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The number N of “hits” in a day on a Web site on the internet is Poisson (80). Suppose the probability is 0.10 that any hit results in a sale, is 0.30that the result is a request for information, and is 0.60 that the inquirer just browses but does not identify an interest. What is the probability of 10 or more sales? Whatis the probability that the number of sales is at least half the number of information requests (use suitable simple approximations)?

X = 0:30; Y = 0:80;PX = ipoisson(80*0.1,X); PY = ipoisson(80*0.3,Y);icalc: X Y PX PY - - - - - - - - - - - -PX10 = (X>=10)*PX' % Approximate calculation PX10 = 0.2834pX10 = cpoisson(8,10) % Direct calculation pX10 = 0.2834M = t>=0.5*u; PM = total(M.*P)PM = 0.1572
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The number N of orders sent to the shipping department of a mail order house is Poisson (700). Orders require one of seven kinds of boxes, which with packing costs havedistribution

Cost (dollars) 0.75 1.25 2.00 2.50 3.00 3.50 4.00
Probability 0.10 0.15 0.15 0.25 0.20 0.10 0.05

What is the probability the total cost of the $2.50 boxes is no greater than $475? What is the probability the cost of the $2.50 boxes is greater than the cost of the $3.00 boxes?What is the probability the cost of the $2.50 boxes is not more than $50.00 greater than the cost of the $3.00 boxes? Suggestion . Truncate the Poisson distributions at about twice the mean value.

X = 0:400; Y = 0:300;PX = ipoisson(700*0.25,X); PY = ipoisson(700*0.20,Y);icalc Enter row matrix of X-values XEnter row matrix of Y-values Y Enter X probabilities PXEnter Y probabilities PY Use array operations on matrices X, Y, PX, PY, t, u, and PP1 = (2.5*X<=475)*PX' P1 = 0.8785M = 2.5*t<=(3*u + 50); PM = total(M.*P)PM = 0.7500
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One car in 5 in a certain community is a Volvo. If the number of cars passing a traffic check point in an hour is Poisson (130), what is the expected number of Volvos? Whatis the probability of at least 30 Volvos? What is the probability the number of Volvos is between 16 and 40 (inclusive)?

P1 = cpoisson(130*0.2,30) = 0.2407 P2 = cpoisson(26,16) - cpoisson(26,41) = 0.9819
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A service center on an interstate highway experiences customers in a one-hour period as follows:

  • Northbound: Total vehicles: Poisson (200). Twenty percent are trucks.
  • Southbound: Total vehicles: Poisson (180). Twenty five percent are trucks.
  • Each truck has one or two persons, with respective probabilities 0.7 and 0.3.
  • Each car has 1, 2, 3, 4, or 5 persons, with probabilities 0.3, 0.3, 0.2, 0.1, 0.1, respectively

Under the usual independence assumptions, let D be the number of persons to be served. Determine E [ D ] , Var [ D ] , and the generating function g D ( s ) .

T Poisson (200*0.2 + 180*0.25 = 85), P Poisson (200*0.8 + 180*0.75 = 295).

a = 85 b = 200*0.8 + 180*0.75b = 295 YT = [1 2]; PYT = [0.7 0.3]; EYT = dot(YT,PYT)EYT = 1.3000 VYT = dot(YT.^2,PYT) - EYT^2VYT = 0.2100 YP = 1:5;PYP = 0.1*[3 3 2 1 1];EYP = dot(YP,PYP) EYP = 2.4000VYP = dot(YP.^2,PYP) - EYP^2 VYP = 1.6400EDT = 85*EYT EDT = 110.5000EDP = 295*EYP EDP = 708.0000ED = EDT + EDP ED = 818.5000VT = 85*(VYT + EYT^2) VT = 161.5000VP = 295*(VYP + EYP^2) VP = 2183VD = VT + VP VD = 2.2705e+03NT = 0:180; % Possible alternativegNT = ipoisson(85,NT); gYT = 0.1*[0 7 3]; [DT,PDT]= gendf(gNT,gYT); EDT = dot(DT,PDT)EDT = 110.5000 VDT = dot(DT.^2,PDT) - EDT^2VDT = 161.5000 NP = 0:500;gNP = ipoisson(295,NP); gYP = 0.1*[0 3 2 2 1 1]; [DP,PDP]= gendf(gNP,gYP); % Requires too much memory
g D T ( s ) = exp ( 85 ( 0 . 7 s + 0 . 3 s 2 - 1 )) g D P ( s ) = exp ( 295 ( 0 . 1 ( 3 s + 3 s 2 2 s 3 + s 4 + s 5 ) - 1 ) )
g D ( s ) = g D T ( s ) g D P ( s )
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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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