91Ó°ÊÓ

The deciles of any lowest distribution$=10\%$

The deciles of any highest distribution$=10\%$

Areas$=0.1$and$0.9$

The figure illustrates what we're looking for:

In the standard Normal Curve, the deciles are $10\%$and $90\%$percent of the distribution on either side of the standard curve.

Due to the curvature's symmetry, the area from $z=0$to either end is $0.5$while the area from $z=0$to the top or bottom is $1$.

As shown here, $0.1$is equal to $0.5-0.4$and $0.9$is equal to $0.5+0.4$.

Based on the Table of Normal Curve, $0.3997$is the nearest value to $0.4$.

This corresponds to a $z=1.28$value.

Therefore, $z=-1.28$is the standardized value with area $0.10$to the left of $z=1.28$and is the standardized value with area $0.90$to the left of $z=1.28$.

Hence, the deciles are$Â\pm 1.28$.

The heights of young women with mean$=64.5inches$

Standard deviation $=2.5inches$

we have to find out the the declines of the heights of young women.

Assuming random variable $X$indicates heights of young women, then we are given the following:

$x~N(\mathrm{μ},{\mathrm{σ}}^2)where\mathrm{μ}=64.5inchesand\mathrm{σ}=2.5inches$

The unknown $x\text{'}s$standardized value is $z=Â\pm 1.28$.

Hence, $X$satisfies the equation:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ Â\pm 1.28=\frac{x-64.5}{2.5} \end{gathered}$$

Find the$x$value:

$$\begin{gathered} x=64.5Â\pm 1.28(2.5) \\ x=61.3inchesandx=67.7inches \end{gathered}$$

The decline for the heights of young women are $61.3inchesand67.7inches$