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Math Help - Probability density function

  1. #1
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    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)
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  2. #2
    Moo
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    Hello,

    In fact, you're looking for this probability :
    \mathbb{P}(0<x<M) (where M is the maximum area in sq. ft. of the pizza), which is, by definition of the pdf :
    \mathbb{P}(0<x<M)=\int_0^M f(x) ~dx

    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.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    In fact, you're looking for this probability :
    \mathbb{P}(0<x<M) (where M is the maximum area in sq. ft. of the pizza), which is, by definition of the pdf :
    \mathbb{P}(0<x<M)=\int_0^M f(x) ~dx

    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.
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