A variable is normally distributed with mean $0$and standard deviation $4$.

Part (a): Determine and interpret the quartiles of the variable.

Part (b): Obtain and interpret the second decile.

Part (c): Find the value that $15\%$of all possible values of the variable exceed.

Part (d): Find the two values that divide the area under the corresponding normal curve into a middle area of $0.8$and two outside areas of$0.1$. Interpret your answer.

The given mean is $0$and standard deviation is$4$.

The first quartile or $25$th percentile of X is defined as $P\left(X\le x\right)=0.25$.

$$\begin{gathered} P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.25 \\ P\left(Z\le \frac{x-0}{4}\right)=0.25 \\ \frac{x-0}{4}=-0.6745 \\ x=-2.698 \end{gathered}$$

On interpreting, we can say,$25\%$of the x-values are below $-2.698$and $75\%$of the x-values are above $-2.698$.

The second quartile or $50$th percentile of X is defined as $P\left(X\le x\right)=0.5$.

$$\begin{gathered} P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.5 \\ P\left(Z\le \frac{x-0}{4}\right)=0.5 \\ \frac{x-0}{4}=0 \\ x=0 \end{gathered}$$

On interpreting, we can say, $50\%$ of the x-values are below $0$and $50\%$ of the x-values are above$0$.

The third quartile or $75$th percentile of X is defined as $P\left(X\le x\right)=0.75$.

$$\begin{gathered} P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.75 \\ P\left(Z\le \frac{x-0}{4}\right)=0.75 \\ \frac{x-0}{4}=0.6745 \\ x=2.698 \end{gathered}$$

On interpreting, we can say, $75\%$of the x-values are below $2.698$and $25\%$of the x-values are above$2.698$.

Shade the region corresponding second decile or $20$th percentile of X is defined as role="math" localid="1652466124094" $P\left(X\le x\right)=0.2$.

role="math" localid="1652466163070" $$\begin{gathered} P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.2 \\ P\left(Z\le \frac{x-0}{4}\right)=0.2 \\ \frac{x-0}{4}=-0.8416 \\ x=-3.3665 \end{gathered}$$

On interpreting, we can say, role="math" localid="1652466244012" $20\%$ of all observations are smaller than$-3.3665$.

On calculating the value,

$$\begin{gathered} P\left(X\ge x\right)=0.15 \\ 1-P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.15 \\ P\left(Z\le \frac{x-0}{4}\right)=1-0.15 \\ \frac{x-0}{4}=1.036 \\ x=4.1457 \end{gathered}$$

On interpreting, approximately $4$ values of X exceeds$15\%$.

The Z-score corresponding to the two outside areas of $0.1$is given below,

$$\begin{gathered} z_1=-1.282, \\ {z}_2=1.282 \end{gathered}$$

Now, first value $x_1$is obtained as,

$$\begin{gathered} x_1=0+\left(-1.282\right)4 \\ =-5.128 \end{gathered}$$

Second value $x_2$is obtained as,

$$\begin{gathered} x_2=0+\left(1.282\right)4 \\ =5.128 \end{gathered}$$

On interpreting, we can say, $80\%$of all observations are between $-5.128$and$5.128$.