Question

4. The time, t, that a butterfly lives after emerging from its chrysalis can be modeled by the probability function 36 20 0, otherwise f(t) , where t is measured in days. a. Sketch the graph of f(t). (What are the x- and y axes labels?) b. Shade the region on your graph that represents the probability that a butterfly will die within 7 days of emerging. c. Compute the probability that a butterfly will die within 7 days of emerging. d. If a large number of butterflies emerge on the same day, after how many days would you expect only 7% to be alive?

Answer 1

$Note~that~expression~of~f(t)~is~not~correct~since\\ \int_0^\infty f(t)dt=\int_0^\infty \frac{36}{(6+t)^2}dt=36\left[-(6+t)^{-1} \right ]^\infty_0=6\neq 1\\ So~I ~have~modified ~it~as~follows:\\ f(t)=\frac{6}{(6+t)^2}~~if~t\geq 0.\\ =0~otherwise$

a. R code:

t=seq(0,10,by=0.01)
f=6/(6+t)^2
plot(t,f,lwd=2,col=1,type="l",ylab="f(t)",xlab="t")

b.

(c)

$P(X\leq 7)=\int_0^7\frac{6}{(6+t)^2}dt=\left[-\frac{6}{(6+t)} \right ]^7_0=6*(1/6-1/(6+7))=0.5384$

$CDF~of~T~is\\ F_T(t)=P(T\leq t)= \int_0^t f(t)dt=\int_0^t \frac{36}{(6+t)^2}dt=6\left[-(6+t)^{-1} \right ]^t_0\\ =6(1/6-1/(6+t))=\frac{6(6+t-6)}{6(6+t)}=\frac{t}{6+t}\\ Now,~\frac{t}{6+t}=1-0.07~or,~t=(6+t)*0.07~or,~0.93t=6*(1-0.07)~\\ or~t=6*(1-0.07)/0.93=6\\ Answer:~6~days.$