Mastering the Poisson Process: Key Concepts and Solutions

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Clemson University**We aren't endorsed by this school

Course

MATH 3020

Subject

Statistics

Date

Dec 12, 2024

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Uploaded by BailiffRoseOpossum17

Activity - Lesson 10 - The Poisson Process - Sample Solutions Note: You must attend class and submit your work before leaving to earn credit for this learning activity. This activity covers Lesson 10 and uses the quiz utility in canvas to record your participation. « Your final grade on this activity will be based solely upon participation and attendance. « The correct answers for any auto-graded questions will be shown immediately upon completing the activity. « Sample solutions are provided in the course schedule on our Canvas home page. Use these to learn the material and skills covered and to help you prepare for exams. + PLEASE ask questions and/or meet with the instructor if there are any concepts you still do not understand! Learning Objectives ¢ CO-2: Apply basic concepts of probability, random variables, and commonly used statistical probability distributions, including sampling distributions and the central limit theorem. o LO-2.2: Define the term random variable and distinguish between discrete and continuous random variables. o LO-2.3: Calculate probabilities. o LO-2.4: Use mathematical functions and graphical displays to represent probability models. o LO-2.5: Find the mean and variance for discrete and continuous random variables. o LO-2.6: Find percentiles for random variables. o LO-2.7: Create and interpret QQ-plots to compare a set of data to a specified distribution. o LO-2.8: Estimate parameters for common continuous distributions. BEST PRACTICES FOR CALCULATIONS AND ROUNDING: ¢ |tis best to use R as your calculator and use the round command to determine the final answer! ¢ Do not round in any intermediate steps. ¢ In R, you must always explicitly specify multiplication using * o Example: You cannot type (3+2.4)(8+1.2), you must enter (3+2.4)*(8+1.2) ¢ You can get a different value than R if you round by hand whenever the digit immediately to the right of your desired decimal place is a 5... R uses an International Rounding Standard which does not follow the standard by-hand approach. o Here is the information from R's help page on the round command: "Note that for rounding off a 5, the IEC 60559 standard (see also ‘IEEE 754’) is expected to be used" o For Exams, if you ever believe this issue caused you to miss a question, please put in a request for leniency which details your calculation. We will gladly check if this was the case and, if so, return your points! Question 1 Definitions and Concepts Choose the correct answer to complete each of the statements below. The concepts in this question are covered in Lesson 10. e Answer numbers are given in parentheses (#) before each required answer to help with reviewing your submission. » The specific sections are indicated in brackets {} after each statement to help you find the relevant information in the lesson. Each answer is worth 3 points. o For a given Poisson process: {Introduction to the Poisson Process} o The number of events over an interval is a (1) [DISCRETE] random variable that follows a/an (2) [POISSON] distribution.

o The length of the interval between events is a (3) [CONTINUOUS] random variable that follows a/an (4) [EXPONENTIAL] distribution. o APoisson process is a model for a series of discrete events where: {Poisson Process — Simplistic Definition and Criteria} o the average time between events is (5) [KNOWN] but the exact timing of the events is (6) [RANDOM]. o The arrival of an event is (7) [INDEPENDENT] of the event before. + A Poisson process satisfies the following criteria: {Poisson Process — Simplistic Definition and Criteria} o Two events (8) [CANNOT] occur at the same point. o The average rate (events per interval) is (9) [CONSTANT]. o The events are (10) [INDEPENDENT] of each other. « The exponential distribution with parameter X is the same as: {Exponential Distribution} o The Weibull distribution with 8 = (11) [1, ONE] and § = 1//\ o The gamma distribution with 7 = (12) [1, ONE] and A\ = A\ Question 2 Note: There is a difference between a Poisson PROCESS and a Poisson DISTRIBUTION. Suppose the random variable X follows a Poisson distribution with \ = 1.6. Find P(X = 2). Round your answer to 3 decimal places. {Review The Poisson Distribution including Definition: Poisson Distribution and Poisson Examples} Hint: You should be using the dpois command in R (or the formula for the PMF of a Poisson distribution). Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details This question can be solved manually using the PMF of a Poisson distribution: e A)\® x! [ f(m) = Substituting the values of and \ we get: e 0. 1.6° Sample Code #Question 2 round(dpois(2, 1.6),3) round(exp(-1.6)*1.6"2/factorial(2),3) Sample Output > #Question 2 > round(dpois(2, 1.6),3) [1] e.258

