CS 3341 (Dr. Baron)

Practice Final Exam. Solutions.

1. Consider an M/M/1 queuing system with the average interarrival time of 5 minutes and the average service time of 4 minutes. Compute

(a) the fraction of time when the server is idle;
(b) the fraction of time when there are at least 4 jobs in the system;
(c) expectation of the number of jobs in the system at any time.

Solution.

(a) r=4/5=0.8, hence, P(X=0)=1-r=0.2

(b) $P(X\ge 4) = r^4 = 0.4096$

(c) E(X)=r/(1-r) = 4

2. Lunar eclipse is a rare event that occurs at the average rate of 3 eclipses in 5 months according to a Poisson process.

(a) Compute the probability of exactly 4 lunar eclipses during the year 2002.
(b) Compute the mean and variance for the time until the 50-th lunar eclipse. Compute (or use a suitable approximation) the probability of at least 50 lunar eclipses during a six-year period. Solution.

(a) The number of lunar eclipses in 2002 has Poisson distribution with the parameter (12)(3/5)=7.2 (don't round it to 7).
P(X=4)=e^-7.2(7.2)⁴/4!=0.0836. or, from a Table, P(4)=F(4)-F(3)=0.084.

(b) The time T until the 50-th lunar eclipse has Gamma distribution with parameters 50 and 3/5=0.6;
E(T)=50/(0.6)=83.33 months;
Var(T)=50/(0.6)²=138.89, the standard deviation is 11.79.
T is a sum of 50 independent Exponential times, and according to the Central Limit Theorem, it is approximately Normal. Therefore,
P(T < 6 years) = P(T < 72 months) = P(Z < (72-83.33)/11.79) = P(Z < -0.96) = 1-0.8315 = 0.1685.

3. Takeoffs and landings at a certain airport are restricted during severe weather conditions. No takeoffs are allowed when lightning is observed around the airport area. Updates are issued every minute. It has been noticed that the probability of lightning is 30% if lightning was observed during the preceding minute. However, this probability is only 10% if there was no lightning a minute ago.

(a) Takeoffs are not allowed at 10 am. Compute the probability that takeoffs will be allowed at 10:03.
(b) Meteorologists predict a 40% chance of lighting at 10 am. Compute the probability that takeoffs will be allowed at 10:03.
(c) Compute the probability that takeoffs will be allowed at 19:33.

Solution.

The transition probability matrix of this Markov chain is $P=\left(\begin{array}{cc} .3 & .7 \\ .1 & .9 \end{array}\right)$ , so that $P^3=\left(\begin{array}{cc} .132 & .868 \\ .124 & .876 \end{array}\right)$ .

(a) P⁽³⁾(lightning -> No lightning )=0.868

(b) If the initial distribution is (.4 .6), then the distribution of (X₃) is $(.4\ .6)P^3=(.1272\ .8728)$ . Answer: 0.8728.

(c) Solve the system:
0.3 p₁ + 0.1 p₂ = p₁
0.7 p₁ + 0.9 p₂ = p₂
p₁ + p₂ = 1

From the first equation, p₂=7p₁.
The second equation is equivalent to this.
Substituting into the third equation, we get 8p₁=1, so that p₁=1/8 and p₂=7/8.

At 19:33, after a large number of transitions, P(lightning)=1/8 and P(no lightning)=7/8.

4. An electronic parts factory produces resistors. Statistical analysis of the output suggests that the resistances follow an approximately normal distribution with the standard deviation of 0.156 ohms. A sample of 60 resistors has the average resistance of 0.55 ohms.

(a) Based on these data, construct a 95% confidence interval for the population mean resistance.
(b) If the actual population mean resistance is exactly 0.5 ohms, what is the probability that an average of 60 resistances is 0.55 ohms or higher?

