In the one sigma interval around the mean (i.e., one standard distribution) fall 68.2% of the observations.
In which interval would exactly 50% of the observations fall?
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asked Dec 24, 2012 at 23:23
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9$\begingroup$ Use a table of the normal distribution function $\Phi(x)$ to figure out the solution to $\Phi(x) = 0.75$. The (shortest of many possible) interval you seek is $(-x,x)$. $\endgroup$– Dilip SarwateCommented Dec 24, 2012 at 23:36
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1$\begingroup$ ... to which the OP will have to add the mean. $\endgroup$– Stephan KolassaCommented Dec 25, 2012 at 19:55
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$\begingroup$ @StephanKolassa Actually, following the sense of the OP's statement: taking "one sigma interval around the mean" to mean that $68.2$% of the probability mass lies within one standard deviation from the mean, the value of $x$ that is the solution to $\Phi(x)=0.75$ should be taken to mean that $50$% of the probability mass lies within $x$ standard deviation from the mean. $\endgroup$– Dilip SarwateCommented Dec 26, 2012 at 1:12
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1 Answer
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First, notice that there are many intervals that will contain $50\%$ of the density. For instance, $(-\infty, \mu)$ contains half of the density.
But I think you want to shrink down the $68\%$ interval evenly on both sides until it contains exactly half of the density.
This could be rephrased as asking for the first and third quartiles of a Gaussian distribution; half of the density will be between those values.
We can explore this with a standard normal distribution to see how many standard deviations (since the standard deviation is just $1$) above the mean the third quartile is and how many standard deviations below the mean the first quartile is. (Because of the symmetry of the Gaussian distribution, those numbers should be equal, but we will check that.)
R software makes this easy.
qnorm(0.25, 0, 1) # Quantile 0.25 is the first quartile.
qnorm(0.75, 0, 1) # Quantile 0.75 is the third quartile.
These numbers are $-0.6744898$ and $+0.6744898$, respectively. This means that $25\%$ of the density is between the mean and $-0.6744898\sigma$ and another $25\%$ of the density is between the mean and $0.6744898\sigma$.
Let's try this for a different Gaussian.
mu <- 7
sigma <- 13
pnorm(mu + 0.6744898 * sigma, mu, sigma) - pnorm(mu - 0.6744898 * sigma, mu, sigma)
As expected, we get $0.5$.
If you replace $0.6744898$ with $1$ in the above code, you will get the familiar $\sim68\%$. Likewise, replacing $0.6744898$ with $2$ will give the familiar $\sim 95\%$.
