MHB Standard Deviation: 10 & 11 Consecutive Positive Multiples of n

AI Thread Summary
The discussion focuses on calculating the standard deviation of consecutive positive multiples of a positive integer n. The standard deviation for 10 consecutive multiples of n is derived using the formula σ = (n/2)√((N^2-1)/3), where N is the number of multiples. Similarly, the standard deviation for 11 consecutive multiples can be calculated using the same formula with N set to 11. The user seeks clarity on whether the relationship can be determined from the provided information. The conversation emphasizes the importance of understanding the standard deviation formula in this context.
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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