Question

Suppose that wait times for customers at a grocery store cashier line are uniformly distributed between one minute and twelve minutes.
a. What are the mean and variance of the waiting time?
b. What is the probability that a customer waits less than seven minutes?
c. What is the probability that a customer waits between four and twenty minutes?
d. Suppose that a customer who waits k minutes in line receives a coupon worth a 0.2 k^(1/4) dollar discount on a future visit. What is the mean of the coupon value for a customer?

Answer 1

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

$1\leq X \leq 12$

And the probability is equal for each value of X, so:

$p(X=1)=p(X=2)=....=p(X=12)$

The mean of a uniform distribution is given by:

$E[X]=\frac{b+a}{2}$

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

$E[X]=\frac{12+1}{2}=6.5$ (minutes)

The variance of a uniform distribution is given by:

$Var[X]=\frac{1}{12}(b-a)^2$

And substituting the values of this problem,

$Var[X]=\frac{1}{12}(12-1)^2=10.1$ (minutes)

b)

Since the distribution is uniform between 1 and 12, we can write the probability density function as

$f(x)=\frac{1}{b-a}$

The cumulative function gives the probability that the values of X is less than a certain value t:

$p(X$ (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:

$p(X$

c)

The probability that a customer waits between four and twenty minutes can be rewritten as

$p(4$

This can be written as:

$p(4$ (1)

However, the probabilty of X>4 can be written as

$p(X>4)=1-p(X$

Also, we notice that

$p(X$ because the maximum value of X is 12; therefore, we can rewrite (1) as

$p(4$

We can calculate $p(X$ by using the same method as in part b:

$p(X$

So, we find

$p(4$

d)

In this part, we know that a customer waits for

X = k

minutes in line, and he receives a coupon worth

$0.2k^{\frac{1}{4}}$ dollars.

Here we want to find the mean of the coupon value.

Here therefore we have a new variables defined as

$Y=0.2X^{\frac{1}{4}}$

Given a variable with standard (between 0 and 1) uniform distribution X, the variable

$Y=X^n$

follows a beta distribution, with parameters $(\frac{1}{n},1)$, and whose mean value is given by

$E[Y]=\frac{1/n}{1+\frac{1}{n}}$

In this case,

$n=\frac{1}{4}$

So the mean value of $X^{1/4}$ is

$E[X^{1/4}]=\frac{1/(1/4)}{1+\frac{1}{1/4}}=\frac{4}{1+4}=\frac{4}{5}=0.8$

However, our variable is distribution is non-standard, because its values are between 1 and 12, so the range is

$Min = 1^{1/4}=1\\Max =12^{1/4}=1.86$

So, the actual mean value of $X^{1/4}$ is

$E[X^{1/4}]=0.8\cdot (1.86-1)+1=1.69$

However, in the  definition of Y we also have a factor 0.2; therefore, the mean value of Y is

$E[Y]=0.2E[X^{1/4}]=0.2\cdot 1.69 =0.34$