91Ó°ÊÓ

A value is randomly selected from a normal distribution. The probability of the value to be an outlier is to be computed.

The interquartile range has the following expression:

$IQR=Q_3-Q_1$

The third quartile$\left(Q_3\right)$ is the percentile; that is, 75% of all the values are less than the third quartile.

Thus, the expression for the third quartile in terms of the probability value is as follows:

$P(Z<z)=0.75$

From the standard normal table,the area of 0.75is observed corresponding to the row value of 0.6and column value of 0.07. Thus, thez-score is 0.67.

The first quartile$\left(Q_1\right)$ is the percentile; that is, 25% of all the values are less than the first quartile.

Thus, the expression for the third quartile in terms of the probability value is as follows:

$P(Z<z)=0.25$

From the standard normal table,the area of 0.25is observed corresponding to the row value -0.6and column value 0.07. Thus,the z-score is -0.67.

The z-score for the IQR becomes:

$$\begin{gathered} IQR=\text{z - score}\text{of}{Q}_3-\text{z - score}\text{of}{Q}_1 \\ =0.67-(-0.67) \\ =1.34 \end{gathered}$$

The upper limit is computed as follows:

$$\begin{gathered} \text{Upper}  \text{Limit}  =Q_3+\left(1.5×IQR\right) \\ =0.67+\left(1.5×1.34\right) \\ =2.68 \end{gathered}$$

This means that values above the z-score of 2.68 can be considered as outliers.

The lower limit is computed as follows:

$$\begin{gathered} \text{Lower}  \text{Limit}  =Q_1-\left(1.5×IQR\right) \\ =-0.67-\left(1.5×1.34\right) \\ =-2.68 \end{gathered}$$

This means that values below the z-score of -2.68 can be considered as outliers.

The probability that a randomly selected value is an outlier is computed below:

$$\begin{gathered} P\left(\text{outlier}\right)=P\left(z<-2.68\right)+P\left(z>2.68\right) \\ =P\left(z<-2.68\right)+\left(1-P\left(z<2.68\right)\right) \\ =0.0037+\left(1-0.9963\right) \\ =0.0074 \end{gathered}$$

Therefore, the probability of a value being an outlier is equal to 0.0074.