Solution

(a) n=60, sample mean = 0.55, population standard deviation = 0.156.
From the Normal distribution table, z_a/2=z_0.025=1.96
Then the 95% confidence interval is 0.55 +- (1.96)(0.156)/(root of 60)
Answer: [0.51; 0.59]
_ (b) P(X > 0.55) = P(Z > (0.55-0.5)/(0.156/root of 60)) = P(Z > 2.48) = 1-0.9934 = 0.0066

5. Based on the data in # 4, is there a significant evidence at a 2% level of significance that the population mean resistance is

(a) greater than 0.5 ohms?
(b) not equal to 0.5 ohms? (a) Test H₀: mean = 0.5 vs H_A: mean > 0.5.
Find z_a=z_0.02=2.05. The acceptance region is Z < 2.05.
The test statistic equals
Z = (0.55 - 0.5) / (0.156 / root of 60) = 2.48
This Z belongs to the rejection region, therefore, reject H₀ in favour of H_A. The data provided significant evidence that the population mean resistance is greater than 0.5 ohms.

(b) Two-sided alternative, that the mean does not equal 0.5 ohms.
Find z_a/2=z_0.01=2.326. The acceptance region is |Z| < 2.326.
The test statistic is the same, Z=2.48.
This Z belongs to the rejection region, therefore, reject H₀ in favour of H_A. The data provided significant evidence that the population mean resistance is not equal 0.5 ohms.

6. Consider an electronic message center that can transfer only one message at a time. Also, it has sufficient memory for only 1 other message. If a message arrives while the memory is full, this message will be returned. Messages arrive according to a Bernoulli counting process with 2-second frames and the average arrival rate of 3 messages per minute. The average time it takes to transfer one message is 10 seconds.

(a) Find the steady-state distribution of the number of messages in the message center.
(b) Compute the average number of messages in the center at any time.

Solution.

(a) Compute p_a=0.1 and p_s=0.2. Then, the transirion probability matrix of this single-server Bernoulli queuing process is

$P=\left(\begin{array}{ccc} .9 & .1 & 0 \\ .18 & .74 & .08 \\ 0 & .18 & .82 \end{array}\right)$ .

The steady-state distribution is (81/146, 45/146, 20,146), or (.555, .308, .137).

(b) $E(X)=(0)\pi_0+(1)\pi_1+(2)\pi_2=85/146$ , or 0.582.

7. The Southeastnorthwest Telephone Company models long-distance calls of its customers by a Bernoulli counting process. On the average, customers place calls every 2 seconds. The probability of a call during any given frame is p.

(a) What frame length provides p=0.05?
(b) Using this frame length, compute the probability of more than 1,700 calls between 6 pm and 7 pm. Use the Normal approximation of the Binomial distribution.

Solution.

(a) $\Delta=p/\lambda=0.1$ sec

(b) X has binomial distribution with n=36,000 frames during 1 hour and p=0.05.

P(X>1700)=P(Z>-2.42) = 0.9922.

For the optimal allocation of disk space to hold a large number of images, five digital images are randomly selected for statistical analysis, with the following results:

Image	Size (Mb)
1	120
2	100
3	100
4	200
5	80

(a) Assuming normal distribution of sizes, construct a 90% confidence interval for the mean image size.
(b) Is there significant evidence, at a 10% level of significance, that the mean size is greater than 100 Megabytes?

Solution.

(a) From the given data, the sizes have a sample mean of 120 and a standard deviation of 46.9. This standard deviation is estimated from data, therefore we use t-distribution with n-1=4 degrees of freedom. From the Table, find t_0.1/2=t_0.05=2.132.
A 90% confidence interval for the mean size is
120 +- (46.9)(2.132)/2.24 = 120 +- 44.6 = [75.4 ; 164.6]

(b) Test H₀: mean size = 100 vs H_A: mean size > 100.
This is a one-sided, right-tail test. Find t_0.10=1.533. The acceptance region is {t < 1.533}.
The test statistic equals
t = (120 - 100) / (46.9 / 2.24) = 0.955
This t belongs to the acceptance region, therefore, do NOT reject H₀ in favor of H_A. The data did not provide significant evidence that the package mean size is greater than 100 Megabytes.

The following is a random sample from Uniform(a,b) distribution 6, 4, 5, 2, 4, 0, 5, 3, 6, 5
(a) Estimate a and b by the method of moments.
(b) Estimate a and b by the method of maximum likelihood.

Solution.

