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Obtaining standard deviation using different equations

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As you can see I used two equations in order to obtain $\sigma$ (standard deviation) but did not obtain the same result.

I think my mistake has to be at b), calculating the average value of $(\Delta j)^2$.

First method to compute $\sigma$ :

Standard deviation is defined:

$$\sigma^2 = <(\Delta j)^2>$$

And we also know:

$$\Delta j = j - < j > $$

So first of all I calculated < j >, the average value of a discrete function:

$$< j > = \sum_{j = 0}^{\infty} j P(j) = \frac{1}{N} \sum_{j = 0}^{\infty} j N(j) -------> EQ1$$

Which equals to 21.

Then I computed each $\Delta j$ for each sample (age 14,15...)

At this point I applied EQ1 in order to obtain $<(\Delta j)^2>$ and subsequently $\sigma^2$

As you can see I got $\sigma$ = 3.117 using method 1

Second method to compute $\sigma$ :

Another equation involving $\sigma$:

$$\sigma^2 = < j^2 > -< j >^2$$

I computed $< j^2 >$ using EQ1 and squared < j >. Afterwards I applied the formula and got:

$\sigma$ = 4.309

There has to be a mistake in method one (I think) but I do not see it

What do you think?

Thank you

asked Aug 18 in Physics Problems by JD_PM (490 points)
edited Aug 19 by JD_PM
Sorry I do not understand your calculation. Can you explain what your 2 different methods were?
Okey Sr I updated the question

1 Answer

1 vote
 
Best answer

I think my mistake has to be at b), calculating the average value of $(\Delta j)^2$.

Yes, this part of your calculation does seem to be incorrect. In calculating the average you have to weight each value of $(\Delta j)^2$ by its frequency $N(j)$. You do not appear to have done that.

answered Aug 19 by sammy gerbil (26,096 points)
edited Aug 19 by sammy gerbil
It was that thanks :)
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