Assignment 11

course MTH 174

õߒ}|ͨassignment #011

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Physics II

10-22-2008

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18:59:20

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 daily catch for the fishing data is:

This integral is broken into two parts:

integral from 2 to t x * 0.04x dx

integral from 2 to t of 0.04 * x^2 dx

0.04 * (t^3/3 - 2^3/3) = 2.773

The second integral is:

integral from 6 to 8 of x * (-0.06x + 0.6) dx

integral from 6 to 8 of -0.06x^2 + 0.6x dx

-0.06 * (8^3/3 - 6^3/3) + 0.6(8^2/2 - 6^2/2) = 2.48

The mean is then the sum of these two integrals

2.48 + 2.773 = 5.253

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18:59:43

what is the mean daily catch?

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RESPONSE -->

This was given above.

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19:00:45

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

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RESPONSE -->

The mean daily catch for the fishing data is:

This integral is broken into two parts:

integral from 2 to t x * 0.04x dx

I neglected to mention that t was six in the first integral.

The second integral is:

integral from 6 to 8 of x * (-0.06x + 0.6) dx

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19:01:28

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

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RESPONSE -->

The mean is just like the first moment.

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19:03:23

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

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

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19:11:53

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

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RESPONSE -->

3e^(-3t) for t >= 0 because as t increases (time increases) it is much more likely that a customer will walk in.

The others are not a good fit because:

p(t) = cost 0 <= t <= 2`pi

e^(t-2`pi) t >= 2`pi

is cyclical from 0 to 2`pi then increases

e^(-3t) for t > 0 is not a good fit because it starts at a fixed time at 0

p(t) = 1/4 for 0 <= t <= 4 is not a good fit because it is a flat line. This does not look like a probability density because it does not converge to a point

** 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 x we get 1 - e^-(3 x), which approaches 1 as x -> infinity. This makes the function a probability density function. Furthermore it is a decreasing function. **

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19:12:00

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

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RESPONSE -->

See above.

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19:12:03

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

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&#Your work looks good. See my notes. Let me know if you have any questions. &#