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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

Introduction about quantum dots in nanotechnology
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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
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s. Reply
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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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