[edited] need on stats question 40 points

answer the new questions.
in advance.

1. The probability is 0.52 that the officer will pull over a driver and then the expected number more. So, if x is the expected number of drivers pulled over until one is not texting, we have ...

... x = 0.52(1+x)

... 0.48x = 0.52

... x = 0.52/0.48 = 13/12 = 1 1/12 ≈ 1 . . . . matches selection A

(Comment on this result: I find it interesting that these are the odds in favor of finding a driver who texts. That is, if the probability of texting is 0.98, the odds are 49:1 that a driver will be texting, and the expected number of pull-overs is 49.)

2. The probability of at most 2 being cured is the probability of 0, 1, or 2 being cured. You need to add up those probabilities. The sum in answer selection B does that.

3. The mean of a binomial distribution is ...

... μx = np = 60·0.2 = 12

... σx = √(np(1-p)) = √(12·0.8) ≈ 3.0984

These match selection D.

4. 20C14(0.8^14)(0.2^6) = 38760·.043980·0.000064 ≈ 0.109100 . . . matches A

5. mean(x) = 0.94; mean(x^2) = 1.58, so ...

... σx = √(1.58 -0.94²) ≈ 0.83 . . . . matches selection B