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
Messages
11
Reaction score
0
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!
 
Mathematics news on Phys.org
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...:)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top