<h3>Answer:</h3>
- A — 1
- B — ∑[k=0..2] 15Ck(.24^k)(.76^(15-k))
- D — 12 and 3.1
- A — 0.1091
- B — 0.83 ...
<h3>Explanation:</h3>
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
(<em>Comment on this result</em>: 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
