15.2 Problems on random selection  (Page 3/7)

A game is played as follows:

1. A wheel is spun, giving one of the integers 0 through 9 on an equally likely basis.
2. A single die is thrown the number of times indicated by the result of the spin of the wheel. The number of points made is the total of the numbers turned up on the sequence ofthrows of the die.
3. A player pays sixteen dollars to play; a dollar is returned for each point made.

Let Y represent the number of points made and $X=Y-16$ be the net gain (possibly negative) of the player. Determine the maximum value of

X , $E\left[X\right]$ , $\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[X\right]$ , $P\left(X>0\right)$ , $P\left(X>=10\right)$ , $P\left(X>=16\right)$ .

gn = 0.1*ones(1,10); gy = (1/6)*[0 ones(1,6)]; [Y,PY]= gendf(gn,gy); [X,PX]= csort(Y-16,PY); M = max(X)M = 38 EX = dot(X,PX) % Check EX = En*Ey - 16 = 4.5*3.5EX = -0.2500 % 4.5*3.5 - 16 = -0.25 VX = dot(X.^2,PX) - EX^2VX = 114.1875 Ppos = (X>0)*PX' Ppos = 0.4667P10 = (X>=10)*PX' P10 = 0.2147P16 = (X>=16)*PX' P16 = 0.0803

Marvin calls on four customers. With probability ${p}_{1}=0.6$ he makes a sale in each case. Geraldine calls on five customers, with probability ${p}_{2}=0.5$ of a sale in each case. Customers who buy do so on an iid basis, and order an amount Y i (in dollars) with common distribution:

$Y=\left[200\phantom{\rule{0.277778em}{0ex}}220\phantom{\rule{0.277778em}{0ex}}240\phantom{\rule{0.277778em}{0ex}}260\phantom{\rule{0.277778em}{0ex}}280\phantom{\rule{0.277778em}{0ex}}300\right]\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}PY=\left[0.10\phantom{\rule{0.277778em}{0ex}}0.15\phantom{\rule{0.277778em}{0ex}}0.25\phantom{\rule{0.277778em}{0ex}}0.25\phantom{\rule{0.277778em}{0ex}}0.15\phantom{\rule{0.277778em}{0ex}}0.10\right]$

Let D 1 be the total sales for Marvin and D 2 the total sales for Geraldine. Let $D={D}_{1}+{D}_{2}$ . Determine the distribution and mean and variance for D 1 , D 2 , and D . Determine $P\left({D}_{1}\ge {D}_{2}\right)$ and $P\left(D\ge 1500\right)$ , $P\left(D\ge 1000\right)$ , and $P\left(D\ge 750\right)$ .

gnM = ibinom(4,0.6,0:4); gnG = ibinom(5,0.5,0:5);Y = 200:20:300; PY = 0.01*[10 15 25 25 15 10]; [D1,PD1]= mgdf(gnM,Y,PY); [D2,PD2]= mgdf(gnG,Y,PY); ED1 = dot(D1,PD1)ED1 = 600.0000 % Check: ED1 = EnM*EY = 2.4*250 VD1 = dot(D1.^2,PD1) - ED1^2VD1 = 6.1968e+04 ED2 = dot(D2,PD2)ED2 = 625.0000 % Check: ED2 = EnG*EY = 2.5*250 VD2 = dot(D2.^2,PD2) - ED2^2VD2 = 8.0175e+04 [D1,D2,t,u,PD1,PD2,P]= icalcf(D1,D2,PD1,PD2); Use array opertions on matrices X, Y, PX, PY, t, u, and P[D,PD] = csort(t+u,P);ED = dot(D,PD) ED = 1.2250e+03eD = ED1 + ED2 % Check: ED = ED1 + ED2 eD = 1.2250e+03 % (Continued next page)VD = dot(D.^2,PD) - ED^2VD = 1.4214e+05 vD = VD1 + VD2 % Check: VD = VD1 + VD2vD = 1.4214e+05 P1g2 = total((t>u).*P) P1g2 = 0.4612k = [1500 1000 750];PDk = zeros(1,3); for i = 1:3PDk(i) = (D>=k(i))*PD'; enddisp(PDk) 0.2556 0.7326 0.8872

A questionnaire is sent to twenty persons. The number who reply is a random number $N\sim$ binomial (20, 0.7). If each respondent has probability $p=0.8$ of favoring a certain proposition, what is the probability of ten or more favorable replies? Of fifteen or more?

gN = ibinom(20,0.7,0:20); gY = [0.2 0.8]; gendDo not forget zero coefficients for missing powers Enter gen fn COEFFICIENTS for gN gNEnter gen fn COEFFICIENTS for gY gY Results are in N, PN, Y, PY, D, PD, PMay use jcalc or jcalcf on N, D, P To view the distribution, call for gD.P10 = (D>=10)*PD' P10 = 0.7788P15 = (D>=15)*PD' P15 = 0.0660pD = ibinom(20,0.7*0.8,0:20); % Alternate: use D binomial (pp0) D = 0:20;p10 = (D>=10)*pD' p10 = 0.7788p15 = (D>=15)*pD' p15 = 0.0660

what is the stm
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brayan
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what king of growth are you checking .?
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research.net
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NANO
how can I make nanorobot?
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Tarell
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