Question

9. (25 points) George keeps track of how often he makes it through stop lights on his way to work every morning and how many times he has to stop. Assume that whether or not he makes it through one light is independent of whether or not he makes it through the next light (there are some reasons why they might not be independent, but for this problem, you can assume that they are). According to George's records, he makes it through a traffic light about 60% of the time. There are 5 traffic lights between George's house and Gray's Bible College, George wants to know the probability of making it through all 5 lights. a. Explain why this is a binomial distribution problem (include all 4 conditions in your explanation!): D FW in the probability distribution for this problem C. What is the average number of lights that George should make it through on his morning commute?

Answer 1

* Solution :- If we define x as No. of traffic lights that oftes he makes it through stop lights. As there are s traffic lights, the possible Values of x are şo, 1, 2, 3, 4,53 Also, p = prob that George makes it through a traffic light = 0.60 & making through one light is independ est of other. Hence the defined random variable x has Binomial disth with parameters n=5&p= 0.6 In this situation, there are only two outcomes that he will have to stop lights & he will not have to stop ie binomial) ir X ~ B(n=s, p=0.6) :. Its pmf is P(x)= x pngna, a= 0,1,2...5 , q=1-8 P(")= SCR (0.6)*(0.4)5-x, x=0, 1, 2, 3, 4,5 0.w. b) The prob disth of X is as follows X 1 o 1 12 13 14 15 Pa 0.01024 0.0768 0.2304 9-7416 0.2592 0.07776 10.3456 The probabilities are calculated as P(o)= 5 (0.6) (0.4 5 = 0.0 1024 PUI) = Sci (0.6)'(0.4)4 = 0.0768 & so on. c) The avg. no. of lights that George should make it through on his morning commute = E(X) EX= np = 5(0-6)= 3 Arg. no. of lights are 3.