Solution - Assignment#3 - 2023

ECI-114 Probabilistic Systems Analysis for Civil Engineers, Summer Session 1 2023 Department of Civil and Environmental Engineering University of California Davis Instructor: Prakash Singh Badal 3160 Ghausi Hall, psinghbadal@ucdavis.edu Homework due: July 17 th , 2023 8:00 pm Homework # 3: Random Variables and Distribution Functions 1) 12 points, 4 each. During the last couple of months, Qualcomm has been supplying Apple with some components of the new iPhone X. Recently, an employee released a statement saying that the company has been having quality control issues with the products. However, Apple sold most of the iPhone X in advance, and the customers are waiting for their devices. The best assessment indicates that there are 2.4% of iPhones in every lot that does not meet the quality standards. Because of time constraints, Apple cannot reject the lots and decides to conduct an inspection process. They ask a group of students from ECI 114 to help in the sampling process. To start the process, define (for the 3 questions below) the random variable of interest, name its distribution, and give the appropriate pmf (in equation form), as well as answer the questions. (a) In a sample lot of 200 iPhones, what is the probability of getting at least 2 defective phones? Random Variable: The number of defective iPhones out of 200 sampled units Distribution: Binomial pmf:              otherwise 0 200 ,..., 3 , 2 , 1 , 0 024 . 1 024 . 0 ) ( 200 200 x C x p x x x Want P(x ≥ 2) so, calculate 1 – P(x < 2) = 1 – P(0) – P(1) = 1 – 0.007762 – 0.038173 = 0.954065 (b) What is the probability that every single phone from the 200-item sample will have to be examined for 1 defective phone to be found? Random Variable: The number of iPhones sampled (number of trials) until the first defective iPhone is found Distribution: Geometric pmf:              otherwise 0 200 ,..., 3 , 2 , 1 , 0 024 . 0 024 . 0 1 ) ( 1 x x p x x        024 . 0 024 . 0 1 ) ( 199 x p 0.000191 (c) What is the expected number of phones that will have to be examined before a defective one is found? E[X] for a Geometrically Distributed RV is: 1/p = 1/0.024 = 41.66667 2) (12 points, 4 each) A testing machine is used to generate pressure to test construction pieces. The pressure generated by the machine is uniformly distributed with an average of 5 MPa and a standard deviation of 250000Pa (1 MPa = 1000000 Pa). (a) What is the range of the pressures generated by the machine? (Hint, the uniform distribution is symmetric about its mean, which should help you calculate the range) Solve using the values for standard deviation and the average (expected value):

Var[X] = (250000/1000000) 2 = (0.25) 2 = 0.0625 = (b – a) 2 /12 (b – a) = (0.0625*12) 1/2 = 0.866025, solve for b in terms of a: b = 0.866025 + a Substitute our equation for b into the equation for the expected value: (b + a)/2 = (2a + 0.866025)/2 = 5 a = 4.566987 b = 0. 866 + a = 5.433013 The interval = [4.567, 5.433] (b) What is the probability the machine will generate between 4.6 and 5.3 MPa? Remember, the pdf of x, f(x) = 1/(b-a) Integrate f(x) between 4.6 and 5.3, and you get the probability, Pr( 4.6<X<5.3) = 0.8083 (c) A certain test is invalidated if more than 5.1 MPa are generated. What is the probability of performing an invalid test using this machine? Similar to the previous problem, you need to find, Pr ( X > 5.1), that means integrating from 5.1 to 5.4333 = 0.3845 3) (15 points, 3 each) Assuming that X~N(12, 3) ( “X is a Normally distributed random variable with a mean μ and standard deviation σ.” ) Determine: μ = 12; σ = √3 (a) Pr[X ≤ 8] = P [Z ≤ (8 – 12)/ √3 ] = P[Z ≤ -4/ √3 ] = 0.01046 (b) Pr[X > 20] = P [Z > (20 – 12)/ √3 ] = P[Z > 8/ √3 ] = 1 – P[Z < 8/ √3 ] ≈ 1 – 1 = 0.000 (c) Pr[ 3 ≤ X ≤ 1 2] = P[(3 – 12)/ √3 ≤ Z ≤ (12 – 12)/ √3 ] = P[Z ≤ 0] – P[Z ≤ -5.19615] ≈ 0.5 – 0.00 = 0.500 (d) Calculate the value of x, given, P[X<x] = 0.33, z = -0.44, z = (x – μ)/σ = ( x – 12)/ √3 x = (-.44*( √ 3))+12= 11.2379 (e) Calculate the value of x, given, P[ x < X < 9] = 0.1 P[X < 9] – (P[X < x]) = 0.1 = P[Z <(9-12)/ √3 ] – (P[X < x]) = 0.0416 – (P[X < x]) = 0.1 -0.058655 = Pr[Z < z] Probability cannot be negative. No such x exists. 4) 8 points . California, on average, deals with 5 fires each year. (a) What is the probability that fewer than three fires will happen in the state in a given year? Use the Poisson distribution. X = the number of fires in a time span, T. T = 1 year, λ = 5/ year. P(X < 3) = P(0) + P(1) + P(2) = = 0.124625 5) 8 points, 2 each . For a continuous random variable Y, the PDF has the form:

