#18(d) on p. 365 says that the number of cars we tow a week has a Poisson distn with mean 2.
We can either pay $25 per car or a flat $50/week. Let use choose the option of $25/car.
Consider new questions:
1. Over the course of a year, how many weeks will we make money and how many weeks will we lose money?
(a) We make money if fewer than 2 cars are abandoned.
From PCDF, this is 0.406. With this p, BCDF says that the probability of making money on 20 or fewer weeks per yr is 0.434884.
Using the CLT and p=0.406, mean=21.11 and std dev=3.541.
Mean-2*s.d.=14. From BCDF, Prob(X<=14) = 0.02878
Also, Mean+2*s.d. = 28 and Prob(X<=28) = 0.9806898.
So there is about 3% chance of making money on 14 or fewer weeks and a 2% chance of making money on more than 28 weeks.
(b) We lose money if more than 2 cars are abandoned.
From PCDF, this is 0.3233236.
The mean is 16.8128 and the s.d. is 4.71686.
Mean + 2*s.d. = 26.247
Mean - 2*s.d. = 7.379
From BCDF, P(X<=26) = 0.997333
From BCDF, P(X<=7) = 0.00164
2. How much money might we make/lose over the course of a year?
Over the course of a year, we have a Poisson distribution with mean 52*2=104. Approximate this with a normal with mean 104 and
s.d. = sqrt(104) = 10.198. The net number of cars (the difference between the flat rate and the weekly rate) has mean 0 and s.d. 10.198.
The probability that the net is over 20 cars (=2*s.d.) is 1-0.9772=0.0228.
Thus, with the option of $25/car, there is a 2% chance that we will "lose" $500 (=20 cars * $25/car). There is also a 2% chance that we will "win"
$500.
With 80% probability, we will be within +/- 1.28 s.d. of the mean. So 80% of the time the "net" cars will be +/- 13, which amounts to $326/yr.
Considering the year-long cost, this is about +/- 12%.
Using PCDF with L=2*52 and x=2*52+13, prob = 0.90534.
Also, with L=2*52 and X=2*52-14, prob=0.09055. The difference is 0.81479, which is about 80%.
(This takes a while on the TI-85.)