[SOLVED] P.D.F: Continuous random variables 2

**Quacky** · March 15th 2010, 03:59 PM

Sorry, but I'm afraid I have two more problems.

The following function represents queue time in minutes of people in a restaurant.

$f(x)=\frac{x}{2500}(100-x^2)$ for $<br /> 0\leq x\leq 10$
x=0 otherwise

a) What is the probability that a traveller will have to queue for more than 2 minutes? a:0.9216

b) 3 people go independantly to the restaurant. Find the probability that one has to queue for less than one minute, another for between one and two minutes, and the third for more than two minutes.
a:0.0064

Any help is appreciated.
I'm not sure where I need to integrate between, and the seond question just looks hideous.

**matheagle** · March 15th 2010, 04:05 PM

I prefer my duck.

(a) ${1\over 2500}\int_2^{10} x(100-x^2)dx$

(b) do we care which waits how long????
because there are 3! rearrangements of this (multinomial)

$P(X<1)P(1<X<2)P(X>2)$

or $6P(X<1)P(1<X<2)P(X>2)$ if we don't care which person does what here

where $P(X<1)={1\over 2500}\int_0^{1} x(100-x^2)dx$

$P(1<X<2)={1\over 2500}\int_1^{2} x(100-x^2)dx$

and $P(X>2)={1\over 2500}\int_2^{10} x(100-x^2)dx$

**Quacky** · March 15th 2010, 04:08 PM

I like guard duck too. But thanks for the helpful response.

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