Assignment 11

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course Mth 174

7/19 7

Question:     ****   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:

The linear functions that are between the two points are y = .04 x and y = .6 - .06 x

 

Probability density = 1

 

Mean value = x * p(x). 

 

int( x*.04 x, x, 2, 6) = .04 / 3 * (6^3-2^3) = .04*208/3=8.32/3 = 2.77

int(-.06x^2 + .6x, x, 6, 8) = [-.02 x^3 + .3 x^2 ] = -.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

 

 

confidence rating #$&*:

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 3

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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:

 OK

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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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

int(3 e^(-3t))t, 0, 4) = 1 - e^-12 = 1   

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

 3

.............................................

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

 

 

------------------------------------------------

Self-critique Rating:

 OK

*********************************************

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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

int(3 e^(-3t))t, 0, 4) = 1 - e^-12 = 1   

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

 3

.............................................

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

 

 

------------------------------------------------

Self-critique Rating:

#*&!

 OK

*********************************************

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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

int(3 e^(-3t))t, 0, 4) = 1 - e^-12 = 1   

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

 3

.............................................

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.  **

 

.........................................

                                                  22:23:53

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Self-critique (if necessary):

 OK

 

 

------------------------------------------------

Self-critique Rating:

#*&!#*&!

 OK

*********************************************

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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

int(3 e^(-3t))t, 0, 4) = 1 - e^-12 = 1   

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

 3

.............................................

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.  **

 

.........................................

                                                  22:23:53

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Self-critique (if necessary):

 OK

 

 

------------------------------------------------

Self-critique Rating:

#*&!#*&!#*&!

 OK

*********************************************

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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

int(3 e^(-3t))t, 0, 4) = 1 - e^-12 = 1   

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

 3

.............................................

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.  **

 

.........................................

                                                  22:23:53

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Self-critique (if necessary):

 OK

 

 

------------------------------------------------

Self-critique Rating:

#*&!#*&!#*&!#*&!

 OK

*********************************************

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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

int(3 e^(-3t))t, 0, 4) = 1 - e^-12 = 1   

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

 3

.............................................

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.  **

 

.........................................

                                                  22:23:53

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Self-critique (if necessary):

 OK

 

 

------------------------------------------------

Self-critique Rating:

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