Question

all the codes should be done in matlab

Part A (Based off week 10 workshop content) Recently a study was performed on a Queeislad highway regarding the quan tity and speed of vehicles that use the road. It is your job to determine the probability of certain eveuts with the use of probability distributions For each question, clearly state which probability distribution the random variable follows. You mast solve each problem by band and werify using a MATLAB function 1. During the testing time, it was found that cars pass by at an average rate of 300 per hour, with the time between cars being independent What is the probability of seing less than 5 cars in a mimate? 2. What is the probability of there being more than 1 minute between cars? 3. The speed of cars was found to follow a normal distribution with a mean of 95 km/h and a variance o 5 k. Given the speed limit is 100km/h, what is the probability that a random car is speeding? 4. At any given point, what is the probability that at least 2 out of the next 5 cars will be speeding?

Answer 1

Answer:

The cars pass by at an average rate of 300 per hour. The average rate for 1 minute interval is 300/60 5 per minute 1) Let X be the number of cars that pass in one minute interval. We can say that X has a Poisson distribution with parameter A5. The Poisson probability of X-x cars pass in 1 minute is given by The probability of seeing less than 5 cars in a minute is 50e-5 51e 0! 1! 2! 3! 4! -0.0067 0.03370.0842 0.14040.1755 0.4405 The probability of seeing less than 5 cars in a minute is 0.4405 Matlab code for PX<5) PX< 4) >> poisscdf (4,5) ans 0.4405

2. Since the number of cars arriving in 1 minute interval has a Poisson distribution with parameter λ the inter arrival time has an exponential distribution with parameter λ 5. 5, Let Y be the time between the arrival of cars. Y has an exponential distribution with parameter μ=-=-=0.2 The pdf of Y is 0.2 The cdf of Y is The probability of there being more than 1 minute between the cars is PlY > 1) = 1-P(Y < 1) = 1-(1-e-5x1)-e-5 = 0.0067 ans: The probability of there being more than 1 minute between the cars is Matlab code, for P(Y>1)1- P(Y<1) >> 1-expcdf (1,0.2) ans 0.0067

95 km/h and variance 3) Let X be the speed of a random car. X has a normal distribution with mean μ the probability that a random car is speeding is same as the probability that the speed of a random car is greater than 100 km/h (the speed limit) P(X > 100)-P (X-μ > 100-95) V5 get the z score of 100 - P (Z 2.24) -1- P(Z 2.24) 1 (0.5+0.4875) using standard normal table for z-2.24 0.0125 ans: the probability that a random car is speeding is 0.0125 MAtlab code for PX 100) 1-P(X<100), and directly using the distribution of X (rather than the Z score) >> 1-normcdf (100,95,sqrt (5)) ans- 0.0127

4) From 3) we know that the probability that a random car is speeding is 0.0125 Let Y be the number of cars out of 5, which are speeding. Y has a Binomial distribution with parameters, number of trials (number of cars observed) n 5 and success probability (the probability that a random car is speeding) p 0.0125 The Binomial probability of Yay out of 5 cars are speeding is 0.01251 0.0125)-v 5! g(5-y)! the probability that at least 2 out of next 5 cars will be speeding is P(Y 2) 1- P(Y <2) = 1-(P(Y = 0) + P(Y = 1)) 0)0.01251 -0.0125)s0+ (1)0.0125'(1 - 0.0125)5-1 5! 0.01251(1-0.0125)5-1 001250(1-0.0125)5-0 1(5- 1)! 1-0.9985 0.0015 ans: the probability that at least 2 out of next 5 cars will be speeding is 0.0015 MATLAB code for P(Y>=2) =1-P(Y(21) >>1-binocdf (1,5,0.0125) ans 0.0015