Equate the sample mean 4 and the population mean (a+b)/2, i.e., a+b=8.
Subtract the sample mean 4 from each X_i: 2, 0, 1, -2, 0, -4, 1, -1, 2, 1, square: 4, 0, 1, 4, 0, 16, 1, 1, 4, 1, add: 32, and compute the sample variance S²=32/(10-1)=3.56. Equate the sample variance and the population variance 3.56 = (b-a)²/12, i.e., b-a = 6.53.

We have a+b=8 and b-a=6.53. Adding these equations, we get b=7.26. Subtracting them, we get a=0.73 (which is not a possible value of a because we observed X=0. So, typically, a=0 is taken as an estimate).

(b) The joint density of X_i is f(X | a,b) = (1/(b-a))¹⁰ where all X_i belong to [a,b]. This means a <= 0 < 6 <= b.
This density increases in a and decreases in b. Therefore, the largest possible value of a (=0) and the smallest possible value of b (=6) are maximum likelihood estimates.

10.

A customer service representative answers telephone calls from customers in the order they are received. On the average, customers call every 15 minutes. After the call is answered, it takes an average of 10 minutes to complete it. The system does not allow customers to be "on hold", thus the system's capacity is limited by 1 job. If this system is modeled by a Bernoulli single-server queuing process with limited capacity and (1/2)-minute frames,

(a): Find the steady-state distribution of the number customers is the system.
(b): Compute the expected number of customers in the system at any time.

Solution. We have $\lambda_a=1/15$ , $\lambda_s=1/10$ and $\Delta=1/2$ . Therefore, $p_a=\lambda_a\Delta=1/30$ and $p_s=\lambda_s\Delta=1/20$ .

(a) Possible states of the system are X=0,1. Compute the transition probability matrix,

$\begin{displaymath}P=\left(\begin{array}{cc} 1-p_a & p_a \\ p_s(1-p_a) & 1-p_s(1... ...n{array}{cc} 29/30 & 1/30 \\ 29/600&571/600\end{array}\right). \end{displaymath}$

Solve $\pi P=\pi$ along with $\pi_0+\pi_1=1$ .

$\pi_0=(29/30)\pi_0 + (29/600)\pi_1$ leads to $(1/30)\pi_0=(29/600)\pi_1$ , or $\pi_0 = (29/20)\pi_1$ .

$\pi_1=(1/30)\pi_0+(571/600)\pi_1$ leads to $(1/30)\pi_0=(29/600)\pi_1$ , same thing (just checking!)

Now, $\pi_0+\pi_1=(29/20)\pi_1+\pi_1=(49/20)\pi_1=1$ , therefore, $\pi_1=20/49$ and $\pi_0=(29/20)\pi_1=29/49$ .

Answer: $\fbox{ $\pi_0=29/49=0.5918$\space and $\pi_1=20/49=0.4082$\space }$ .

(b) $E(X)=0\pi_0+1\pi_1=\pi_1=\fbox{ 20/49=0.4082 }$ .

11.

We have to accept or reject a large shipment of items. For quality control purposes, we collect a sample of 200 items and find 24 defective items in it.

(a): Construct a 96% confidence interval for the proportion of defective items in the whole shipment.
(b): The manufacturer claims that at most one in 10 items in the shipment is defective. At the 4% level of significance, do we have sufficient evidence to disprove this claim? Do we have it at the 15% level? How about the 20% level?

Solution.

(a) Find the sample proportion 24/200=0.12. Then for alpha=1-0.96=0.04, find z_0.02=2.054
The confidence interval is 0.12 +- (2.054)(Root of (0.12)(1-0.12)/200) = 0.12 +- 0.047 = [0.073, 0.167]

(b) Test H₀ : p = 0.1 vs H_A : p>0.1. Disproving the manufacturer's claim means rejecting H₀ in favor of this H_A. The observed test statistic is Z = (0.12-0.1)(Root of (0.12)(1-0.12)/200) = 0.8704. The critical values, from the table of Normal distribution, are z_0.04=1.75, z_0.15=1.04, z_0.20=0.84. Therefore, we DO NOT have a significance evidence at the 4% and 15% levels to disprove the manufacturer's claim. However, since Z > 0.84 belongs to the rejection region for the 20% level, we DO have a significance evidence against the manufacturer's claim at the 20% level of significance.

Michael Baron