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Suppose binomial (12, 0.375), Poisson (4.5), and exponential (1/4.5). For each random variable, calculate and tabulate the probability of a value at least k , for integer values .
k = 3:8;
Px = cbinom(12,0.375,k);Py = cpoisson(4.5,k);
Pz = exp(-k/4.5);disp([k;Px;Py;Pz]')3.0000 0.8865 0.8264 0.5134
4.0000 0.7176 0.6577 0.41115.0000 0.4897 0.4679 0.3292
6.0000 0.2709 0.2971 0.26367.0000 0.1178 0.1689 0.2111
8.0000 0.0390 0.0866 0.1690 The number of noise pulses arriving on a power circuit in an hour is a random quantity having Poisson (7) distribution. What is the probability of having at least10 pulses in an hour? What is the probability of having at most 15 pulses in an hour?
P1 = cpoisson(7,10) = 0.1695 P2 = 1 - cpoisson(7,16) = 0.9976
The number of customers arriving in a small specialty store in an hour is a random quantity having Poisson (5) distribution. What is the probability the numberarriving in an hour will be between three and seven, inclusive? What is the probability of no more than ten?
P1 = cpoisson(5,3) - cpoisson(5,8) = 0.7420
P2 = 1 - cpoisson(5,11) = 0.9863 Random variable binomial (1000, 0.1).
k = [80 100 120];P = cbinom(1000,0.1,k)
P = 0.9867 0.5154 0.0220P1 = cpoisson(100,k)
P1 = 0.9825 0.5133 0.0282 The time to failure, in hours of operating time, of a televesion set subject to random voltage surges has the exponential (0.002) distribution. Suppose the unithas operated successfully for 500 hours. What is the (conditional) probability it will operate for another 500 hours?
Twenty “identical” units are put into operation. They fail independently. The times to failure (in hours) form an iid class, exponential (0.0002). This means the“expected” life is 5000 hours. Determine the probabilities that at least k , for , will survive for 5000 hours.
p = exp(-0.0002*5000)
p = 0.3679k = [5 8 10 12 15];P = cbinom(20,p,k)
P = 0.9110 0.4655 0.1601 0.0294 0.0006 Let gamma (20, 0.0002) be the total operating time for the unitsdescribed in [link] .
P1 = gammadbn(20,0.0002,100000) = 0.5297 P2 = cpoisson(0.0002*100000,20) = 0.5297
The sum of the times to failure for five independent units is a random variable gamma . Without using tables or m-programs, determine .
Interarrival times (in minutes) for fax messages on a terminal are independent, exponential ( ). This means the time X for the arrival of the fourth message is gamma(4, 0.1). Without using tables or m-programs,utilize the relation of the gamma to the Poisson distribution to determine .
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