> round(exp(-1.6)*1.6~2/factorial(2),3) [1] e.258 Question 3 Note: There is a difference between a Poisson PROCESS and a Poisson DISTRIBUTION. Suppose the random variable X follows a Poisson distribution with A = 3.72. Find P(X > 3). Round your answer to 3 decimal places. {Review The Poisson Distribution including Definition: Poisson Distribution and Poisson Examples} Hint: You should be using the sum and dpois commands in R (or the formula for the PMF of a Poisson distribution) and the complement rule. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details This question can be solved manually using the PMF of a Poisson distribution: e A )\® z! . flz) = Applying the complement rule and substituting the appropriate values of and A\ we get: ~3.72 , g o0 -3.72 , 3 791 —8.72 | 3 792 —3.72 . 3 798 «PX>3)=1- |5 = + £ = + £ : 4 £ = ~ 0.51 Sample Code #Question 3 round(1-sum(dpois(@:3, 3.72)),3) round(1-sum(exp(-3.72)*(3.72)~(0:3)/factorial(0:3)),3) Sample Output > #Question 3 > round(1-sum(dpois(@:3, 3.72)),3) [1] e.51 > round(1-sum(exp(-3.72)*(3.72)"(0:3)/factorial(@:3)),3) [1] e.51 Question 4 Suppose the random variable X follows an exponential distribution with parameter A = 1.97. Find P(X > 1). Round your answer to 3 decimal places. {Review The Exponential Distribution including Definition: Exponential Distribution and Exponential Examples} Hint: You should be using either the pexp command in R or the formula for the CDF of an exponential distribution.

Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details This question can be solved manually using the CDF of an exponential distribution: e Fiz)=1—-¢e* Applying the complement rule and substituting the values of = and A we get: e PX>1)=1-P(X<1)=1-F(1)=1—(1-¢e19"1) x~0.139 Sample Code #Question 4 round(pexp(1, 1.97, lower.tail=F),3) round(1-pexp(1, 1.97),3) round(1-(1-exp(-1.97*1)),3) Sample Output > #Question 4 > round(pexp(1, 1.97, lower.tail=F),3) [1] ©.139 > round(1-pexp(1, 1.97),3) [1] @.139 > round(1-(1-exp(-1.97%1)),3) [1] e.139 Question 5 Suppose the random variable X follows an exponential distribution with parameter A = 31.9. Find Q3 (the third quartile) for this random variable. Round your answer to 3 decimal places. {Review The Exponential Distribution including Definition: Exponential Distribution and Exponential Examples} Hint: You should be using either the gexp command in R or solving for & by setting the CDF of an exponential distribution equal to the appropriate value. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details This question can be solved manually using the CDF of an exponential distribution: ¢« Flz)=1—¢e™ « Setting the CDF equal to 0.75, substituting the value of )\, and solving for z we get: —1n(0.25) ¢ 0.75=1—e3% =025 =39 = —31.92 =In(0.25) => = = 319 =~ 0.043 Sample Code #Question 5 round(qexp(0.75, 31.9),3)

round(qexp(@.25, 31.9, lower.tail=F),3) round(-log(@.25)/31.9,3) Sample Output > #Question 5 > round(qexp(@.75, 31.9),3) [1] @.843 > round(qexp(@.25, 31.9, lower.tail=F),3) [1] ©.043 > round(-log(@.25)/31.9,3) [1] ©.043 Question 6 Suppose the random variable X follows an exponential distribution with parameter A = 3.6. The mean and standard deviation of X are both equal to what value? Round your answer to 3 decimal places. {Review The Exponential Distribution including Definition: Exponential Distribution and Exponential Examples} Hint: You should be using the formula for the mean of an exponential random variable (and realizing that the standard deviation is the same value). Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details Cp=B(X) =5 Sample Code #Question 6 round(1/3.6,3) Sample Output > #Question 6 > round(1/3.6,3) [1] ©.278 Question 7 Note: There is a difference between a Poisson PROCESS and a Poisson DISTRIBUTION. e There are TWO possible random variables in any Poisson Process. o For this question, you need to correctly identify which distribution applies. Consider a Poisson process with an average rate of 8 events per hour.

What is the standard deviation of X = the number of events per hour? Round your answer to 3 decimal places. {Review Lesson 10 carefully paying particular attention to the section Introduction to the Poisson Process} Hint: First consider which distribution is involved. What distribution does the number of events per hour have? Determine whether this random variable is a Poisson or exponential. Then use the appropriate equation to calculate the standard deviation. Remember that Poisson random variables are discrete and exponential random variables are continuo Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details e O — \/X Sample Code #Question 7 round(sqrt(8),3) Sample Output > #Question 7 > round(sqrt(8),3) [1] 2.828 Question 8 Suppose X follows a Poisson distribution. If the P(X = 0) = 0.329. Find A = the mean of X. Round your answer to 3 decimal places. Hint: Set up an appropriate equation using the Poisson PMF and solve for A. (R is not helpful for this problem, except as a calculator.) Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. {Review Definition: Poisson Distribution} Mathematical Details e A \? fla) = » We have z = 0 and f(0) = 0.329. Substituting and solving for z, we find: e 2 \0 Y + 0320 = =—— = ¢ = —A = In(0.329) > A = — In(0.329) ~ 1.112 Sample Code #Question 8 round(-log(©.329), 3)

