Answer:
(a) <em>X</em> = number of dealerships the customer needs to call before she finds a used red Miata car.
(b) <em>X</em> = 1, 2, 3, 4,...
(c) The distribution of <em>X</em> is, .
(d) The expected number of dealerships would we expect her to call until she finds one that has the car is 3.57.
(e) The probability that she must call at most 4 dealerships is 0.7313.
(f) The probability that she must call 3 or 4 dealerships is 0.2497.
Step-by-step explanation:
(a)
The random variable <em>X</em> can be defined as the number of dealerships the customer needs to call before she finds a used red Miata car.
(b)
The values that the random variable <em>X</em> can assume are:
<em>X</em> = 1, 2, 3, 4,...
(c)
The random variable <em>X</em> is defined as the number of trials before the first success. Each trial is independent of the others. And each trial has a similar probability of success.
This implies that the random variable <em>X</em> follows a Geometric distribution.
The probability of success, i.e. the probability that any independent dealership will have the car is <em>p</em> = 0.28.
Thus, the distribution of <em>X</em> is, .
(d)
The expected value of a Geometric distribution is:
Compute the expected number of dealerships would we expect her to call until she finds one that has the car as follows:
Thus, the expected number of dealerships would we expect her to call until she finds one that has the car is 3.57.
(e)
Compute the probability that she must call at most 4 dealerships as follows:
P (X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
Thus, the probability that she must call at most 4 dealerships is 0.7313.
(f)
Compute the probability that she must call 3 or 4 dealerships as follows:
P (X = 3 or X = 4) = P (X = 3) + P (X = 4)
Thus, the probability that she must call 3 or 4 dealerships is 0.2497.