2, 40 pts Suppose that the percentage of American drivers who are multitaskers (

Question

2, 40 pts Suppose that the percentage of American drivers who are multitaskers (eg talk on cell phones, eat a snack or text message at the same time they are driving) is approximately 75%. In a random sample of 15 drivers, let X equal the number of drivers who are not multitaskers. 1. How is X distributed?, Determine the probability density function of X. 2. Give the mean and the variance and Standard deviation of X 3. Find the probability that at most X is at least 4. l. Find the probability that at most X is at most 3

Explanation / Answer

a) X has a binomial distribution with probability density function
P(x;0.25,15)=(15cx)(0.25)x(10.25)(nx)

b) Mean = n*p = 15*0.25 = 3.75

Variance = n*p*(1-p) = 15*0.25*0.75 = 2.8125

SD = sqrt(Var) = sqrt(2.8125) = 1.6771

c) P(X >= 4) = P(X > 3.5) = P(Z > (3.5-3.75)/1.6771) = P(Z > -0.15) = 0.5596
Note: This is using binomial to normal approximation

d) P(X <= 3) = 1 - P(X >= 4) = 1 - 0.5596 = 0.4404

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