question 10

#$&*

course Mth 174

Question: problem 8.6.8 (3d edition 8.5.8, formerly 8.4.6) $1000/yr continuous deposit at 5%

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Your solution:

10000 = int(1000e^(0.05t) dt, 0,t)

20000e^(0.05t) - 20000 = 10000

ln(3/2)/0.05= 8.11

confidence rating #$&*: 3

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Given Solution:

In a short time interval `dt at t years after the start the amount deposited will be 1000 * `dt.

Suppose that T stands for the time required to grow to $10,000. Then the $1000 * `dt will be able to grow for T - t years at a 5% continuous rate. It will therefore grow to $1000 * `dt * e^(.05 ( T - t) ).

Adding up all the contributions and taking the limit as `dt -> 0 we get the integral of 1000 e^(.05 ( T - t) ) with respect to t, integrated from t = 0 to t = T.

An antiderivative is -1000 / .05 e^(.05 T - t)); evaluating at the limits and subtracting we get 20,000 ( e^(.05 T) - e^0) = 20,000 (e^(.05 T) - 1)

Setting this equal to 10,000 we get e^(.05 T) - 1 = .5, or e^(.05 T) = 1.5.

Taking ln of both sides gives .05 T = ln(1.5). Thus T = ln(1.5) / .05 = 8.1093.

It takes 8.1093 years for the principle to reach $10,000 with initial principle $1000.

If initial principle is $2000 then the equation becomes

2000 e^(.05 T) + 20,000 ( e^(.05 T) - 1 ) = 10,000, or

22,000 e^(.05 T) = 30,000 so

e^(.05 T) = 30/22,

.05 T = ln(30/22)

T = ln(30/22) / .05 = 6.2.

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Question:

problem 8.7.21 (4th edition 8.7.20 3d edition 8.6.20 formerly page 414 #12)

death density function f(t) = c t e^-(kt). What is c in terms of k?

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Your solution:

int(cte^(-kt)`dt 0, inf) = 1

c[1/-k te^(-kt) + 1/k^2 e^(-kt)] = 1

c = 1/[-1/k te^(-kt) + 1/k^2 e^(-kt)]

c = 1/[1/k^2]= k^2

int(k^2te^(-kt) dt, 0, 5) = 0.6

-5ke^(-5k) - e^(-5k) = 0.6

k = 0.275, c = k^2 = 0.756

CDF = int(k^2te^(-kt) dt, 0, x) = 1-kte^(-kt) - e^(-kt)

confidence rating #$&*: 3

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Given Solution:

** The integral of any probability density function should be 1, which is equivalent for the present problem to saying that every sick individual will eventually die. Thus the integral of c t e^(-kt), from t=0 to infinity, is 1. This is the relationship you solve for c.

An antiderivative of c t e^(-kt) is F(t) = - c (t e^(-kt) / k + e^(-kt) / k^2) = -c e^(-kt) ( k t + 1) / k^2.

lim{t -> infinity)(F(t)) = 0.

F(0) = c(- 1/k^2) = -c/k^2.

So the integral from 0 to infinity is c / k^2.

This integral must be 1.

So c / k^2 = 1 and

c = k^2 . **

** See also the previous note, where we see that in order for this function to be a probability distribution, the constant c must take the value k^2.

We now have the information that 40% die within 5 years, so that the integral of f(t) from 0 to 5 is .4. This integral is in terms of c and k and yields an equation relating c and k. Combining this information with our previously found relationship between c and k you can find both c and k.

We have for the proportion dying in the first 5 years:

integral ( k^2 t e^-(kt) dt, t from 0 to 5) = .4.

Using the antiderivative

F(t) = -c e^(-kt) ( k t + 1) / k^2 = - k^2 e^(-kt) ( k t + 1) / k^2

= -e^(-kt) ( kt + 1)

we get

F(5) - F(0) = .4

1 - e^(-5 k) ( 5 k + 1) = .4

e^(-5 k) ( 5 k + 1) = .6.

This equation presents a problem because it can't be solved exactly.

If we graph the left-hand side as a function of k we see that there are positive and negative solutions. We are interested only in positive solutions because otherwise the limit of the original antiderivative at infinity won't be 0 and the integral will be divergent.

Solving approximately using Derive, with trial interval starting at 0, we get k = .4045.

the cumulative function is just the integral of the density function--the integral from 0 to t of f(x), where x is our 'dummy' integration variable. We have

P(t) = cumulative distribution function = integral ( k^2 x e^-(kx) dx, x from 0 to t).

Using the same antiderivative function as before this integral is

P(t) = F(t) - F(0) = -e^(-kt) ( kt + 1) - (- e^(-k*0) ( k*0 + 1) ) = 1 - e^(-kt) ( kt + 1).

