The number of traffic accidents that occur on a particular stretch of road during a month
follows a Poisson distribution with a mean of 7.6. Find the probability that less than three
accidents will occur next month on this stretch of road. (0.018757)
9. The number of traffic accidents that occur on a particular stretch of road during a month
follows a Poisson distribution with a mean of 7. Find the probability of observing exactly
three accidents on this stretch of road next month. (0.052129)
10. The number of traffic accidents that occur on a particular stretch of road during a month
follows a Poisson distribution with a mean of 6.8. Find the probability that the next two
months will both result in four accidents each occurring on this stretch of road. (0.00985)
11. Suppose the number of babies born during an 8-hour shift at a hospital's maternity wing
follows a Poisson distribution with a mean of 6 an hour. Find the probability that five
babies are born during a particular 1-hour period in this maternity wing. (0.160623)
12. The university policy department must write, on average, five tickets per day to keep
department revenues at budgeted levels. Suppose the number of tickets written per day
follows a Poisson distribution with a mean of 8.8 tickets per day. Find the probability that
less than six tickets are written on a randomly selected day from this distribution.
(0.128387)
13. A taxi firm has two cars which it hires out day by day. The number of demands for a car
on each day is distributed as Poisson distribution with mean 1.5. Calculate the proportion
of days on which neither car is used and the proportion of days on which some demands
is refused
14. If calls to your cell phone are a Poisson process with a constant rate ?=0.5 calls per hour,
what's the probability that, if you forget to turn your phone off in a 3 hour lecture, your
phone rings during that time? How many phone calls do you expect to get during this
lecture?
15. The average number of defects per wafer (defect density) is 3. The redundancy built into
the design allows for up to 4 defects per wafer. What is the probability that the
redundancy will not be sufficient if the defects follow a Poisson distribution?
16. The mean number of errors due to a particular bug occurring in a minute is 0.0001
a) What is the probability that no error will occur in 20 minutes?
b) How long would the program need to run to ensure that there will be a 99.95% chance
that an error wills showup to highlight this bug
1. Calculate the Poisson distribution whose ? (Average Rate of Success)) is 3 & X (Poisson
Random Variable) is 6.
2. Customers arrive at a checkout counter according to a Poisson distribution at an average
of 7 per hour. During a given hour, what are the probabilities that
a) No more than 3 customers arrive?
b) At least 2 customers arrive?
c) Exactly 5 customers arrive?
3. Manufacturer of television set knows that on an average 5% of their product is defective.
They sells television sets in consignment of 100 and guarantees that not more than 2 set
will be defective. What is the probability that the TV set will fail to meet the guaranteed
quality?
4. It is known from the past experience that in a certain plant there are on the average of 4
industrial accidents per month. Find the probability that in a given year will be less that 3
accidents.
5. Suppose that the change of an individual coal miner being killed in a mining accident
during a year is 1.1499. Use the Poisson distribution to calculate the probability that in
the mine employing 350 miners- there will be at least one accident in a year.
6. The number of road construction projects that take place at any one time in a certain city
follows a Poisson distribution with a mean of 3. Find the probability that exactly five road
construction projects are currently taking place in this city. (0.100819)
7. The number of road construction projects that take place at any one time in a certain city
follows a Poisson distribution with a mean of 7. Find the probability that more than four
road construction projects are currently taking place in the city. (0.827008)
EEE 304, Communication System I, Spring 2018 (Deadline: 27.03.2018) - 263.2018 mert . comer 10.00- 12.00 pm / @stu.HALL Generation of random sequences: While deterministic signals such as square pulses, sinc Soft waveforms, sinuses and cosines are used in specific applications, almost all other real-life NO+ 10 signals from econometric series to radar returns, from genetic codes to multimedia signals in cost consumer electronics are information-bearing random signals. re Time sequences: Generate a sequence of 1000 equally spaced samples of a Gauss- - je Markov process using the recursive relation: Xn = axn-1+ Win = 1,2, ..., 1000. Here ab int assume that X0 = 0, and {w,) is a sequence of independent identically distributed Gaussian random variables. You can use the randn function in MATLAB to generate re zero-mean, unit-variance random variables. Below is given a low-pass filter (a>0) excited by a white noise sequence, and the filter has a real pole at z- a. Plot the output waveform for the following values of a: a= 