91Ó°ÊÓ

The given mean is $10$and standard deviation is$3$.

Sketch a normal curve using $$\begin{gathered} \mathrm{μ}=10, \\ \mathrm{σ}=3 \end{gathered}$$

Shade the region corresponding to first quartile or $25$th percentile.

We need to find the z-score corresponding to $P_{25}$is the one having an area of $0.25$ to its left under the standard normal curve. From the normal distribution tables, the z-score is $-0.67$.

On finding the value of x,

$$\begin{gathered} x=\mathrm{μ}+\mathrm{σ}z \\ =10+3\left(-0.67\right) \\ =7.98 \end{gathered}$$

On interpreting, we can say, $25\%$ of the x-values below $7.98$and $75\%$values above $7.98$.

Shade the region corresponding to second quartile or $50$th percentile.

We need to find the z-score corresponding to $P_{50}$is the one having an area of $0.5$ to its left under the standard normal curve. From the normal distribution tables, the z-score is $0$.

On finding the value of x,

$$\begin{gathered} x=\mathrm{μ}+\mathrm{σ}z \\ =10+3\left(0\right) \\ =10 \end{gathered}$$

On interpreting, we can say, $50\%$ of the x-values below $10$and $50\%$values above$10$.

Shade the region corresponding to second quartile or $75$th percentile.

We need to find the z-score corresponding to $P_{75}$is the one having an area of $0.75$ to its left under the standard normal curve. From the normal distribution tables, the z-score is $0.67$.

On finding the value of x,

$$\begin{gathered} x=\mathrm{μ}+\mathrm{σ}z \\ =10+3\left(0.67\right) \\ =12.02 \end{gathered}$$

On interpreting, we can say,$75\%$of the x-values below$12.02$andlocalid="1652465263979" $25\%$values above$12.02$.

Shade the region corresponding seventh percentile.

We need to find the z-score corresponding to $P_{70}$is the one having an area of $0.52$ to its left under the standard normal curve. From the normal distribution tables, the z-score is $0.52$.

On finding the value of x,

$$\begin{gathered} x=\mathrm{μ}+\mathrm{σ}z \\ =10+3\left(0.52\right) \\ =11.57 \end{gathered}$$

On interpreting, we can say, $70\%$ of the x-values below $11.57$and $30\%$values above$11.57$.

First, we find the z-score corresponding to $P_{65}$is the one having an area of $0.39$to its left under the standard normal curve. From the normal distribution tables, the z-score is $0.39$.

On finding the value of x,

$$\begin{gathered} x=\mathrm{μ}+\mathrm{σ}z \\ =10+3\left(0.39\right) \\ =11.16 \end{gathered}$$

Shade the region corresponding to $0.5$th percentile.

We need to find the z-score corresponding to $P_{0.5}$is the one having an area of $0.005$ to its left under the standard normal curve. From the normal distribution tables, the z-score is $-2.576$.

On finding the value of x,

$$\begin{gathered} x=\mathrm{μ}+\mathrm{σ}z \\ =10+3\left(-2.576\right) \\ =2.27 \end{gathered}$$

Shade the region corresponding to second quartile or $99.5$th percentile.

We need to find the z-score corresponding to $P_{99.5}$is the one having an area of $0.995$ to its left under the standard normal curve. From the normal distribution tables, the z-score is $2.576$.

On finding the value of x,

$$\begin{gathered} x=\mathrm{μ}+\mathrm{σ}z \\ =10+3\left(2.576\right) \\ =17.73 \end{gathered}$$

On interpreting, we can say, $99\%$of all observations are between$2.27$and$17.73$.