Note that for k = .4045 this function is

P(t) = 1 - e^(-.4045 t) ( -.4045 t + 1).

You can check that for this function, P(5) = .4 (40% die within 5 years) and lim{t -> infinity}(P(t)) = 1.

**

STUDENT QUESTION

Can you walk me thru how you obtained your antiderivative???????????????????????

INSTRUCTOR REPSONSE

To get an antiderivative of t e^(-k t) you use integration by parts, with u = t and dv = e^(-k t) dt.

This gives you du = dt and v = -1/k e^(-kt).

Thus u v - int(v du) = -t / k e^(-k t) - int(-1/k e^(-k t) dt) = -t / k e^(-k t) - 1 / k^2 e^(-k t).

This algebraically rearranges to the given forms.

Note: The questions below are 'followup questions'. You need not insert answers to these questions, but you may do so. If you do, please precede your answer with **** and add #$&* in the line immediately after your insertion. You will be able to recognize similar 'followup questions' on many questions in subsequent assignments.

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What is the cumulative death distribution function?

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What integral did you use to obtain the cumulative death distribution

function and why?

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Self-critique (if necessary): ok

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Question:

**** query problem 8.7.4 (3d edition 8.6.4) fn y = 4 for 0 < x < c. **** Is the function a probability distribution function or a cumulative probability function? What is the value of c?

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Your solution:

PDF, because a CDF must be increasing and ass original.

int(4 dx, 0, c) = 1, then c = 0.25

CDF = int(4 dx 0,x)/1 = 4x, on [0, 1/4] and CDF = 1 otherwise

confidence rating #$&*: 3

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Given Solution:

** A cumulative distribution function must be increasing, and it starts at 0.

This is a prob distribution function.

The integral of the function must be 1.

The integral of y = 4 from x = 0 to x = c is 4 c.

So 4c = 1 and c = ¼. **

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Self-critique (if necessary):

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Question:

**** query problem 8.7.9 (4th edition 8.7.8 3d edition 8.6.8) Function linear from (0,0) - (2,c) - (4, 3c) - (?, 3c) **** Is the function a probability distribution function or a cumulative probability function? What is the value of c?

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Your solution:

CDF, it’s increasing and pass the origin.

1=3c, c=1/3

int(p(x) dx, 0, x)/int(p(x) dx, 0, 2) = 1/6 x, p(x) = 1/6 on [0,2]

int(p(x) dx, 0, x)/int(p(x) dx, 2, 4) = 1/3x, p(x) = 1/3 on [2,4]

p(x) = 0 on[4,inf]

confidence rating #$&*: 3

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Given Solution:

** This function would have infinite area and can’t be a probability distribution function. It’s a cumulative probability function.

Cumulative probability functions start at 0 and end up with value 1. This one starts at 0 and ends up with value 3c. So 3c = 1 and c = 1/3. **

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question:

**** query problem 8.7.13 (4th edition 8.7.12 3d edition 8.6.12). F(x) cum distr fn heights of trees in meters **** What is the meaning of F(7) = .6?

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Your solution:

F(7) = 0.6 means 60% of trees is under 7 meters.

F(7)=>F(6), because F(x) is CDF, which is increasing (at least nondecreasing)

confidence rating #$&*: 3

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Given Solution:

** The cumulative distribution function tells you for any x the probability that the quantity being governed by the function will be less than or equal to x. It is the cumulative probability of all occurrences up to that value.

Thus F(7) = .6 means that .6 of the trees are up to and including 7 meters in height. **

**** Which is greater, F(6) or F(7)? Justify in terms of trees.

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20:35:10

** There are more trees up to and including 7 meters tall than trees up to and including 6 meters tall, since the first characterization includes the second. **

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question:

problem page 415 #18 probability distribution function for the position of a pendulum bob

This question has been omitted from recent editions of the text. You should, however, answer it and self-critique your answer.

If the position of the bob of a swinging pendulum is observed at a large number of random times, we can use our observations to construct a graph of the probability distribution for the position of the pendulum.

Describe the expected characteristics of this graph.

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01:37:23

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Your solution:

confidence rating #$&*:

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Given Solution:

** The pendulum bob is most likely to be found where it is moving most slowly, and least likely where it is moving most quickly.

It's slowest near its ends and quickest near the center.

So the probability density would be low in the middle and high near the ends.

Velocity increases smoothly as we approach the center, and decreases smoothly as we move away from the center, so the probability would decrease smoothly from x = -a to x = 0, then increase smoothly as we move from x = 0 to x = a. **

the curve is a parabola opening upward

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