91Ó°ÊÓ

Observations are classified as outliners by$1.5×IQR$ rule.

The typical normal table in the appendix contains probability for values smaller than $z-$score as well as probabilities to the left of $z-$score.

The $1st$quartile's characteristic demonstrates that $25\%$of the data values are below it.

The $3rd$quartile's characteristic demonstrates that $75\%$of the data values are below it.

In the normal probability table of the appendix, the $z−$score corresponds to the likelihood of $25\%$(or $0.25$) and $75\%$(or 0.75):

$z=Â\pm 0.67$

Then

The quartile is the mean increased by the product of $z−$ score and standard deviation.

$Q=\mathrm{μ}Â\pm z\mathrm{σ}=\mathrm{μ}Â\pm 0.67\mathrm{σ}$

The interquartile range is the difference between the $3rd$ and $1st$ quartile.

$$\begin{gathered} IQR=Q_3−Q_1 \\ =\mathrm{μ}+0.67\mathrm{σ}−(\mathrm{μ}−0.67\mathrm{σ}) \\ =1.34\mathrm{σ} \end{gathered}$$

Multiply both sides by $1.5:$

$1.5IQR=1.5(1.34\mathrm{σ})=2.01\mathrm{σ}$

Now,

Outliers are more than $1.5IQR$ below the 1st quartile$\left(Q_1\right)$

Or

Outliers are more than $1.5IQR$ above the 3rd quartile $\left(Q_3\right)$

Thus,

$$\begin{gathered} Q_1−1.5IQR=\mathrm{μ}−0.67\mathrm{σ}−2.01\mathrm{σ} \\ =\mathrm{μ}−2.68\mathrm{σ} \end{gathered}$$

Or

$$\begin{gathered} Q_3+1.5IQR=\mathrm{μ}+0.67\mathrm{σ}+2.01\mathrm{σ} \\ =\mathrm{μ}+2.68\mathrm{σ} \end{gathered}$$

Then

The corresponding probability,

$$\begin{gathered} P(z<−2.68orz>2.68)=2×P(z<−2.68) \\ =2×0.0037 \\ =0.0074 \\ =0.74\% \end{gathered}$$

Therefore,

$0.74\%$ of the observations are classified as outliners by the $1.5×IQR$ rule is the same in any Normal distribution.