Final answer:
i) The probability density function (pdf) of X is f(x) = 8e^(-8x), x ≥ 0. ii) The probability of waiting less than 12 minutes between successive speeders is 0.982. iii) The probability of waiting more than 12 minutes between successive speeders is 0.018. iv) The population mean and variance of X are 7.5 minutes and 1.1719 minutes^2, respectively. v) The population mean and variance of Y = 3X - 1 are 21.5 minutes and 10.5471 minutes^2, respectively.
Step-by-step explanation:
i) To find the probability density function (pdf) of X, we can differentiate the cumulative distribution function (cdf) with respect to x. So, the pdf of X is f(x) = d/dx[F(x)] = d/dx[1-e^(-8x)] = 8e^(-8x), x ≥ 0.
ii) The probability of waiting less than 12 minutes between successive speeders can be found by evaluating the cdf of X at x = 12/60 = 0.2 hours. So, P(X < 0.2) = 1 - e^(-8*0.2) = 0.982.
iii) The probability of waiting more than 12 minutes between successive speeders can be found by integrating the pdf of X from x = 12/60 = 0.2 to infinity. So, P(X > 0.2) = ∫[8e^(-8x)]dx from 0.2 to infinity = 1 - P(X < 0.2) = 1 - 0.982 = 0.018.
iv) The population mean of X can be found by integrating x * pdf of X from x = 0 to infinity. So, E(X) = ∫[x * 8e^(-8x)]dx from 0 to infinity = 1/8 hours = 7.5 minutes. The population variance can be found by integrating (x - mean)^2 * pdf of X from x = 0 to infinity. So, Var(X) = ∫[(x - mean)^2 * 8e^(-8x)]dx from 0 to infinity = 1/64 hours^2 = 1.1719 minutes^2.
v) Let Y = 3X - 1. The population mean of Y can be found by substituting the population mean of X into the equation for Y. So, E(Y) = 3 * E(X) - 1 = 3 * 7.5 - 1 = 22.5 - 1 = 21.5 minutes. The population variance of Y can be found by substituting the population variance of X into the equation for Y. So, Var(Y) = 3^2 * Var(X) = 9 * 1.1719 = 10.5471 minutes^2.