A car has five traffic lights on its route. Independently of other traffic lights, each traffic light turns red as the car approaches the light (and thus forces the car to stop at the light) with a probability of 0.4. a Let K he a random variable that denotes the number of lights at which the car stops with parameters (5,0.4). What is the PMF of K? b. What is the probability that the car stops at exactly two lights? c. What is the probability that the car stops at more than two lights? d.What is the expected value and the variance of K?

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A car has five traffic lights on its route. Independently of other traffic lights, each traffic light turns red as the car approaches the light (and thus forces the car to stop at the light) with a probability of 0.4.   a Let K he a random variable that denotes the number of lights at which the car stops with parameters (5,0.4). What is the PMF of K?   b. What is the probability that the car stops at exactly two lights?   c. What is the probability that the car stops at more than two lights?   d.What is the expected value and the variance of K?

redhotdevil13l3 · Accepted Answer

Step 1 Let, K be the random variable that denotes number of lights at which the car stops Where, K∼Binom(n=5,p=0.4) a. Probability mass function of binomial distribution is as follows, P(K=k)=nCk×pk×(1−p)n−k;k=0,1,…n =0 ; otherwise Where, n is the number of trials K is random variable k is number of success p is the probability of getting a success. Therefore, for the given example pmf is as follows, P(K=k)=5Ck×0.4k×(1−0.4)5−k; k=0,1,2,3,4,5 =0 ; otherwise Step 2 b. The probability that the car stops at exactly two lights is as follows, P(K=2)=5C2×0.42×(1−0.4)5−2 =5!2!(5−2)!×0.16×0.216 =10×0.16×0.216 =0.3456 Therefore, the probability that the car stops at exactly two lights is 0.3456. c. The probability that the car stops at more than two lights is as follows, P(K&amp;gt;2)=1−P(K≤2) =1−P(K=0)+P(K=1)+P(K=2) =1−({5}C0×0.40×(1−0.4)5−0)+({5}C1×0.41×(1−0.4)5−1)+({5}C2×0.42×(1−0.4)5−2) =1−(1×1×0.0776)+(10×0.4×0.1296)+(10×0.16×0.216) =1−0.9416 =0.0584 Therefore, the probability that the car stops at more than two lights is 0.0584. d. Expected value and variance of K: E(K)=n×p =5×0.4 =2 Therefore, the expected value of random variable K is 2. Var(K)=n×p×(1−p) =5×0.4×0.6 =1.2 Therefore, the variance of random variable K is 1.2.

Ana Robertson · Accepted Answer

Step 1  Suppose it is known from large amounts of historical data that X, the number of cars that arrive at a specific intersection during a 20-second time period, is characterized by the following discrete probability function:  сf(x)=e−44xx!, for x=0, 1, 2,сs˙  a) Find the probability that in a specific 20 -second time period, more than 8 cars arrive at the intersection.  b) Find the probability that only 2 cars arrive, 3.36 For a laboratory assignment, if the equipment is working, the density function of the observed outcome, X, is  f(x)={2(1−x),0&amp;lt;x&amp;lt;10,otherwise

A car has five traffic lights on its route. Independently of

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