A variable is normally distributed with mean $68$and standard deviation $10$.

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

Part (b): Obtain and interpret the $99$thpercentile.

Part (c): Find the value that $85th$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 $0.90$and two outside areas of $0.05$. Interpret your answer.

The given mean is $68$and standard deviation is $10$.

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

$$\begin{gathered} P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.25 \\ P\left(Z\le \frac{x-68}{10}\right)=0.25 \\ \frac{x-68}{10}=-0.6745 \\ x=61.255 \end{gathered}$$

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

$$\begin{gathered} P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.5 \\ P\left(Z\le \frac{x-68}{10}\right)=0.5 \\ \frac{x-68}{10}=0 \\ x=68 \end{gathered}$$

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

$$\begin{gathered} P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.75 \\ P\left(Z\le \frac{x-68}{10}\right)=0.75 \\ \frac{x-68}{10}=0.6745 \\ x=74.745 \end{gathered}$$

On interpreting, the first, second and third quartiles are $61.225,68$and$74.745$.

The $99th$ percentile of X is defined as $P\left(X\le x\right)=0.99$

$$\begin{gathered} P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.99 \\ P\left(Z\le \frac{x-68}{10}\right)=0.99 \\ \frac{x-68}{10}=2.326 \\ x=91.26 \end{gathered}$$

On interpreting, we get, $99\%$of all observations are smaller than$91.26$.

The required value,

$$\begin{gathered} P\left(X\ge x\right)=0.85 \\ 1-P\left(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\right)=0.85 \\ P\left(Z-\frac{x-68}{10}\right)=0.15 \\ \frac{x-68}{10}=-1.036 \\ x=57.64 \end{gathered}$$

On interpreting, approximately $58$ values of X exceeds $85\%$.

The Z-score corresponding to the two outside areas of $0.05$is given below,

$$\begin{gathered} z_1=-1.645 \\ {z}_2=1.645 \end{gathered}$$

Now, the first value is obtained,

$$\begin{gathered} x_1=\mu +z_1\sigma \\ =68+\left(-1.645\right)10 \\ =51.55 \end{gathered}$$

The second value is obtained,

$$\begin{gathered} x_2=\mu +z_2\sigma \\ =68+\left(1.645\right)10 \\ =84.449 \end{gathered}$$

On interpreting, we can say, $90\%$of all observations are between $51.551$and $84.449$.