Sample Output > #Question 8 > round(-log(@.329), 3) [1] 1.112 Question 9 Flaws occur in a certain Mylar material according to a Poisson process with an average rate of 0.049 flaw per square yard. If 32 square yards are inspected, what is the probability that there are no flaws? Round your answer to 3 decimal places. Remember: You must carefully consider the proper unit conversion. The rate is given per square yard but the question is asked about a large piece of material. You must convert the average rate per square yard to the average rate need for the size of material being inspected. Hint: First determine the appropriate distribution (Poisson, or exponential) and then use the appropriate R command to find this probability. Remember there is a difference between a Poisson Process and a Poisson distribution. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details This question can be solved manually using the PMF of a Poisson distribution: e AN® z! [ f(w) = « Substituting the values of £ and A we get: e—0.049-32 . (_0049 . 32)0 o ~ 0.208 - f(0) = Sample Code #Question 9 gq9.rate <- 0.049%32 round(dpois(®, g9.rate),3) round(sum(exp(-q9.rate)*(q9.rate)”0/factorial(e)),3) Sample Output > #Question 9 > gq9.rate <- 9.049%*32 > round(dpois(@, q9.rate),3) [1] ©.208 > round(sum(exp(-q9.rate)*(q9.rate)”0/factorial(e)),3) [1] ©.208 Question 10 The time to failure of a certain type of electrical component is assumed to follow a Poisson process with an average rate of 1 failure every 6.4 years. What is the probability that a component will fail in the first year?

Round your answer to 3 decimal places. Remember: You must carefully consider the proper unit conversion. The question is asked in terms of years so your rate must be converted to "per year". Hint: First determine the appropriate distribution (Poisson, or exponential) and then use the appropriate R command to find this probability. Remember there is a difference between a Poisson Process and a Poisson distribution. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details This question can be solved manually using the CDF of an exponential distribution: e Flz)=1—e e Substituting the values of and A\ we get: e PX<1)=F(1)=1—-e /690 ~ 0.145 Sample Code #Question 10 gle.rate <- 1/6.4 round(pexp(1,q10.rate),3) round(1- exp(-ql@.rate*1),3) Sample Output > #Question 10 > gqlo@.rate <- 1/6.4 > round(pexp(1,ql@.rate),3) [1] ©.145 > round(1- exp(-ql@.rate*1),3) [1] e.145 Question 11 Maijor cracks in a highway are modeled by a Poisson process with an average rate of 0.231 cracks per mile. What is the probability that the first major crack occurs between mile 12 and mile 15 after the start of inspection? Round your answer to 3 decimal place. Hint: First determine the appropriate distribution (Poisson, or exponential) and then use the appropriate R command to find this probability. Remember there is a difference between a Poisson Process and a Poisson distribution. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details This question can be solved manually using the CDF of an exponential distribution: e« Flz)=1—-¢e* « Substituting the values of £ and A we get: « P(12 < X <15) = F(15) — F(12) = (1 — e~ (©281)(15)) _ (1 — ¢~ (0-231)(12)) ~ 0.031

Sample Code #Question 11 round(pexp(15, ©.231) - pexp(12, ©.231),3) round((1- exp(-0.231*15))-(1- exp(-0.231%¥12)),3) Sample Output > #Question 11 > round(pexp(15, ©.231) - pexp(12, 0.231),3) [1] e.@31 > round((1- exp(-6.231*15))-(1- exp(-0.231*12)),3) [1] e.831 Question 12 Suppose the counts recorded by a Geiger counter follow a Poisson process with an average rate of 1.00 counts per minute. Let X represent the time (in minutes) between successive counts. Find the median value of X. Round your answer to 3 decimal places. Hints: o The median represents what percentile? « First determine the appropriate distribution (Poisson, or exponential) and then use the appropriate R command to find this value. Remember there is a difference between a Poisson Process and a Poisson distribution. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details This question can be solved manually using the CDF of an exponential distribution: e« Flz)=1—-e™ « Setting the CDF equal to 0.5, substituting the value of A, and solving for we get: c05=1-e®*=05=¢%=2z=—In(0.5) =~ 0.693 Sample Code #Question 12 round(qexp(@.5, 1),3) round(-log(@.5),3) Sample Output > #Question 12 > round(qexp(@.5, 1),3) [1] ©.693 > round(-log(@.5),3) [1] ©.693

