91Ó°ÊÓ

The distances of such nests to the nearest marshland are normally distributed with mean $4.66\mathrm{km}$ and standard deviation $0.75\mathrm{km}$. Let $Y$ be the distance of a randomly selected nest to the nearest marshland.

we have $\mathrm{μ}=4.66\mathrm{km}$and $\mathrm{σ}=0.75\mathrm{km}$

$P(Y>5)$Indicate the probability that a randomly chosen nest is more than $5$miles away from the nearest marshland.

The figure shows the required shaded region.

We need to compute the z-score for the $y$-value $5$:

$y=5→$

$z=\frac{5-\mathrm{μ}}{\mathrm{σ}}$

$=\frac{5-4.66}{0.75}$

$=0.45$

We have to determine the standard normal curve that lies below $0.45$. The area to the left of $0.45$it $0.6736$.The area to the right of $0.45$is $1-0.6736=0.3264$. The required area, shaded in the figure, is $0.3264$

The distances of such nests to the nearest marshland are normally distributed with mean $4.66\mathrm{km}$ and standard deviation $0.75\mathrm{km}$. Let $Y$ be the distance of a randomly selected nest to the nearest marshland.

The normal curve for the variable is depicted in the following diagram. The tick marks are separated by units, which means that the distance between each tick mark is equal to the standard deviation.

The required shaded zone and its delimiting $y$-values, $3$and $6$, are shown in the diagram.

We need to compute the $z$-scores for the $y$-values $3$and $6$:

$y=3→$

$z=\frac{3-\mathrm{μ}}{\mathrm{σ}}$

$=\frac{3-4.66}{0.75}$

$=-2.21$

And

$y=6→$

$=\frac{6-\mathrm{μ}}{\mathrm{σ}}$

$=\frac{6-4.66}{0.75}$

$=1.79$

The z- scores are marked beneath the $y$-values in the figure.

We need to find the area between the two points on the standard normal curve $-2.21$and $1.79$. The area to the left of $-2.21$is $0.0136$and the area to the left of $1.79$it $0.9633$. The required area, shaded in the figure, is therefore $0.9633-0.0136=0.9497$.