Respuesta :
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
[tex]1\leq X \leq 12[/tex]
And the probability is equal for each value of X, so:
[tex]p(X=1)=p(X=2)=....=p(X=12)[/tex]
The mean of a uniform distribution is given by:
[tex]E[X]=\frac{b+a}{2}[/tex]
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
[tex]E[X]=\frac{12+1}{2}=6.5[/tex] (minutes)
The variance of a uniform distribution is given by:
[tex]Var[X]=\frac{1}{12}(b-a)^2[/tex]
And substituting the values of this problem,
[tex]Var[X]=\frac{1}{12}(12-1)^2=10.1[/tex] (minutes)
b)
Since the distribution is uniform between 1 and 12, we can write the probability density function as
[tex]f(x)=\frac{1}{b-a}[/tex]
The cumulative function gives the probability that the values of X is less than a certain value t:
[tex]p(X<t)=F(t)=\int\limits^t_a {\frac{1}{b-a}} \, dx[/tex] (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:
[tex]p(X<7)=\int\limits^7_a {\frac{1}{b-a}} \, dx =\frac{7-a}{b-a}=\frac{7-1}{12-1}=0.54[/tex]
c)
The probability that a customer waits between four and twenty minutes can be rewritten as
[tex]p(4<X<20)[/tex]
This can be written as:
[tex]p(4<X<20)=p(X>4)\cdot p(X<20)[/tex] (1)
However, the probabilty of X>4 can be written as
[tex]p(X>4)=1-p(X<4)[/tex]
Also, we notice that
[tex]p(X<20)=1[/tex] because the maximum value of X is 12; therefore, we can rewrite (1) as
[tex]p(4<X<20)=1-p(X<4)[/tex]
We can calculate [tex]p(X<4)[/tex] by using the same method as in part b:
[tex]p(X<4)=\int\limits^4_a {\frac{1}{b-a}} \, dx =\frac{4-a}{b-a}=\frac{4-1}{12-1}=0.27[/tex]
So, we find
[tex]p(4<X<20)=1-0.27=0.73[/tex]
d)
In this part, we know that a customer waits for
X = k
minutes in line, and he receives a coupon worth
[tex]0.2k^{\frac{1}{4}}[/tex] dollars.
Here we want to find the mean of the coupon value.
Here therefore we have a new variables defined as
[tex]Y=0.2X^{\frac{1}{4}}[/tex]
Given a variable with standard (between 0 and 1) uniform distribution X, the variable
[tex]Y=X^n[/tex]
follows a beta distribution, with parameters [tex](\frac{1}{n},1)[/tex], and whose mean value is given by
[tex]E[Y]=\frac{1/n}{1+\frac{1}{n}}[/tex]
In this case,
[tex]n=\frac{1}{4}[/tex]
So the mean value of [tex]X^{1/4}[/tex] is
[tex]E[X^{1/4}]=\frac{1/(1/4)}{1+\frac{1}{1/4}}=\frac{4}{1+4}=\frac{4}{5}=0.8[/tex]
However, our variable is distribution is non-standard, because its values are between 1 and 12, so the range is
[tex]Min = 1^{1/4}=1\\Max =12^{1/4}=1.86[/tex]
So, the actual mean value of [tex]X^{1/4}[/tex] is
[tex]E[X^{1/4}]=0.8\cdot (1.86-1)+1=1.69[/tex]
However, in the definition of Y we also have a factor 0.2; therefore, the mean value of Y is
[tex]E[Y]=0.2E[X^{1/4}]=0.2\cdot 1.69 =0.34[/tex]
