MHB Problem of the Week # 159 - April 14, 2015

[Ackbach]

Here is this week's POTW:

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What is the $\text{IQR}$ of the standard normal distribution? Using the $1.5\times\text{IQR}$ rule, what $z$ scores would be considered outliers?

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[Ackbach]

Congratulations to jacobi for his correct solution. Here it is below:

Spoiler

The IQR of the standard normal distribution is $Q_3-Q_1$, where $Q_n$ is the nth quartile. To find the first and third quartiles, we have to find the points where the CDF of the normal distribution is $1 \over 4$ and $3 \over 4$, respectively, since $25\%$ and $75\%$ of the data are below the first and third quartiles, respectively.
The CDF of the normal distribution is \[ \frac{1}{\sigma \sqrt{2 \pi} }\int_{- \infty}^{x} e^{- \frac{1}{2} \left ( \frac{x-\mu}{\sigma} \right )^2} dx = \frac{1}{2} \left ( 1+ \operatorname{erf} \left ( \frac{x-\mu}{\sigma \sqrt{2}} \right ) \right ). \]
Therefore, the inverse of this is \[\mu + \sigma \sqrt{2} \operatorname{erf}^{-1} ( 2p-1), \] where p is the probability required. Evaluating this from $p=\frac{1}{4}$ to $p=\frac{3}{4}$, we get \[ 2 \sigma \sqrt{2} \operatorname{erf}^{-1} \left ( \frac{1}{2} \right ) \] as the IQR.
Therefore, the z scores considered outliers satisfy the condition \[ |z-\mu| > 3 \sigma \sqrt{2} \operatorname{erf}^{-1} \left ( \frac{1}{2} \right ). \]

[EDITOR'S NOTE] The standard deviation for the standard normal distribution is 1, and the mean is 0. Hence, the outliers will be \[ |z| > 3 \sqrt{2} \operatorname{erf}^{-1} \left ( \frac{1}{2} \right ). \]