What is the root mean square average?

  • #1
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Uing a probability distribution for values obtained in throwing 3 dice together, find the uncertainty associated with throwing th 3 dice togehter, that is, the root mean square average of the deviation of a given thrown from the average throw.

What exactly is "the average throw" ? I know what the rms is but what is the "root mean square average"?
 

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  • #2
Galileo
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The rms is the Root Mean Square average. It is by definition the root of the average of the squares of the deviations from the average. (phew)

The average throw is simply the average result you get when tossing lots of 3d6's. It's somewhere between 10 and 11.
 
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  • #3
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here's what I did to calculate the average throw, I first found out all the probabilities to throw the numbers 3-17. So for example, there are 15 ways to roll a sum of 7, which means that the probability is 15/207 (total number of ways). Then I multiplied the probability by the number (so in the previous case 15/207 * 7). I added all of these numbers up to get 10.75362339. Would this be the average?

Also, what exactly is "the deviations from the average"?
 
  • #4
nrqed
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UrbanXrisis said:
here's what I did to calculate the average throw, I first found out all the probabilities to throw the numbers 3-17. So for example, there are 15 ways to roll a sum of 7, which means that the probability is 15/207 (total number of ways). Then I multiplied the probability by the number (so in the previous case 15/207 * 7). I added all of these numbers up to get 10.75362339. Would this be the average?

Also, what exactly is "the deviations from the average"?

Yes, that's the average. For deviation, they just want you to calculate the RMS value. You need to calculate [itex] {\sqrt{ {{<x^2>}} - {{< x>}}^2}}[/itex]. So calculate the average value of the *square* of the values given by the dice. Take that result minus the square of 10.75362339. Take the square root of the result.
 
  • #5
Galileo
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UrbanXrisis said:
here's what I did to calculate the average throw, I first found out all the probabilities to throw the numbers 3-17. So for example, there are 15 ways to roll a sum of 7, which means that the probability is 15/207 (total number of ways). Then I multiplied the probability by the number (so in the previous case 15/207 * 7). I added all of these numbers up to get 10.75362339. Would this be the average?

Also, what exactly is "the deviations from the average"?
Am I missing something? :confused: You are throwing three dice, right? So it's possible to throw 18 and the number of possible outcomes is 6x6x6=216 (six possible outcomes for each die).

By deviation from the average, I just mean [itex]x-\langle x\rangle[/itex]. Where <x> is the average. So the square of the deviation from the average is [itex](x-\langle x\rangle)^2[/itex] (it's just another random variable). Its average value is called the variance:
[tex]\sigma^2=\langle(x-\langle x\rangle)^2\rangle=\langle x^2\rangle-\langle x\rangle^2[/tex]
And the square root is the standard-deviation, which is what you are asked to find.
 
  • #6
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I thought RMS is [tex]=\sqrt { \frac{(x_1 - <x>)^2 + ... + (x_n -<x>)^2}{n}} [/tex]

the formula [tex]\sqrt{\langle x^2\rangle-\langle x\rangle^2}[/tex] would only work for one of one of the values... say... I rolled a 10, then the RMS would be [tex]\sqrt{\langle 10^2\rangle-\langle 10.5\rangle^2}[/tex]

but the uncertainty associated with throwing the 3 dice together, would mean that I would have to take into account all the rolls right? so that the RMS is [tex]=\sqrt { \frac{(x_1 - <x>)^2 + ... + (x_n -<x>)^2}{n}} =\sqrt { \frac{(3- 10.5)^2 + ... + (18 -10.5)^2}{15}} [/tex]
 
  • #7
nrqed
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UrbanXrisis said:
I thought RMS is [tex]=\sqrt { \frac{(x_1 - <x>)^2 + ... + (x_n -<x>)^2}{n}} [/tex]

the formula [tex]\sqrt{\langle x^2\rangle-\langle x\rangle^2}[/tex] would only work for one of one of the values... say... I rolled a 10, then the RMS would be [tex]\sqrt{\langle 10^2\rangle-\langle 10.5\rangle^2}[/tex]
No, the standard deviation [tex]\sqrt{\langle x^2\rangle-\langle x\rangle^2}[/tex] is defined for the set of all possible throws. The xpression [itex] \langle x^2 \rangle[/itex] is the average of the squared of the values.
but the uncertainty associated with throwing the 3 dice together, would mean that I would have to take into account all the rolls right? so that the RMS is [tex]=\sqrt { \frac{(x_1 - <x>)^2 + ... + (x_n -<x>)^2}{n}} =\sqrt { \frac{(3- 10.5)^2 + ... + (18 -10.5)^2}{15}} [/tex]
This gives the same thing as the expression I gave above (assuming that you take into account how many permutations give a certain value, right? For exampple, for the result 4, there are 3 possibilities).

Notice that the average [itex] \langle (x - \langle x \rangle)^2 \rangle [/itex] is equal to [itex] \langle x^2 - 2 x \langle x \rangle +\langle x \rangle^2 \rangle [/itex] which is [itex] \langle x^2 \rangle - \langle x \rangle^2 [/itex].

Patrick
 

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