Pr(-3 <= X <= 3) = 0.5 + 0.01 + 0.04 + 0.03 + 0.23 + 0.1 + 0.09 = 1 Using the values defining p(x) in the table above, determine the following probabilities (b) Pr(X = 0 or X = 2) = 0.03 + 0.1 = 0.13 (c) Pr(X ≤ 2.9 ) = 1 – 0.09 = 0.91 (d) Pr(X ≥ 0 or X= 1 ) = 0.03 + 0.23 + 0.1 + 0.9 = 0.45 (e) Pr(-1 < X < 0) = 0 7) 15 points, 5 each . You walk to campus every morning. Your route passes through the W Cowell Blvd. and Anderson intersection. Sometimes you arrive at the intersection and have to wait for the “pedestrian walk” signal, others you encounter a “no - walk” signal. Given your interest in statistics and probability, you recorded the status of the walk signal (pass/no pass) when you arrive at the intersection. After many and many days of walking up to the intersection, you realize that the walk signal will be in the no-walk mode 80% of the time. Assume every walk is an independent trial and we are starting from tomorrow. Define the random variable, the distribution, and write the equation of the pmf for the following questions. (a) What is the probability that the first morning you arrive to a walk signal is the 5 th morning you approaches it? (0.80)(0.80)(0.80)(0.80)(0.20) = 0.08192 Random Variable: The number of walks (if you said “mornings” that is OK too) before you arrive to your first walk signal Distribution: Geometric pmf: ?(?) = { 0.8 𝑥−1 · 0.2 ? = 1, 2, 3,4 … 0 ??ℎ???𝑖?? (b) What is the probability that you have to wait at the intersection 10 days in a row? You can solve the problem under 2 considerations: Alt. 1 . You are assuming that you wait 10 days in a row without the expectation that you will arrive at a walk/pass signal on the 11 th morning. This would be a binomial distribution. Therefore, you are finding the probability of (success) waiting 10 days out of 10 days. ?(?) = 𝐶 10 10 0.8 10 · 0.2 0 = 0.107374 Alt. 2 . You expect to wait 10 mornings but get the pass signal on the 11 th morning. This would be a geometric distribution. ?(?) = 0.8 10 · 0.2 = 0.0214 (c) What is the expected value and standard deviation for the random variable defined in part (b)? For Alt. 2. E(X) = 1/p = 1/0.2 = 5 V(X) = : q/p 2 = 0.80/(0.2 2 ) = 20 Standard deviation = (20) 1/2 = 4.472 8) (20 points, 5 each) I drive to school every morning and take the same route. I have found that I get a red light at the corner of Russell and La Rue 40% of the time and get a green light the rest of the time. Assume that each of my commutes is independent. For problems (a), (b), and (c) define (i) the random variable of interest (1 point) and (ii) name its distribution (1 point) before answering the question posed (3 points).

(a) What is the probability that I have to wait 4 morning commutes until the next red light (red light happening on the 4th morning)? (i) Random Variable: The number of morning commutes until the next red light. (ii) Distribution: Geometric Distribution p(4) = (1-0.4) 3 (0.4) = 0.0864 (b) How many commutes can I expect to wait to encounter my 3 rd red light? (i) Random Variable: The number of morning commutes until the 3 rd red light. (ii) Distribution: Negative Binomial Distribution Use the expected value equation: r/p, where r = 3 and p = 0.4 3/0.4 = 7.5 (c) In a week I drive to work 5 times. What is the likelihood that I encounter more than 2 red lights during the morning commute in 2 weeks? (i) Random Variable: The number of red lights in 10 commutes (ii) Distribution: Binomial We want the probability that x (the number of successes) is greater than 2: P(x>2) = 1 – (p(0) + p(1) + p(2)) p(0) = (1)(1)(0.6 10 ) = 0.00605 p(1) = 10(0.4)(0.6 9 ) = 0.0403 p(2) = (45)(0.4 2 )(0.6 8 ) = 0.121 P(x>2) = 1 – 0.0065 – 0.0403 – 0.121 = 0.8322 (d) What is the probability that I get a green light the next morning I drive? q = 1 – p = 0.6