I'm new to studying z-scores and I've been told that for a gaussian statistic, around 95% of the values lie within the area two standard deviations above and below the mean, which (in accordance to my interpretation) would imply, ∫ μ − 2 σ μ + 2 σ A e − ( ( x − μ ) / σ ) 2 d x = 0.95 ∗ ∫ − ∞ + ∞ A e − ( ( x − μ ) / σ ) 2 d xFirstly, am I correct in my presumption? and secondly, is there any way to calculate the integral on the left to prove this point mathematically?

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I&#039;m new to studying z-scores and I&#039;ve been told that for a gaussian statistic, around 95% of the values lie within the area two standard deviations above and below the mean, which (in accordance to my interpretation) would imply,      ∫          μ      −      2      σ              μ      +      2      σ        A      e          −      (      (      x      −      μ      )              /            σ              )        2                    d    x  =  0.95  ∗      ∫          −      ∞              +      ∞        A      e          −      (      (      x      −      μ      )              /            σ              )        2                    d    xFirstly, am I correct in my presumption? and secondly, is there any way to calculate the integral on the left to prove this point mathematically?

Caiden Barrett · Accepted Answer

Using a change of variables   x  =  μ  +  σ  t, and inserting the missing 1/2 in the exponentials, the left side can be evaluated using the error function  A      ∫          −      2        2        e          −              t        2                    /            2        d  t  =      2    π    A  erf  (      2    )while the right side is       2    π    A. In fact   erf  (      2    )  ≈  0.9544997360.

I'm new to studying z-scores and I've been told that for a gaussian statistic, around 95% of the val

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