a) Mean: 6.5 min, variance: 10.1 min
b) 0.54 (54%)
c) 0.73 (73%)
d) 0.34
Step-by-step explanation:
a)
Here we can call X the variable indicating the waiting time for the customers:
X = waiting time
We are told that the waiting time is distributed uniformly between 1 and 12; this means that
And the probability is equal for each value of X, so:
The mean of a uniform distribution is given by:
where a and b are the minimum and maximum values of the variable X. In this case,
a = 1
b = 12
So the mean value of X is
(minutes)
The variance of a uniform distribution is given by:
And substituting the values of this problem,
(minutes)
b)
Since the distribution is uniform between 1 and 12, we can write the probability density function as
The cumulative function gives the probability that the values of X is less than a certain value t:
(1)
In this case, we want to find the probability that the waiting time is less than 7 minutes, so
t = 7
We also have:
a = 1
b = 12
Therefore, calculating (1) and substituting, we find:
c)
The probability that a customer waits between four and twenty minutes can be rewritten as
This can be written as:
(1)
However, the probabilty of X>4 can be written as
Also, we notice that
because the maximum value of X is 12; therefore, we can rewrite (1) as
We can calculate by using the same method as in part b:
So, we find
d)
In this part, we know that a customer waits for
X = k
minutes in line, and he receives a coupon worth
dollars.
Here we want to find the mean of the coupon value.
Here therefore we have a new variables defined as
Given a variable with standard (between 0 and 1) uniform distribution X, the variable
follows a beta distribution, with parameters , and whose mean value is given by
In this case,
So the mean value of is
However, our variable is distribution is non-standard, because its values are between 1 and 12, so the range is
So, the actual mean value of is
However, in the definition of Y we also have a factor 0.2; therefore, the mean value of Y is