In a series of 2n observations, half of them equal a and remaining half -a.If the standard deviation of the observations is 2,then |a| equals ?

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In a series of 2n observations, half of them equal a and remaining half -a.If the standard deviation of the observations is 2,then |a| equals ?

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Answer 1 · 2018-01-12T00:18:04

$| a | = 2$ $absa=2$

Explanation:

Recall the definition of standard deviation:

$σ = \sqrt{\frac{\sum {(x_{i} - μ)}_{i}^{2}}{N}}$ $sigma = sqrt((sum(x_i-mu)^2)/N)$

Where $μ$ $mu$ is the mean, $x_{i}$ $x_i$ are the observations, $\sum$ $sum$ is the sum, and $N$ $N$ is the number of observations.

Let's first find the mean. By definition:

$μ = \frac{\sum x_{i}}{N}$ $mu=(sumx_i)/N$

Recall that $\sum x_{i}$ $sumx_i$ represents the sum of all the observations. Since we have $n$ $n$ observations that are $a$ $a$ and $n$ $n$ observations that are $- a$ $-a$ , we write:

$mu=(overbrace(a+a+...+a)^(n" times")+overbrace(-a -a-...-a)^(n " times"))/(2n)$

Which is equivalent to:

$mu=(na+n(-a))/(2n)$

And, simplifying:

$mu=0$

This makes our calculation for the standard deviation much simpler as well:

$sigma=sqrt((sum(x_i-0)^2)/N)=sqrt((sumx_i^2)/(2n))$

Recall that $sumx_i^2$ means to take the sum of the square of every observation we have. This translates into:

$sigma=sqrt((overbrace(a^2+a^2+...+a^2)^(n" times")+overbrace((-a)^2+(-a)^2+...+(-a)^2)^(n" times"))/(2n))$

Which we can rewrite with more mathematical precision as:

$sigma=sqrt((n(a^2)+n(a^2))/(2n))$

Then, we simplify:

$sigma = sqrt((2na^2)/(2n))=sqrt(a^2)=absa$

We are told that $sigma=2$ , so

$absa=2$

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In a series of 2n observations, half of them equal a and remaining half -a.If the standard deviation of the observations is 2,then |a| equals ?

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