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A property is offered for sale. Experience indicates the number N of bids is a random variable having values 0 through 8, with respective probabilities
Number | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Probability | 0.05 | 0.15 | 0.15 | 0.20 | 0.15 | 0.10 | 0.10 | 0.05 | 0.05 |
The market is such that bids (in thousands of dollars) are iid symmetric triangular on [150 250].Determine the probability of at least one bid of $210,000 or more.
gN = 0.01*[5 15 15 20 15 10 10 5 5];PY = 0.5 + 0.5*(1 - (4/5)^2)
PY = 0.6800>>PW = 1 - polyval(fliplr(gN),PY)
PW = 0.6536%alternate
gY = [0.68 0.32];
[D,PD]= gendf(gN,gY);
P = (D>0)*PD'
P = 0.6536
Suppose $N\sim $ binomial $(10,\phantom{\rule{0.166667em}{0ex}}0.3)$ and the Y _{i} are iid, uniform on $[10,\phantom{\rule{0.166667em}{0ex}}20]$ . Let V be the minimum of the N values of the Y _{i} . Determine $P(V>t)$ for integer values from 10 to 20.
gN = ibinom(10,0.3,0:10);
t = 10:20;p = 0.1*(20 - t);
P = polyval(fliplr(gN),p) - 0.7^10P =
Columns 1 through 70.9718 0.7092 0.5104 0.3612 0.2503 0.1686 0.1092
Columns 8 through 110.0664 0.0360 0.0147 0
Pa = (0.7 + 0.3*p).^10 - 0.7^10 % Alternate form of gNPa =
Columns 1 through 70.9718 0.7092 0.5104 0.3612 0.2503 0.1686 0.1092
Columns 8 through 110.0664 0.0360 0.0147 0
Suppose a teacher is equally likely to have 0, 1, 2, 3 or 4 students come in during office hours on a given day. If the lengths of the individual visits, in minutes, areiid exponential (0.1), what is the probability that no visit will last more than 20 minutes.
gN = 0.2*ones(1,5);
p = 1 - exp(-2);FW = polyval(fliplr(gN),p)
FW = 0.7635gY = [p 1-p]; % Alternate[D,PD] = gendf(gN,gY);PW = (D==0)*PD'
PW = 0.7635
Twelve solid-state modules are installed in a control system. If the modules are not defective, they have practically unlimited life. However, with probability $p=0.05$ any unit could have a defect which results in a lifetime (in hours) exponential (0.0025). Under the usual independence assumptions, what is the probability the unit does not failbecause of a defective module in the first 500 hours after installation?
p = 1 - exp(-0.0025*500);
FW = (0.95 + 0.05*p)^12FW = 0.8410
gN = ibinom(12,0.05,0:12);gY = [p 1-p];[D,PD] = gendf(gN,gY);PW = (D==0)*PD'
PW = 0.8410
The number N of bids on a painting is binomial $(10,\phantom{\rule{0.166667em}{0ex}}0.3)$ . The bid amounts (in thousands of dollars) Y _{i} form an iid class, with common density function ${f}_{Y}\left(t\right)=0.005(37-2t)$ $2\le t\le 10$ . What is the probability that the maximum amount bid is greater than $5,000?
p = 0.45;
P = 1 - (0.7 + 0.3*p)^10P = 0.8352
gN = ibinom(10,0.3,0:10);gY = [p 1-p];[D,PD] = gendf(gN,gY); % D is number of "successes"Pa = (D>0)*PD'
Pa = 0.8352
A computer store offers each customer who makes a purchase of $500 or more a free chance at a drawing for a prize. The probability of winning on a draw is 0.05.Suppose the times, in hours, between sales qualifying for a drawing is exponential (4). Under the usual independence assumptions, what is the expected time between a winning draw?What is the probability of three or more winners in a ten hour day? Of five or more?
${N}_{t}\sim $ Poisson $\left(\lambda t\right)$ , ${N}_{Dt}\sim $ Poisson $\left(\lambda pt\right)$ , ${W}_{Dt}$ exponential $\left(\lambda p\right)$ .
p = 0.05;
t = 10;lambda = 4;
EW = 1/(lambda*p)EW = 5
PND10 = cpoisson(lambda*p*t,[3 5])
PND10 = 0.3233 0.0527
Noise pulses arrrive on a data phone line according to an arrival process such that for each $t>0$ the number N _{t} of arrivals in time interval $(0,t]$ , in hours, is Poisson $\left(7t\right)$ . The i th pulse has an “intensity” Y _{i} such that the class $\{{Y}_{i}:1\le i\}$ is iid, with the common distribution function ${F}_{Y}\left(u\right)=1-{e}^{-2{u}^{2}}$ for $u\ge 0$ . Determine the probability that in an eight-hour day the intensity will not exceed two.
N _{8} is Poisson (7*8 = 56) ${g}_{N}\left(s\right)={e}^{56(s-1)}$ .
t = 2;
FW2 = exp(56*(1 - exp(-t^2) - 1))FW2 = 0.3586
The number N of noise bursts on a data transmission line in a period $(0,t]$ is Poisson $\left(\mu t\right)$ . The number of digit errors caused by the i th burst is Y _{i} , with the class $\{{Y}_{i}:1\le i\}$ iid, ${Y}_{i}-1\sim $ geometric $\left(p\right)$ . An error correcting system is capable or correcting five or fewer errors in any burst. Suppose $\mu =12$ and $p=0.35$ . What is the probability of no uncorrected error in two hours of operation?
${F}_{W}\left(k\right)={g}_{N}\left[P(Y\le k)\right]$ $P(Y\le k)-1-{q}^{k-1}$ ${N}_{t}\sim $ Poisson $\left(12t\right)$
q = 1 - 0.35;
k = 5;t = 2;
mu = 12;FW = exp(mu*t*(1 - q^(k-1) - 1))
FW = 0.0138
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