Assignment 11

course MTH 174

1:56 p.m. 7/1/2010 Not as bad as the previous assignment so that's good. I was just getting used to the old QA formats, the programs are temperamental and even though I don't rely on them the given solution sections are at least a good type of immediate feedback.

›ŽäÔ›pöSäëð´ût¾ðs‘êhŠRðassignment #011

011. `query 11

Physics II

07-01-2010

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13:38:08

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 -->

Here I evaluated the integral Int(xp(x) dx) piecwise. I used 0.04x for 2 <= x<= 6 and -0.06x + .6 for 6< x<= 8. I setup the integral like this: Int(x(0.04x dx, 2,6)+ Int(x(-.06x+.6) dx, 6,8). Evaluating this gives a mean of 5.25. Looking at the graph we see that this seems to be a good value.

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13:38:39

what is the mean daily catch?

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RESPONSE -->

The mean daily catch was: 5.25 tons.

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13:40:30

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

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RESPONSE -->

Int(x(.04x) dx, 2,6) + Int(x(-.06+.6) dx, 6,8)

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13:42:17

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

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RESPONSE -->

The formula is exaclty the same and you could almost say that the moment is the mean distance of the weight function values.

Good. Just for reference here is my standard solution:

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) = .04208/3=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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13:51:21

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

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RESPONSE -->

3e^-3t seems the most likely as it gives the most realistic values. All of the others have something about them that doesn't fit with the model. e^-3t never returns a value greater than 1. Cos t actually gives negative values. and the 1/4 functions interval seems to rule it out.

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13:51:50

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

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Like I repsonded above: 3e^-3t.

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13:53:10

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

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RESPONSE -->

I put this in my response above. Basically the other three either give unrealistic values.

Again for your reference:

Our function must be a probability 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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13:54:00

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

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RESPONSE -->

This section wasn't as bad for me as the previous two. Maybe I'll get the hang of this material yet.

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Good work.

I agree about the benefits of immediate feedback. In your case I'm not concerned about using the open query as a crutch.

I have placed reasonbly well-edited versions of the ueries through #17 in the same folder as the rest. I'm still not sure about making them available to all students, and I haven't closely checked the edits so there might be some errors, though most of the solutions have been closely checked and verified.

You can access these documents at

http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/cal2/query_&&.htm