Combine two standard deviations?

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Calculating the mean of two combined standard deviations is straightforward, but determining the variance is complex due to the squaring of values. For uncorrelated jointly distributed random variables, the variance of their sum can be expressed as the sum of their individual variances. The discussion highlights the ambiguity of the term "standard deviation," which can refer to a population parameter, an estimator, or a descriptive statistic. It suggests looking for specific formulas in statistical texts that address pooled means and variances for accurate calculations. Understanding these distinctions is crucial for correctly solving problems involving combined standard deviations and variances.
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Homework Statement
Two materials, x1,..,x10 and y1,..y5 (note the 5!) have the following stats:
x_m = 5313
s_x = 5.2
y_m = 5309
s_y = 3

Assume both x and y are the same material, what is the new mean and standard deviation?
Relevant Equations
$$s^2 = \frac {1}{n - 1} * \sum (x_i - x_m)^2$$
$$s^2 = \frac {1}{n - 1} * ( \sum x_i^2 -1/n (\sum x_i)^2)$$
The mean is easy to calculate:
(x_m * 10 + y_m * 5) /15 = 5312
Which is correct.

But when you're suppose to calculate the variance it's impossible.
The values are squared so none of the equations will really help me..
 
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It can be shown that for uncorrelated jointly distributed random variables the variance of the sum of those variables is

σ2 = ∑σi2
 
Addez123 said:
Assume both x and y are the same material, what is the new mean and standard deviation?
In statistics, terms like "standard deviation" are ambiguous. "Standard deviation" may refer to:

1)A parameter of the distribution of a population

2)An "estimator", i.e. a formula for estimating the parameter from the values in a sample

3)A descriptive statistic. For example, some texts on descriptive statistics define the standard deviation of a sample to be ##\sqrt{ s^2}## where ##s^2 = \frac{1}{n} \sum (x_i - x_m)^2##. They use the factor ##\frac{1}{n}## instead of ##\frac{1}{n-1}##.

What @gleem wrote applies to interpreting "standard deviation" as 1) a parameter of a distribution. However, I suspect the problem you stated refers to "standard deviation" in the sense of 2) or 3).

If your problem occurs in a part of the text that deals with "pooled" means, standard deviations, and variances, you should look for formulas that the text uses to solve such problems. It would surprise me if a text expects you to discover the formulas by yourself.

I have in mind formulas like those on https://www.statisticshowto.com/pooled-standard-deviation/, but you will have to see what your own text expects you to use.
 
Last edited:
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I'll proceed like you did for the mean.
For a given sample
$$s^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-x_m)^2=\frac{1}{n-1}\left(\sum_{i=1}^nx_i^2-nx_m^2\right)$$
Suppose that ##z_i## represents an element of your new sample (combination of the first two), then you have$$s_z^2=\frac{1}{n_z-1}\left(\sum_{i=1}^{n_z}z_i^2-n_zz_m^2\right)$$
Notice that ##\sum z_i^2=(x_1^2+...+x_{10}^2+y_1^2+...+y_5^2)=\sum x_i^2+\sum y_i^2##, which gives you$$s_z^2=\frac{1}{n_z-1}\left(\sum_{i=1}^{10}x_i^2+\sum_{i=1}^{5}y_i^2-n_zz_m^2\right)$$
You have already found ##z_m##. Now, you need to find ##n_z##, ##\sum x_i^2## and ##\sum y_i^2##.
 
Last edited:
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