Assignment 11

course MTH 174

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Physics II

12-03-2007

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12:26:54

Query 8.8.2 (3d edition 8.7.2) 8.7.2. Probability and More On Distributions, p. 421 daily catch density function piecewise linear (2,.08) to (6.,24) to (8,.12)

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RESPONSE -->

The mean value of the cumulative distribution function = the mean catch = the integral from negative infinity to infinity of (the density function * x) where the density function = p(x).

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12:33:33

what is the mean daily catch?

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RESPONSE -->

the integral from 2 to 6 of (x(0.04x) dx) + the integral from 6 to 8 of (x(-0.06x + 0.6) dx) = the mean daily catch = 2.77333 + 2.48 =

approximately 5.25

Very good. Details for your reference:

** You are asked here to find the mean value of a probability density function.

The linear functions that fit between the two points are y = .04 x and y = .6 - .06 x.

You should check to be sure that the integral of the probability density function is indeed 1, which does turn out to be the case here.

The mean value of a distribution is the integral of x * p(x). In this case this gives us the integral of x * .04 x from x = 2 to x = 6, and x = x(.6 - .06 x) from x = 6 to x = 8.

int( x*.04 x, x, 2, 6) = .04 / 3 * (6^3-2^3) = .04*208/3=8.32/3 = 2.77 approx.

int(-.06x^2 + .6x, x, 6, 8) = [-.02 x^3 + .3 x^2 ] eval at limits = -.02 * (8^3 - 6^3) + .3 ( 8^2 - 6^2) = -.02 * 296 + .3 * 28= -5.92 + 8.4 = 2.48.

2.77 + 2.48 = 5.25.

The first moment of the probability function p(x) is the integral of x * p(x), which is identical to the integral used here. The mean value of a probability distribution is therefore its first moment. **

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12:33:45

What integral(s) did you perform to compute a mean daily catch?

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RESPONSE -->

the integral from 2 to 6 of (x(0.04x) dx) + the integral from 6 to 8 of (x(-0.06x + 0.6) dx) = the mean daily catch = 2.77333 + 2.48 =

approximately 5.25

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12:46:50

What does this integral have to do with the moment integrals calculated in Section 8.3?

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RESPONSE -->

The integral gives a point where the area to the right and left of the point underneath the density fuction is equal. A mass of that shape would balance at that point, assuming a constant density.

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12:57:45

Query 8.8.13 (3d edition 8.7.13). Probability and More On Distributions, p. 423 cos t, 0

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RESPONSE -->

(b) p(t) = 3e^(-3t) for t >= 0

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12:57:52

which function might best represent the probability for the time the next customer walks in?

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RESPONSE -->

(b) p(t) = 3e^(-3t) for t >= 0

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13:01:45

for each of the given functions, explain why it is either appropriate or inappropriate to the situation?

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RESPONSE -->

(a) has a nagative probability; inappropriate

(b) area from 0 to infinity = 1; appropriate

(c) area from 0 to infinity = < 1; inappropriate

(d) Inappropriate. Probability should change with time in this situation.

Very good. The details are probably unnecessary but just in case they are of use to you:

** Our function must be a probabiity density function, which is the case for most but not all of the functions.

It must also fit the situation.

If we choose the 1/4 function then the probability of the next customer walking in sometime during the next 4 minutes is 100%. That's not the case--it might be 5 or 10 minutes before the next customer shows up. Nothing can guarantee a customer in the next 4 minutes.

The cosine function fluctuates between positive and negative, decreasing and increasing. A probability density function cannot be negative, which eliminates this choice.

This leaves us with the choice between the two exponential functions.

If we integrate e^(-3t) from t = 0 to t = 4 we get -1/3 e^-12 - (-1/3 e^0 ) = 1/3 (1 - e^-12), which is slightly less than 1/3. This integral from 0 to infinity will in fact converge to 1/3, not to 1 as a probability distribution must do.

We have therefore eliminated three of the possibilities.

If we integrate 3 e^(-3t) from 0 to 4 we get 1 - e^-12, which is almost 1. This makes the function a probability density function. Furthermore it is a decreasing function. **

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13:01:49

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

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