In: Statistics and Probability
The American Housing Survey reported the following data on the number of times that owner-occupied and renter-occupied units had a water supply stoppage lasting 6 or more hours in the past 3 months.
Number of Houses (1000s) | ||||
Number of Times | Owner Occupied | Renter Occupied | ||
0 | 549 | 23 | ||
1 | 5,012 | 542 | ||
2 | 6,100 | 3,734 | ||
3 | 2,544 | 8,690 | ||
4 times or more | 558 | 3,783 |
Do not round intermediate calculations. Round your answers to two decimal places.
a. Define a random variable x = number of times that owner-occupied units had a water supply stoppage lasting 6 or more hours in the past 3 months and develop a probability distribution for the random variable. (Let x = 4 represent 4 or more times.)
x | f(x) |
0 | |
1 | |
2 | |
3 | |
4 | |
Total |
b. Compute the expected value and variance for x.
Total | |
E(x) | |
Var(x) |
c. Define a random variable y = number of times that renter-occupied units had a water supply stoppage lasting 6 or more hours in the past 3 months and develop a probability distribution for the random variable. (Let y = 4 represent 4 or more times.)
y | f(y) |
0 | |
1 | |
2 | |
3 | |
4 | |
Total |
d. Compute the expected value and variance for y.
Total | |
E(y) | |
Var(y) |
(a) Let, X be a random variable number of times that owner occupied units had a water supply stoppage lasting 6 or more hours in the past 3 months.
A probability distribution for the random variable. (Let x = 4 represent 4 or more times.)
X | Number of Houses (1000s) Owner Occupied | f(x) |
0 | 549 | 0.037188 |
1 | 5012 | 0.339497 |
2 | 6100 | 0.413195 |
3 | 2544 | 0.172323 |
4 | 558 | 0.037797 |
total | 14763 | 1 |
The f(x) column has been constructed by dividing the number of
Houses (1000s) Owner Occupied for each X by the total number of
Houses (1000s) Owner Occupied
(b) Expected value of X and variance of X:
Necessary calculation is shown in the table beloiw:
X | f(x) | x*f(x) | x2*f(x) |
0 | 0.037188 | 0 | 0 |
1 | 0.339497 | 0.339497 | 0.339497 |
2 | 0.413195 | 0.82639 | 1.652781 |
3 | 0.172323 | 0.516968 | 1.550904 |
4 | 0.037797 | 0.151189 | 0.604755 |
total | 1 | 1.834045 | 4.147937 |
E(X)=1.834045
Var(X)=4.147937-(1.834045)2
=0.7842159
(c) Let: Y= number of times that renter-occupied units had a water supply stoppage lasting 6 or more hours in the past 3 months .
A probability distribution for the random variable. (Let y = 4 represent 4 or more times.)
Y | Renter Occcupied | f(y) |
0 | 23 | 0.001371 |
1 | 542 | 0.032316 |
2 | 3734 | 0.222633 |
3 | 8690 | 0.518125 |
4 | 3783 | 0.225554 |
Total | 16772 | 1 |
The f(y) column has been constructed by dividing the number of Houses (1000s) Renter Occupied for each Y by the total number of Houses (1000s) Renter Occupied.
(d) Expected value of Y and variance of Y:
Necessary calculation is shown in the table beloiw:
Y | Renter Occcupied | f(y) | y*f(y) | y^2*f(y) | |
0 | 23 | 0.001371 | 0 | 0 | |
1 | 542 | 0.032316 | 0.032316
|
0.032316 | |
2 | 3734 | 0.222633 | 0.445266 | 0.890532 | |
3 | 8690 | 0.518125 | 1.554376 | 4.663129 | |
4 | 3783 | 0.225554 | 0.902218 | 3.608872 | |
Total | 16772 | 1 | 2.934176 | 9.194849 |
E(X)=2.934176
Var(X)=9.194849-(2.934176)2
=0.58546
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