Question

(a) Repeat Problem $6$ where the “circular” area is now on the curved surface of the earth, say all points at distance s from Chicago (measured along a great circle on the earth’s surface) with $s\le \frac{\pi R}{3}$where R = radius of the earth. The seeds could be replaced by, say, radioactive fallout particles (assuming these to be uniformly distributed over the surface of the earth). Find $F\left(s\right)$and$f\left(s\right)$ .

(b) Also find $F\left(s\right)$and$f\left(s\right)$ if$s\le 1<<R$ (say $s\le 1$mile where $R=4000$miles). Do your answers then reduce to those in Problem $6$?

Short answer

The required value is given below.

$\begin{array}{c}\left(a\right)F\left(s\right)=2(1-\cos (s/R\left)\right)\\ f\left(s\right)=\frac{2\sin (s/R)}{R}\\ \left(b\right)F\left(s\right)=\frac{2}{R^2}{s}^2\\ f\left(s\right)=\frac{2}{R^2}s\end{array}$

Step-by-step solution

Given Information

A circular garden bed of radius $1m$ is to be planted so that N seeds are uniformly distributed over the circular area.

Definition of the probability density function

A continuous random variable, whose integral across an interval offers the likelihood that the variable's value falls inside the same interval.

Find the values for part (a).

Let the distance be s and the inner angle from the pole be $\frac{s}{R}$.

Minimum distance is $0$.

Maximum distance is $\frac{\pi R}{3}$.

The area of the maximal spherical cap is $2\pi R^2(1-\cos ((\pi R/3)/R\left)\right)=\pi R^2$.

So, the distribution function and density function becomes as follows.

$\begin{array}{c}F\left(s\right)=\frac{2\pi R^2(1-\cos (s/R\left)\right)}{\pi R^2}\\ F\left(s\right)=2(1-\cos (s/R\left)\right)\\ f\left(s\right)=F^{\text{'}}\left(s\right)\\ f\left(s\right)=\frac{2\sin (s/R)}{R}\end{array}$

Find the values for part (b).

If s is relatively small and R is relatively large, then$\frac{s}{R}$becomes close to $0$.

$\begin{array}{c}\cos (s/R)\approx 1-{\left(\frac{s}{R}\right)}^2\\ \sin (s/R)\approx \frac{s}{R}\end{array}$

So, the distribution function and density function becomes as follows.

$\begin{array}{c}F\left(s\right)\approx 2\left(1-1+{\left(\frac{s}{R}\right)}^2\right)\\ F\left(s\right)=\frac{2}{R^2}{s}^2\\ f\left(s\right)\approx \frac{2s/R}{R}\\ f\left(s\right)=\frac{2}{R^2}s\end{array}$

The required value is given below.

$\begin{array}{c}\left(a\right)F\left(s\right)=2(1-\cos (s/R\left)\right)\\ f\left(s\right)=\frac{2\sin (s/R)}{R}\\ \left(b\right)F\left(s\right)=\frac{2}{R^2}{s}^2\\ f\left(s\right)=\frac{2}{R^2}s\end{array}$