Standard Deviation: 10 & 11 Consecutive Positive Multiples of n

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SUMMARY

The discussion focuses on calculating the standard deviation of consecutive positive multiples of a positive integer n. The population standard deviation formula is provided as $$\sigma=\sqrt{\frac{\sum(x-\mu)^2}{N}}$$. For 10 consecutive multiples of n, the standard deviation is derived as $$\sigma=\frac{n}{2}\sqrt{\frac{10^2-1}{3}}$$, and for 11 consecutive multiples, it is $$\sigma=\frac{n}{2}\sqrt{\frac{11^2-1}{3}}$$. The calculations confirm that the standard deviation can be determined using the specified formulas.

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greprep
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Hi, All. I'm trying to re-familiarize myself with standard deviations. Any resources? I'm reading through the threads here and trying to figure out the following:

"n is a positive integer.
What is the standard deviation of 10 consecutive positive multiples of n.
And what is the standard deviation of 11 consecutive positive multiples of n?"

Can the relationship not be determined from the information given? Many Thanks!
 
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I would begin with the following formula for population standard deviation:

$$\sigma=\sqrt{\frac{\sum(x-\mu)^2}{N}}$$

Next, I would look at:

$$\mu=\frac{1}{N}\sum_{k=m}^{m+(N-1)}(kn)=\frac{n}{2}(2m+N-1)$$

And then:

$$\sum_{k=m}^{m+(N-1)}\left(kn-\frac{n}{2}(2m+N-1)\right)^2=\frac{n^2N\left(N^2-1\right)}{12}$$

And thus:

$$\sigma=\frac{n}{2}\sqrt{\frac{N^2-1}{3}}$$

Now you can use the above formula to answer the questions...:)
 

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