Chapter 2.2, Problem 66E

To determine

Percentage of observations that are classified as outliners by the $1.5\times IQR$ rule.

Answer to Problem 66E

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

Explanation of Solution

Given information:

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

In the appendix, the standard normal table consists of the probabilities for the values smaller than z − score and also the probabilities to the left of z − score.

The property of 1^st quartile reveals that 25% of the data values are below it.

The property of 3^rd quartile reveals that 75% of the data values are below it.

The z − score corresponding to the probability of 25% (or 0.25) and 75% (or 0.75) in the normal probability table of the appendix:

  $z=\pm 0.67$

Then

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

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

Interquartile range is the difference between the 3^rd and 1^st quartile.

  $\begin{array}{l}IQR=Q_3-Q_1\\ =\mu +0.67\sigma -\left(\mu -0.67\sigma \right)\\ =1.34\sigma \end{array}$

Multiply both sides by 1.5:

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

Now,

Outliers are more than 1. 5IQR below the 1^st quartile $\left(Q_1\right)$ .

Or

Outliers are more than 1. 5IQR above the 3^rd quartile $\left(Q_3\right)$ .

Thus,

  $Q_1-1.5IQR=\mu -0.67\sigma -2.01\sigma =\mu -2.68\sigma$

Or

  $Q_3+1.5IQR=\mu +0.67\sigma +2.01\sigma =\mu +2.68\sigma$

Then

The corresponding probability,

  $\begin{array}{l}P\left(z<-2.68\text{or}z>2.68\right)=2\times P\left(z<-2.68\right)\\ =2\times 0.0037\\ =0.0074\\ =0.74\%\end{array}$

Therefore,

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