Question

Suppose your waiting time for a bus in the morning is uniformly distributed on [0,

8], whereas waiting time in the evening is uniformly distributed on [0, 10] independent

of morning waiting time.

a. If you take the bus each morning and evening for a week, what is your total

expected waiting time? [Hint: Define rv's ?1, … , ?10 and use a rule of expected

value.]

b. What is the variance of your total waiting time?

c. What are the expected value and variance of the difference between morning

and evening waiting times on a given day?

d. What are the expected value and variance of the difference between total

morning waiting time and total evening waiting time for a particular week?

Answer 1

Answer

A bit theory

Solution

Here by the problem let us assume  be the random variables representing waitimg times of the morning and evenings respectively where  $X_i\sim Uniform(0,8) \ \ ,i=1(1)5 \ \ and \ \ Y_j\sim Uniform(0,10) \ \ ,j=1(1)5$  and all Xi's and Yj's are independent.

So then 

and  $,Cov(X_i,X_{i'})=0 \ \ , for \ i\neq i'$  , due to independence

Similarly,

and  $,Cov(Y_j,Y_{j'})=0 \ \ , for \ j\neq j'$  due to independence  

And on the other hand,  $Cov(X_i,Y_j)=0$  for all i and j

(a) So the total waiting time be,

$T=X_1+X_2+...+X_5+Y_1+Y_2+...+Y_5=\sum_{i=1}^5X_i+\sum_{j=1}^5Y_j$

So the expected total waiting time be,

$E(T)=E\left ( \sum_{i=1}^5X_i+\sum_{j=1}^5Y_j \right )=\sum_{i=1}^5E(X_i)+\sum_{j=1}^5E(Y_j)=5\times 4+ 5\times 5=45$

(b) And the variance of total waiting time in 5 days be,

$Var(T)=Var\left ( \sum_{i=1}^5X_i+\sum_{j=1}^5Y_j \right )=\sum_{i=1}^5Var(X_i)+\sum_{j=1}^5Var(Y_j)$

since the mutual covariances are all zero due to independence.

$=5\times (16/3)+5\times (25/3)=205/3\approx 68.33$

hence the answer...............

Thank you............