Assignment 11 Query

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course Mth 174

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:

Int. .04x^2, x, 2, 6

.04/3 * x^3, x, 2, 6

(.04/3 * 6^3) – (.04/3 * 2^3)

2.773

Int. .6x - .06x^2, x, 6, 8

.3x^2 - .02x^3 , x, 6, 8

(19.2 – 10.24) – (10.8 – 4.32)

2.48

2.773+2.48=5.253 is our mean value of our function.

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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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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: Just from eyeballing the choices, it cannot be cos t since that has the potential to be negative, and ¼ is just making a massive assumption about someone coming in over 4 minutes. So, I will integrate the middle 2 functions from 0 to 4.

Int. e^-.3t , t, 0, 4

=-1/3e^-12 - -1/3e^0

=1/3

Int. 3e^(-3t), t, 0, 4

-e^(-3t), t, 0 , 4

-e^(-12) - -e^(0)

1

This means 3e^(-3t) is our solution since the int. of the function is 1 which is a trait of a probability distribution function.

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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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