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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 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 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 A service center on an interstate highway experiences customers in a one-hour period as follows:
Under the usual independence assumptions, let D be the number of persons to be served. Determine , , and the generating function .
Poisson (200*0.2 + 180*0.25 = 85), 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
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