Question

1. UPS ships millions of packages in a specific 1-ft^3 packing container. Let Y = amount of space (in ft^3 ) occupied in this container. The probability density function (pdf) of Y is given by fY (y) = 90y 8 (1 − y), 0 < y < 1. 0, otherwise.

(a) Graph the pdf of Y . If you want, you can use this R code below: y = seq(0,1,0.01) pdf = 90*y^8*(1-y) plot(y,pdf,type="l",xlab="y",ylab="PDF") abline(h=0) # puts horizontal line at 0 abline(v=1,lty=2) # puts vertical line at 1; "lty" option alters line type

(b) Find the probability that the contents of a randomly selected 1-ft3 packing container will occupy less than 1/2 of the volume of the container; i.e., calculate P(Y < 0.5). Shade in the corresponding region on your pdf in part (a) showing this probability.

(c) Find the mean and variance of Y . Indicate the units of each quantity. Place an “×” on the horizontal axis of your pdf in part (a) indicating where E(Y ) falls.

(d) I calculated the median of this distribution to be 0.8377 ft3 . Write out an equation (involving an integral) that when solved will give this answer. Make sure you say “what is being solved for” in your equation. You don’t have to try to solve the equation.

Answer 1

a)

丈一 0.0 0.2 0.4 0.6 0.8 1.0

b)

$\int _0^{0.5}90y^8\left(1-y\right)dy=0.0107422$

$\int \:90y^8\left(1-y\right)dy=10y^9-9y^{10}+C$

c)

$E(Y)= \int _0^1y90y^8\left(1-y\right)dy=\frac{9}{11}\quad \left(\mathrm{Decimal:\quad }\:0.81818\dots \right)$

$\int \:y\cdot \:90y^8\left(1-y\right)dy=90\left(\frac{y^{10}}{10}-\frac{y^{11}}{11}\right)+C$

$E(Y^2) = \int _0^1y^290y^8\left(1-y\right)dy=\frac{15}{22}\quad \left(\mathrm{Decimal:\quad }\:0.68182\dots \right)$

var(Y) = E(Y^2) - E(Y)^2

= 15/22 -(9/11)^2

= 0.012396

d)

$\int _0^c90y^8\left(1-y\right)dy=\left(10-9c\right)c^9 = 0.5$