# Probability density function

• January 14th 2009, 07:10 AM
rebirthflame
Probability density function
i have the following pdf which relates to X square ft of a (round) pizza

f(x)= 6(x-x^2) for 0 < x < 1
0 elsewhere

i had to find the mean and variance which i got as 0.5 and 0.05 respectivly

now i have been asked to find the proportion of pizzas that wont fit in a box of size 1ft square

the pizzas are circular so i found the maximum size of pizza that would fit in the box which is 0.25pi so i need to find the are of the graph which is above that value the trouble is integrating between these two values just gives me a silly answer so i know i am doing something wrong i just cant place what (by silly answer i mean negative)
• January 14th 2009, 08:30 AM
Moo
Hello,

In fact, you're looking for this probability :
$\mathbb{P}(0 (where M is the maximum area in sq. ft. of the pizza), which is, by definition of the pdf :
$\mathbb{P}(0

Assuming that the box is a square, you're correct in finding the value $M=0.25 \pi$

So now just calculate $\int_0^{0.25 \pi} 6(x-x^2) ~dx$ (because $0.25 \pi <1$, so you're in an interval contained in (0,1) and hence $f(x)=6(x-x^2)$)
and you'll have the proportion (or probability) you're looking for.
• January 15th 2009, 03:28 AM
mr fantastic
Quote:

Originally Posted by Moo
Hello,

In fact, you're looking for this probability :
$\mathbb{P}(0 (where M is the maximum area in sq. ft. of the pizza), which is, by definition of the pdf :
$\mathbb{P}(0

Assuming that the box is a square, you're correct in finding the value $M=0.25 \pi$

So now just calculate $\int_0^{0.25 \pi} 6(x-x^2) ~dx$ (because $0.25 \pi <1$, so you're in an interval contained in (0,1) and hence $f(x)=6(x-x^2)$)
and you'll have the proportion (or probability) you're looking for.

NB: This integral gives the proportion of pizas that do fit in the box. The proportion that don't is $\int_{0.25 \pi}^1 6(x-x^2) ~dx$ which is the same as $1 - \int_0^{0.25 \pi} 6(x-x^2) ~dx$.