Question 13 Consider a Poisson process with an average rate of 0.169 events per hour. What is the mean of X = the time (in hours) between successive events? Round your answer to 2 decimal places. Hint: First determine the appropriate distribution (Poisson, or exponential) and then use the appropriate formula to find this value. Remember there is a difference between a Poisson Process and a Poisson distribution. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details . N:E(X)=% Sample Code #Question 13 round(1/0.169,2) Sample Output > #Question 13 > round(1/0.169,2) [1] 5.92 Question 14 Consider a Poisson process with an average rate of 0.532 events per hour. What is the variance of X = the time (in hours) between successive events? Round your answer to 2 decimal places. Hint: First determine the appropriate distribution (Poisson, or exponential) and then use the appropriate formula to find this value. Remember there is a difference between a Poisson Process and a Poisson distribution. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details 1 | ] JZZV(X):F Sample Code #Question 14 round(1/0.532/2,2)

Sample Output > #Question 14 > round(1/0.532/2,2) [1] 3.53 Question 15 Earthquake tremors in a certain area follow a Poisson process with an average rate of 7.22 per 12-month period. Assume the number of tremors from one 12-month period is independent of the number in the next 12-month period. Find the standard deviation of the X = number of tremors which occur during any 15-year period. Round your answer to 2 decimal places. Remember: You must carefully consider the proper unit conversion. You are given the average rate per year (12 months) but the question is asked about a multi-year period. You need to convert the yearly rate into a multi-year rate. Hint: First determine the appropriate distribution (Poisson, or exponential) and then use the appropriate formula to find this value. Remember there is a difference between a Poisson Process and a Poisson distribution. Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Mathematical Details « o=+ Sample Code #Question 15 gql5.rate = 15%7.22 round(sqrt(ql5.rate),2) Sample Output > #Question 15 > q15.rate = 15%7.22 > round(sqrt(qls.rate),2) [1] 10.41 Question 16 This is a difficult question! Pay close attention to the hints below which walk you through the correct solution! Minor flaws occur in a certain indoor-outdoor carpet according to a Poisson process with an average rate of 0.36 minor flaws per square yards. A large hardware store sells this carpet in rolls of 6ft by 9ft (2 yds by 3 yds). If there are 19 rolls in stock, what is the probability at least one of the rolls has no minor flaws? Round your answer to 3 decimal places. Hint:

1. Find the probability that a particular roll of carpet will have no flaws. This follows a Poisson distribution but you must convert the rate to the appropriate sized roll. 2. Find the probability that at least one of the rolls has no flaws using an appropriate binomial distribution. 3. AVOID rounding the first probability when finding the second. (Either copy the exact value from R's output or use the appropriate dpois command directly INSIDE the appropriate dbinom command). Note: | have allowed a small margin of error on this question but if you do not get the EXACT answer, please consider carefully whether you rounded correctly. Review the examples in Lesson 9 involving binomial probabilities and in Lesson 10 involving Poisson probabilities. Mathematical Details Using the complement rule and the PMF for binomial random variables along with the PMF of a Poisson random variable, we can solve this question manually. The PMF for a Poisson is: —AZ2a e A ¢ fl@) = — The PMF for a binomial is: « f(z) = (Z) p* - (L=p)** First we must convert the given rate per square yard to a 6 square yard carpet. « We get 6(0.36) = 2.16. Then, using that rate, we must find the probability that there are NO flaws in a particular carpet using the Poisson distribution. —2162-160 . We get £(0) = GT — 0.1153251 To reasonably find P(X > 1) for the associated binomial distribution, we must use the complement rule. e WegetP(X>1)=1-P(X<1)=1-P(X =0) The probability that any individual carpet has no flaws was found to be 0.1153251 and we have 19 carpets. Thus we have: 1 e P(X>1)=1-P(X=0)=1- [( 09) -(0.1153251)° - (1 — 0.1153251)°~% | ~ 0.903 Sample Code #Question 16 gql6.rate <- 6*0.36 prob.® <- dpois(@, ql6.rate) prob.1 <- round(sum(dbinom(1:19, 19, prob.0)),3) prob.1 #Manually prob.@a <- exp(-2.16)*2.16"0/factorial(e) prob.@a prob.la <- round(1-(choose(19,0)*prob.0a”e*(1-prob.08a)»(19-8)), 3) prob.1la Sample Output > #Question 16 > ql6.rate <- 6*0.36 > prob.@® <- dpois(@, qlé.rate) > prob.1 <- round(sum(dbinom(1:19, 19, prob.e)),3) > prob.1 [1] @.903 > #Manually > prob.@a <- exp(-2.16)*2.1679/factorial(Q)