Assignment 11 Query

course Mth 174

DόQʃassignment #011

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Physics II

10-30-2007

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10:29:17

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 problem is to find the mean daily catch here. To do that, I use the integral of the density function multiplied by the variable, like:

int( xp(x) )

In this function, there are two parts to the formula, from 2 to 6 and from 6 to 8. The rates are given below:

p(x) = 0.04x for 2 <= x <=6

p(x) = -0.06x - 0.6 for 6 <= x <= 8

To find the mean of these two functions, I just need to integrate them according to the formula above and add them together:

int (x*0.04x, x, 2, 6) = 2.773

int (x*(-0.06x - 0.6), x, 6, 8) = 2.48

Overall mean = 2.773 + 2.48 = 5.2533

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10:29:39

what is the mean daily catch?

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RESPONSE -->

5.2533 tons of fish.

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10:30:32

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

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RESPONSE -->

int (x*0.04x, x, 2, 6) and

int (x*(-0.06x - 0.6), x, 6, 8)

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10:31:04

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

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RESPONSE -->

The formula is the same and so is the idea: to find the balance point of the graph.

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10:32:55

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

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RESPONSE -->

Ok.

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10:34:40

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)

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12:11:45

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

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RESPONSE -->

(a) p(x) = cos t for 0<= t <= 2'pi and e^(t-2'pi) for t>= 2'pi.

This is unacceptable because cos t becomes negative for part of 0 to 2'pi and a probability density function can never be less than 0.

(b) p(x) = 3e^(-3t)

This works because it is positive for all values of x and the integral from 0 to infinity is 1. So this is the most likely to be a probability density function.

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

Values are positive for all values of x since 0 is the asymptote for this function, but the integral does not approach 1, staying rather at 1/3. Since a probability density function must equal 1 for the period, this isn't an acceptable function.

(d) p(x) = 1/4 for 0 <= t <= 4

First off, is the store open only 4 minutes? Besides which, the integral diverges and never approaches 1, so it can't be a probability density function.

Good. For comparison:

** 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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12:12:07

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

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RESPONSE -->

More applications of the same basic functions to different types of needs. Very interesting.

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Very good work. Let me know if you have questions. &#