Assignment 11

assignment #011

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Physics II

06-15-2009

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18:41:35

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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18:47:41

what is the mean daily catch?

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Mean = sum of how many tons of fish caught a day/ total # of tons of fish

p(t) =

0 t is less than 2

.04x 2 is less than or = to t is less than 6

-.06x + .6 6 is less than or = to t is less than 8

0 t is greater than 8

(I got these expressions from an example in the book and I understand why p(t) = 0, but I could not figure out how they solved for the middle two (.04x and -.06x + 6)

Mean amount is the integral from 8 to 2 tp(t) dt

= integral from 6 to 2 t(.04)dt + integral from 8 to 6 t(-.06 + .6)dt

= t^2/2 (.04) from 6 to 2 + t^1.06/1.06 + t^1.6/1.6 from 8 to 6

= .72 - .08 + 25.96 - 17.28

= 9.32

I realize this cannot be the case because my answer is greater than 8 but I'm not sure what I've done wrong here.

** 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 is 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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18:47:55

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

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Mean amount is the integral from 8 to 2 tp(t) dt

= integral from 6 to 2 t(.04)dt + integral from 8 to 6 t(-.06 + .6)dt

= t^2/2 (.04) from 6 to 2 + t^1.06/1.06 + t^1.6/1.6 from 8 to 6

= .72 - .08 + 25.96 - 17.28

= 9.32

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18:48:53

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

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This integral represents each moment so that the total of what happens in each moment can be averaged.

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18:55:09

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

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18:59:49

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

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I would think (a) would be the pest representation that the next customer walks in because it includes an - exponential function and I assume that as the day goes on fewer and fewer people would come in. It also includes a cos(t) funciton which ossilates so the total function woud ossilate while decreasing.

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19:02:05

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

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(b) and(c) would be inappropriate because they represent a decreasing exonential expression, which would not account for the possible times that many people come in the store later in the day.

(d) is inappropriate because it represents a function that is constant, and this would also be less likely.

** 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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19:03:43

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

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The main thing I could not figure out in this assignment was how to find what p(x) equals in order to find the cumulative distribution or mean, such as in the first question on this assignment.

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&#Good responses. See my notes and let me know if you have questions. &#