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the probability density of heights h of people in Chicago is exponential. Assume that the people can be any (positive) height. The mean height is W. Determine the normalized distribution and calculate the mean square height.

Sketch P(h). Indicate the dimensions of P(h). Calculate the probability that a person is taller than a but shorter than b.

I know the probability density for an exponential distribution is ue^(-ux) for x greater than or equal to 0.

I don't know where to go from here though. How can I normalize without having specific numbers?

I know the mean square height is equal to -^2, but where can i go from here?

Is P(h) simply an exponential decay? What are dimensions? is the probability of a person being taller than a but shorter than b simply the integral from a to b of the probability density function ue^(-ux)?

Thanks for your help and for clarifying!

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can you just confirm the following: so lambda=(1/W). how do i find the normalized distribution? And is P(h) simply an exponential decay? What is meant by dimensions? Thanks for your help! –  Alex Trent Oct 23 '12 at 16:09
    
and is the result of the integration by parts = 2/(r^2) –  Alex Trent Oct 23 '12 at 16:23
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1 Answer 1

The standard notation is that the density function is $\lambda e^{-\lambda x}$ (for $x\ge 0$, and $0$ elsewhere).

The mean is $\int_0^\infty \lambda x e^{-\lambda x}\,dx$. This turns out to be $\dfrac{1}{\lambda}$. You are told that the mean is $W$, so now you know $\lambda$.

As you conjectured, the probability that $a\le X\le b$, where $a\ge 0$, is $$\int_a^b \lambda e^{-\lambda x}\,dx.$$ You know $\lambda$. the integration is not difficult. We get $e^{-\lambda b}-e^{-\lambda a}$.

For the mean square height, that is, $E(X^2)$, you need to find $$\int_0^\infty \lambda x^2e^{-\lambda x}\,dx.$$ One integration by parts should get you to something familiar.

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can you just confirm the following: so lambda=(1/W). how do i find the normalized distribution? And is P(h) simply an exponential decay? What is meant by dimensions? Thanks for your help! –  Alex Trent Oct 23 '12 at 16:37
    
and is the result of the integration by parts = 2/(r^2) ? –  Alex Trent Oct 23 '12 at 16:38
    
@AlexTren: Sorry for delay, I was away. Yes, $\lambda=\frac{1}{W}$. The result of the integration by parts is $\frac{2}{\lambda^2}$, so $2W^2$. By the way, the variance is $W^2$. Yes, it is exponential decay. Undoubtedly lousy model for heights! In the Physics sense, the dimension of the density function is height$^{-1}$. I am not sure what they mean by normalized. Probably it is to measure heights so that $\lambda=1$. So normalized distribution would have density $e^{-z}$ where $z=\lambda x=x/W$. We measure heights in units that make the mean height $1$. –  André Nicolas Oct 23 '12 at 17:14
    
thank you for clarifying! –  Alex Trent Oct 23 '12 at 19:12
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