0.5, a = 0.95, a = 0.995. Comment on the effect of the selection of a on the resulting time sequence.EEE 304, Communication System I, Spring 2018 (Deadline: 27.03.2018) Generation of random sequences: While deterministic signals such as square pulses, since waveforms, sinuses and cosines are used in specific applications. almost all other real-life signals from econometric series to radar returns, from genetic codes to multimedia signals in consumer electronics are information-bearing random signals. Time sequences: Generate a sequence of 1000 equally spaced samples of a Gauss- Markov process using the recursive relation: 1, = ax,-1 + win = 1,2, .., 1000. Here assume that No = 0. and (w,) is a sequence of independent identically distributed Gaussian random variables. You can use the randn function in MATLAB to generate zero-mean. unit-variance random variables. Below is given a low-pass filter (a>0) excited by a white noise sequence, and the filter has a real pole at 2- a. Plot the output waveform for the following values of a: a= 0.5. a - 0.95. a = 0.995. Comment on the effect of the selection of a on the resulting time sequence. Autocorrelation function: In order to estimate the autocorrelation function of this discrete process, one can apply the formula: Ry(m) = lim - EN- W 2n=1*n In-m = E(X,In-m). However. since we have a finite number of samples an approximation would be Ry(m) = - -mx N-m Zal Andn-mim = 0,1. ....64. Notice we shorten the data and consider only the overlapping samples in the window starting at nel and the window starting at nem+1. The theoretical autocorrelation for such a signal model is also known as: R,(m) = ona", where of is the input process variance. Plot the autocorrelation sequence in three separate graphs for the values of a= 0.5. a = 0.95. a = 0.995, but superpose in each graph the analytical and estimated R,(m) and R, (m) correlation functions. Power Spectrum: Using the Wiener-Khinchine theorem, we can find the power spectrum in two different ways. Then do the following: i. Estimate T(() = (x(/]P. where X (f = ) = EN=x(nje /x*. for a=0.95 and N=512 ii. Estimate ,(f) = [X(f)I again as above, but this time averaging ten estimates. each obtained from a different time sequence. In other words, you must use the recursion In = 0.95x1-1 + Whin = 1,2,....512. ten times, and average the results, i.e., P.(f) = ,|x,(/)|". The estimates are noisy and random, and to improve and smooth the estimates, we must calculate the quantities several times (let us say 10 times) and average them. iii. We also know the analytical value of 4,() = [H(]I 4,(f). where 4w(f) is the power spectral density of the input white noise. hence, 4.(f) = 1, and |1-ae-121 1.9025-1.9cos(271)" Plot the power spectra in these three cases and comment on each of them.Consider a simple linear regression (SLR) model which satises Gauss Markov assumptions: Y = .30 + 61 X + u, where the error term it has an unconditional variance of oz. Let B; be a linear estimator of 31, i.e. B; can be expressed as 2&1 wii'i, where n is the sample size and the w; coefcients are functions of sample data of the independent variable X. Let n = 3 in this question for simplicity. (a) (6 points) Derive an expression for the conditional expectation of E . What condition(s) must W1, W2, and W3 satisfy for B: to be an unbiased estimator? (see note below) (b) (6 points) Derive an expression for the conditional variance of 3:. (see note below) (c) (2 points) Prove the identity 23'; X} Xk = 3 (X! X), where ican be 1, 2, or 3, andj and I: will take on the other two values. For example, if i = 1, then j and I: take on values of 2 and 3, hence 2X1 X2 X3 = 3 (X1 X). Orifi = 2, thenj and I: take on values of] and 3, hence 2X2 X1 X3 = (d) (12 points) Make use of your answer in parts (a), (b), and (0), nd the formula of wl, W2. and W3 that makes 3: the estimator with the minimum variance among all linear unbiased estimators. Try your best to simplify the expressions for W1, W2, and W3. Hint for part (d): This is essentially an optimization problem under constraint in which you want to minimize the conditional variance under the constraint that the linear k estimator is unbiased. Use of basic calculus is needed (e.g. % =k X \"'1). Although you will encounter some long expressions in the process of solving this problem, you will discover that many terms cancel out and the solution has a rather simple structure. Be patient. D Question 1 2.5 pts Find the mean, sample standard deviation, sample variance and median for these data points. Data 2.746 5,550 2.810 7.604 8.614 9.778 8,090 8,215 3,216 5.310 mean = 6,193.30, standard deviation = 6,177,364.41, variance = 2,485.43, median = 6.577.00 mean = 6.577.00, standard deviation = 2.619.87. variance = 6.863.738.23, median = 6.193.30 mean = 6,577.00, standard deviation = 2,485.43, variance = 6,177.364.41, median = 6,193.30 mean = 6.193.30. standard deviation = 2,485.43, variance = 6,177,364.41, median = 6,577.00 mean = 6,193.30, standard deviation = 6.863,738.23, variance = 2,619.87, median = 6.577.00 O mean = 6,193.30, standard deviation = 2,619.87, variance = 6,863.738.23, median = 6,577.00
