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The given distribution is a normal distribution

a.

The lower quartile is 25^th percentile

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

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

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

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

So the upper quartile is 0.67449

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

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

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

d.

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

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

So, the probability is 0.006977

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

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

So the probability is 0

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.