Question

Outliers The percent of the observations that are classified as outliers by the $1.5\times IQR$rule is the same in any Normal distribution. What is this percent? Show your method clearly.

Short answer

$0.74\%$ of the observations are classified as outliners by the $1.5\times IQR$rule.

Step-by-step solution

Given information

Observations are classified as outliners by$1.5\times IQR$ rule.

Calculation

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):

$z=\pm 0.67$

Then

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

$Q=\mu \pm z\sigma =\mu \pm 0.67\sigma$

Calculation

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

$$\begin{gathered} IQR=Q_3-Q_1 \\ =\mu +0.67\sigma -(\mu -0.67\sigma ) \\ =1.34\sigma \end{gathered}$$

Multiply both sides by $1.5:$

$1.5IQR=1.5(1.34\sigma )=2.01\sigma$

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,

$$\begin{gathered} Q_1-1.5IQR=\mu -0.67\sigma -2.01\sigma \\ =\mu -2.68\sigma \end{gathered}$$

Or

$$\begin{gathered} Q_3+1.5IQR=\mu +0.67\sigma +2.01\sigma \\ =\mu +2.68\sigma \end{gathered}$$

Then

The corresponding probability,

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

Therefore,

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