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Suppose X is a nonnegative, absolutely continuous random variable. Let Z = g ( X ) = C e - a X , where a > 0 , C > 0 . Then 0 < Z C . Use properties of the exponential and natural log function to show that

F Z ( v ) = 1 - F X - ln ( v / C ) a for 0 < v C

Z = C e - a X v iff e - a X v / C iff - a X ln ( v / C ) iff X - ln ( v / C ) / a , so that

F Z ( v ) = P ( Z v ) = P ( X - ln ( v / C ) / a ) = 1 - F X - ln ( v / C ) a
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Use the result of [link] to show that if X exponential ( λ ) , then

F Z ( v ) = v C λ / a 0 < v C
F Z ( v ) = 1 - 1 - e x p - λ a · ln ( v / C ) = v C λ / a
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Present value of future costs. Suppose money may be invested at an annual rate a , compounded continually. Then one dollar in hand now, has a value e a x at the end of x years. Hence, one dollar spent x years in the future has a present value e - a x . Suppose a device put into operation has time to failure (in years) X exponential ( λ ) . If the cost of replacement at failure is C dollars, then the present value of the replacement is Z = C e - a X . Suppose λ = 1 / 10 , a = 0 . 07 , and C = $ 1000 .

  1. Use the result of [link] to determine the probability Z 700 , 500 , 200 .
  2. Use a discrete approximation for the exponential density to approximate the probabilities in part (a). Truncate X at 1000 and use 10,000 approximation points.
P ( Z v ) = v 1000 10 / 7
v = [700 500 200];P = (v/1000).^(10/7) P = 0.6008 0.3715 0.1003tappr Enter matrix [a b]of x-range endpoints [0 1000] Enter number of x approximation points 10000Enter density as a function of t 0.1*exp(-t/10) Use row matrices X and PX as in the simple caseG = 1000*exp(-0.07*t); PM1 = (G<=700)*PX' PM1 = 0.6005PM2 = (G<=500)*PX' PM2 = 0.3716PM3 = (G<=200)*PX' PM3 = 0.1003
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Optimal stocking of merchandise. A merchant is planning for the Christmas season. He intends to stock m units of a certain item at a cost of c per unit. Experience indicates demand can be represented by a random variable D Poisson ( μ ) . If units remain in stock at the end of the season, they may be returned with recovery of r per unit. If demand exceeds the number originally ordered, extra units may be ordered at a cost of s each. Units are sold at a price p per unit. If Z = g ( D ) is the gain from the sales, then

  • For t m , g ( t ) = ( p - c ) t - ( c - r ) ( m - t ) = ( p - r ) t + ( r - c ) m
  • For t > m , g ( t ) = ( p - c ) m + ( t - m ) ( p - s ) = ( p - s ) t + ( s - c ) m

Let M = ( - , m ] . Then

g ( t ) = I M ( t ) [ ( p - r ) t + ( r - c ) m ] + I M ( t ) [ ( p - s ) t + ( s - c ) m ]
= ( p - s ) t + ( s - c ) m + I M ( t ) ( s - r ) ( t - m )

Suppose μ = 50 m = 50 c = 30 p = 50 r = 20 s = 40. .
Approximate the Poisson random variable D by truncating at 100. Determine P ( 500 Z 1100 ) .

mu = 50; D = 0:100;c = 30; p = 50;r = 20; s = 40;m = 50; PD = ipoisson(mu,D);G = (p - s)*D + (s - c)*m +(s - r)*(D - m).*(D<= m); M = (500<=G)&(G<=1100); PM = M*PD'PM = 0.9209[Z,PZ] = csort(G,PD); % Alternate: use dbn for Zm = (500<=Z)&(Z<=1100); pm = m*PZ'pm = 0.9209
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(See Example 2 from "Functions of a Random Variable") The cultural committee of a student organization has arranged a special deal for tickets to a concert. The agreement is that the organization will purchase tentickets at $20 each (regardless of the number of individual buyers). Additional tickets are available according to the following schedule:

  • 11-20, $18 each
  • 21-30, $16 each
  • 31-50, $15 each
  • 51-100, $13 each

If the number of purchasers is a random variable X , the total cost (in dollars) is a random quantity Z = g ( X ) described by

g ( X ) = 200 + 18 I M 1 ( X ) ( X - 10 ) + ( 16 - 18 ) I M 2 ( X ) ( X - 20 ) +
( 15 - 16 ) I M 3 ( X ) ( X - 30 ) + ( 13 - 15 ) I M 4 ( X ) ( X - 50 )
where M 1 = [ 10 , ) , M 2 = [ 20 , ) , M 3 = [ 30 , ) , M 4 = [ 50 , )

Suppose X Poisson (75). Approximate the Poisson distribution by truncating at 150. Determine P ( Z 1000 ) , P ( Z 1300 ) , and P ( 900 Z 1400 ) .

X = 0:150; PX = ipoisson(75,X);G = 200 + 18*(X - 10).*(X>=10) + (16 - 18)*(X - 20).*(X>=20) + ... (15 - 16)*(X- 30).*(X>=30) + (13 - 15)*(X - 50).*(X>=50); P1 = (G>=1000)*PX' P1 = 0.9288P2 = (G>=1300)*PX' P2 = 0.1142P3 = ((900<=G)&(G<=1400))*PX' P3 = 0.9742[Z,PZ] = csort(G,PX); % Alternate: use dbn for Zp1 = (Z>=1000)*PZ' p1 = 0.9288
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Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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