91影视

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

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

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

Minimum distance is $0$.

Maximum distance is $\frac{蟺R}{3}$.

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

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

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

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

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

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

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

The required value is given below.

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