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
![1\leq X \leq 12](https://tex.z-dn.net/?f=1%5Cleq%20X%20%5Cleq%2012)
And the probability is equal for each value of X, so:
![p(X=1)=p(X=2)=....=p(X=12)](https://tex.z-dn.net/?f=p%28X%3D1%29%3Dp%28X%3D2%29%3D....%3Dp%28X%3D12%29)
The mean of a uniform distribution is given by:
![E[X]=\frac{b+a}{2}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cfrac%7Bb%2Ba%7D%7B2%7D)
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:
![Var[X]=\frac{1}{12}(b-a)^2](https://tex.z-dn.net/?f=Var%5BX%5D%3D%5Cfrac%7B1%7D%7B12%7D%28b-a%29%5E2)
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
![f(x)=\frac{1}{b-a}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7Bb-a%7D)
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:
![p(X](https://tex.z-dn.net/?f=p%28X%3C7%29%3D%5Cint%5Climits%5E7_a%20%7B%5Cfrac%7B1%7D%7Bb-a%7D%7D%20%5C%2C%20dx%20%3D%5Cfrac%7B7-a%7D%7Bb-a%7D%3D%5Cfrac%7B7-1%7D%7B12-1%7D%3D0.54)
c)
The probability that a customer waits between four and twenty minutes can be rewritten as
![p(4](https://tex.z-dn.net/?f=p%284%3CX%3C20%29)
This can be written as:
(1)
However, the probabilty of X>4 can be written as
![p(X>4)=1-p(X](https://tex.z-dn.net/?f=p%28X%3E4%29%3D1-p%28X%3C4%29)
Also, we notice that
because the maximum value of X is 12; therefore, we can rewrite (1) as
![p(4](https://tex.z-dn.net/?f=p%284%3CX%3C20%29%3D1-p%28X%3C4%29)
We can calculate
by using the same method as in part b:
![p(X](https://tex.z-dn.net/?f=p%28X%3C4%29%3D%5Cint%5Climits%5E4_a%20%7B%5Cfrac%7B1%7D%7Bb-a%7D%7D%20%5C%2C%20dx%20%3D%5Cfrac%7B4-a%7D%7Bb-a%7D%3D%5Cfrac%7B4-1%7D%7B12-1%7D%3D0.27)
So, we find
![p(4](https://tex.z-dn.net/?f=p%284%3CX%3C20%29%3D1-0.27%3D0.73)
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
![Y=0.2X^{\frac{1}{4}}](https://tex.z-dn.net/?f=Y%3D0.2X%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D)
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
![E[Y]=\frac{1/n}{1+\frac{1}{n}}](https://tex.z-dn.net/?f=E%5BY%5D%3D%5Cfrac%7B1%2Fn%7D%7B1%2B%5Cfrac%7B1%7D%7Bn%7D%7D)
In this case,
![n=\frac{1}{4}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B1%7D%7B4%7D)
So the mean value of
is
![E[X^{1/4}]=\frac{1/(1/4)}{1+\frac{1}{1/4}}=\frac{4}{1+4}=\frac{4}{5}=0.8](https://tex.z-dn.net/?f=E%5BX%5E%7B1%2F4%7D%5D%3D%5Cfrac%7B1%2F%281%2F4%29%7D%7B1%2B%5Cfrac%7B1%7D%7B1%2F4%7D%7D%3D%5Cfrac%7B4%7D%7B1%2B4%7D%3D%5Cfrac%7B4%7D%7B5%7D%3D0.8)
However, our variable is distribution is non-standard, because its values are between 1 and 12, so the range is
![Min = 1^{1/4}=1\\Max =12^{1/4}=1.86](https://tex.z-dn.net/?f=Min%20%3D%201%5E%7B1%2F4%7D%3D1%5C%5CMax%20%3D12%5E%7B1%2F4%7D%3D1.86)
So, the actual mean value of
is
![E[X^{1/4}]=0.8\cdot (1.86-1)+1=1.69](https://tex.z-dn.net/?f=E%5BX%5E%7B1%2F4%7D%5D%3D0.8%5Ccdot%20%281.86-1%29%2B1%3D1.69)
However, in the definition of Y we also have a factor 0.2; therefore, the mean value of Y is
![E[Y]=0.2E[X^{1/4}]=0.2\cdot 1.69 =0.34](https://tex.z-dn.net/?f=E%5BY%5D%3D0.2E%5BX%5E%7B1%2F4%7D%5D%3D0.2%5Ccdot%201.69%20%3D0.34)