Question

In order to estimate the proportion of residents in a town who drive alternative- fuel cars (e.g., electric, hybrid, or natural gas), a person records the number of cars passing through a certain intersection for a one-half hour time period. A total of 125 cars passed through the intersection at this time, where 14 were alternative-fuel vehicles.

(a) Discuss the 5 conditions for using a binomial distributional model for this problem. Is each one obviously satisfied, obviously not satisfied, or possibly satisfied under certain assumptions (and what are the assumptions?)

(c) Suppose that, nationwide, 8% of cars use alternative fuels. Do the cars using this intersection during this time appear to have a similar propbability of alternative fuel use? Explain

Answer 1

a)
The 5 conditions/assumptions for using a binomial distribution model are as follows,
1) There are fixed number of trials i.e. n is fixed.
2) Each trials has only two outcomes i.e. success and failure.
3) The n trials are independent and repeated under identical condition.
4) For each individual trial, the probability of success is the same.
5) The central problem is to find the probability of r success out of n trials.
In our case,
1) n = 125 which is fixed. (obviously satisfied)
2) If passed car is alternative fuel vehicles then it is considered as success otherwise failure. (obviously satisfied)
3) n = 125 cars are independent and repeated under identical conditions. (obviously satisfied)
4) The probability of success is same because there are only two possibilities that passed car is alternative fuel vehicles or Not i.e. P(Success) = 1/2 (obviously satisfied)
5) The central problem is to find the probability of r success out of n trials. (obviously satisfied) r takes any value between 1 to 14 and n = 125.
c)
Suppose that, nationwide, 8% of cars use alternative fuels means P(success) = 8 % = 0.08.
Cars using this intersection during this time does not appear to have a similar probability of alternative fuel use as the P(success) = 1/2 = 0.5