query 11

**** Query problem 8.7.2, Probability and More On Distributions, p. 421

daily catch density function piecewise linear (2,.08) to (6.,24) to (8,.12)

What is the mean daily catch?

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Your solution:

We are given that p(x) is .04 x for values of x between 2 and 6 and

-.06x +.6 for values of x between 6 and 8

So by adding the integrals of xp(x) we can get the median

So integral from 2 to 6 of x(.04x) + integral from 6 to 8 of x(-.06x +.6)

.04 * (1/3) * x^3 .06*(1/3)*x^3 + (1/2)*.6*x^2

.013x^3 is this antiderivative .02x^3 + .3x^2 is this antiderivative

Now we evaluate and add the 2 together to find the median

2.808-1.07 + 8.96-6.48

2.701 + 2.48 = 5.181

5.181 is the mean

confidence rating #$&*: 2

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Given 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) = .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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Self-critique (if necessary):

Ok

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Self-critique Rating:

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Question:

**** Query problem 8.8.13 (3d edition 8.7.13) (formerly #12, Probability and More On Distributions, p. 423 )

cos t, 0

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

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Your solution:

We are given 4 choices to decide which function would best represent the probability for the next customer walks in.

We will need to evaluate each choice as an integral to see if we can eliminate any choices

Beginning with the integral from 0 to 4 of cos t which would be sin t evaluated at 4 and 0 would give us -.76 - 0 which is -.76 we know our value cannot be negative so this choice is eliminated.

Next we evaluate the integral from 0 to 4 of (1/4) which would have an antiderivative of (1/4)x evaluated at 0 and 4 we get 1 which would indicate 100% probability which means this choice is also eliminated.

Next we evaluate the integral from 0 to 4 pf 3e^(-3t) using u=-3t and -1/3du=dt we get the antiderivative to be -e^(-3t) we evaluate at 0 and 4 and get -.000006+1 which is .99. This could be an answer but we still have one more integral to evaluate

The integral from 0 to 4 of e^(-3t) use u substitution to get u=-3t and -(1/3)du=dt so the antiderivative is e^(-3t) evaluated at 0 and 4 we get .33 so this could also be a choice.

At this point we have eliminated 2 choice and are left with 2 choices so we should see if both are actually probability functions in hopes of eliminating one choice. We do this by checking for convergence to 1 from 0 to infinity which is a property of a probability function so

Integral from 0 to infinity of 36e^(-3t) and integral from 0 to infinity of e^(-3t)

Antiderivative -e^(-3t) antiderivative -(1/3)e^(-3t)

Evaluating we get 0+1=1 evaluating we get 0+(1/3) = 1/3

Yes, probability function converges to 1 no this isn’t a probability function it coverges to 1/3

So the most likely function is 3e^-(3t)

confidence rating #$&*: 2

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Given Solution:

** 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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22:23:53

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Self-critique (if necessary):

ok

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Self-critique Rating:3

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