Question

Bouted Eagles. The rare booted eagle of western Europe was the focus of a study by S. Suarez et al, to identify the optimal nesting habitat for this raptor. According to their paper "Nesting Habitat Selection by Booted Eagles (Hienieefus pennatus) and Implications for Management" (Jamal of Applied Ecolesy, VoL. 37. pp, 215-223). the distances of such nests to the nearest marshland are normally distributed with a mean of 4.66km and a standard deviation of 0.75km. Let Y be the distance of a randomly selected nest to the nearest marshland.

Determine and interpret

a. $P(Y>5).$

b. $P(3\le Y\le 6)$.

Short answer

a) The probability that the distance of a randomly selected nest to the nearest marshland is more than 5km is $0.3264$

b) The probability that the distance of the randomly selected nest to the nearest marshland is between km and 6km is $0.9497$

Step-by-step solution

Given Information (Part a)

To evaluate the probability that the distance between the randomly chosen nest and the nearest wetland is greater than 5 kilometres.

Explanation (Part a)

The mean of the distances to the nearest marshland is:$4.66\mathrm{km}$

The standard deviation is:$0.75km$

In particular, we have $\mu =4.66\mathrm{km}$and $\sigma =0.75\mathrm{km}$

$P(Y>5)$Raise the possibility that a randomly selected nest is more than $5$kilometers away from the nearest wetland.

The figure shows the required shaded region.

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

$y=5\to$

$z=\frac{5-\mu }{\sigma }$

$=\frac{5-4.66}{075}$

$=0.45$

We must locate the region beneath the standard normal curve that is less than $0.45$. $0.6736$is the area to the left of $0.45$. $1-0.6736=0.3264$is the area to the right of $0.45$. The needed area is $0.3264$, which is shaded in the diagram.

Given Information (Part b)

The probability that a randomly selected nest is between 3km and 6km away from the nearest wetland.

Explanation (Part b)

The normal curve associated with the variable is shown in the following figure. Note that the tick marks are units apart; that is, the distance between successive tick marks is equal to the standard deviation.

The figure shows the required shaded region and its delimiting $y-$values, which are 3 and 6.

The z-scores for the $y$-values 3 and 6 must be computed as follows:

$y=3\to$

$z=\frac{3-\mu }{\sigma }$

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

$=-2.21$

And

$y=6\to$

$z=\frac{6-\mu }{z}$

$=\frac{6-4.66}{n75}$

$=1.79$

In the graphic, the $z$-scores are indicated beneath the $y$-values.

Between $-2.21$and $1.79$, we need to find the area under the standard normal curve. $0.0136$is the area to the left of $-2.21$, and $0.9633$is the region to the left of $1.79$. As a result, the required area, shaded in the diagram, is $0.9633-0.0136=0.9497$.

As a result, there's a $0.9497$chance that the distance between a randomly selected nest and the nearest marshland is between$3and6km$