Question

Box plots and the standard normal distribution. What relationship exists between the standard normal distribution and the box-plot methodology (Section 2.8) for describing distributions of data using quartiles? The answer depends on the true underlying probability distribution of the data. Assume for the remainder of this exercise that the distribution is normal.

a. Calculate the values of the standard normal random variable z, call them z_L and z_U, that correspond to the hinges of the box plot—that is, the lower and upper quartiles, Q_L and Q_U—of the probability distribution.

b. Calculate the zvalues that correspond to the inner fences of the box plot for a normal probability distribution.

c. Calculate the zvalues that correspond to the outer fences of the box plot for a normal probability distribution.

d. What is the probability that observation lies beyond the inner fences of a normal probability distribution? The outer fences?

e. Can you better understand why the inner and outer fences of a box plot are used to detect outliers in a distribution? Explain.

Short answer

a. The lower and upper quartile is -0.67449 and 0.67449

b. The zvalues that correspond to the inner fences of the box plot are -2.697959 and 2.697959

c. The zvalues that correspond to the outer fences of the box plot are -4.72143 and 4.72143

d. The probability of an observation falling outside of inner fences is 0.006977 and outer fences is 0

e. The probability is very low for an observation to fall outside of these fences

Step-by-step solution

Given information

The given distribution is a normal distribution

 Calculating the lower and upper quantile

a.

The lower quartile is 25^th percentile

Let z_L be the standard normal random variable that corresponds to Q_L

$$\begin{gathered} i.e,P\left(z<z_L\right)=0.25 \\ \Phi \left(z_L\right)=0.25 \\ {z}_L={\Phi }^{-1}\left(0.25\right) \\ {z}_L=-0.67449 \end{gathered}$$

So the lower quantile is -0.67449

The upper quantile is 75^th percentile

Let z_Q be the standard normal random variable corresponds to Q_U

$$\begin{gathered} i.e,P\left(z<z_U\right)=0.75 \\ \Phi \left(z_U\right)=0.75 \\ {z}_U={\Phi }^{-1}\left(0.75\right) \\ {z}_U=0.67449 \end{gathered}$$

So the upper quartile is 0.67449

 Calculating the inner fences of the box

$$\begin{gathered} \mathrm{b}. \\ IQR=Q_U-Q_L \\ =0.67449-\left(-0.67449\right) \\ =1.34898 \\ \mathrm{The}\mathrm{lower}\mathrm{inner}\mathrm{fence}\mathrm{is}LIF=Q_L-1.5xIQR \\ LIF=Q_L-1.5xIQR \\ =Q_L-1.5x\left(Q_U-Q_L\right) \\ =-0.67449-1.5x1.34898 \\ =-2.697959 \\ \mathrm{The}\mathrm{upper}\mathrm{inner}\mathrm{fence}\mathrm{is}UIF=Q_U+1.5xIQR \\ UIF=Q_U+1.5xIQR \\ =Q_U+1.5\left(Q_U-Q_L\right) \\ =0.67449+1.5\mathrm{x}1.34898 \\ =2.697959 \\ \mathrm{So}\mathrm{the}\mathrm{lower}\mathrm{inner}\mathrm{fence}\mathrm{is}-2.697959\mathrm{and}\mathrm{the}\mathrm{upper}\mathrm{inner}\mathrm{fence}\mathrm{is}2.697959 \end{gathered}$$

 Calculating the outer fences of the box

$$\begin{gathered} \mathrm{C}. \\ \mathrm{IQR}={\mathrm{Q}}_{\mathrm{U}}-{\mathrm{Q}}_{\mathrm{L}} \\ =0.67449-\left(-0.67449\right) \\ =1.34898 \end{gathered}$$

The lower outer fence is $LOF=Q_1-3xIQR$

$$\begin{gathered} LOF=Q_L-3xIQR \\ =Q_L-3x\left(Q_U-Q_L\right) \\ =-0.67449-3x1.34898 \\ =-4.72143 \end{gathered}$$

The upper outer fence is$UOF=Q_U+3xIQR$

$$\begin{gathered} UOF=Q_L+3xIQR \\ =Q_U+3x\left(Q_U-Q_L\right) \\ =0.67449+3x1.34898 \\ =4.72143 \end{gathered}$$

So the lower outer fence is -4.72143 and upper outer fence is 4.72143

 Calculating the probabilities

d.

The probability that observation lies beyond the inner fences of a normal probability distribution is,

$$\begin{gathered} I.E,P\left(z<-2.697959\right)+P\left(z<-2.697959\right)=1-P\left(z<-2.697959\right)+1-\left(z<-2.697959\right) \\ =2-2P\left(z<-2.697959\right) \\ =2-2\Phi \left(2.697959\right) \\ =2-2X0.996512 \\ =0.006977 \end{gathered}$$

So, the probability is 0.006977

The probability that an observation lies beyond the outer fences of a normal probability distribution is,

$$\begin{gathered} I.E,P\left(z<-2.697959\right)+P\left(z>4.72143\right)=1-P\left(z<4.72143\right)+1-\left(z<4.72143\right) \\ =2-2P\left(z<4.72143\right) \\ =2-2\Phi \left(4.72143\right) \\ =2-2X0.999999 \\ 𝆏0 \end{gathered}$$

So the probability is 0

Explanation

The inner and outer fences of box plot are used to detect outliers in a distribution in the following ways:

Values that are beyond the inner fences are deemed potential outliers because they are extreme values that represent relatively rare occurrences. In fact, for a normal probability distribution, less than 1% of the observations are expected to fall outside of inner fences.

Measurements that fall beyond the outer fences are very extreme measurements that require special analysis. Since less than one-hundredth of 1% (0.1% or 0.001) of the measurements from a normal distribution are expected to fall beyond the outer fences, these measurements are considered to be outliers.

From part(d) we clearly understand why the inner and outer fences of the box plot are used to detect outliers in a distribution.