1 would be the mean of the weight of wool obtained from sheep numbers 1-10, data point 2 would be the mean of the weights obtained from sheep numbers 11-20 and so on.

The mean given would then be the mean of these data points (which would be the same as the mean of the individual numbers), but the standard deviation would be lower than that of the individual numbers by a factor of SQRT(10). To get the variance we need to find that of the sum we therefore need to multiply the given variance by SQRT(10), which becomes 10 when you bring it inside the radical in formula B, and hence the 10^2 factor. The same applies to the Western farm except that that farm would have to be quoting the mean of 15 replicates rather than 10.

Very helpful, thank you.