Respuesta :
Answer:
The probability distribution for x is:
[tex]P(x=0)=0.037\\\\P(x=1)=0.339\\\\P(x=2)=0.414\\\\P(x=3)=0.172\\\\P(x=4)=0.038\\\\[/tex]
Step-by-step explanation:
The question is incomplete.
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)
# Owner Occupied Renter Occupied
0 547 23
1 5,012 541
2 6,110 3,734
3 2,544 8,660
4+ 557 3,784
The sample space of x is [0,1,2,3,4].
We can calculate the probabilities of x as the relative frequency of each value of x.
The total sample size of owner-occupied houses is:
[tex]N=547+5,012+6,110+2,544+557=14,770[/tex]
Then, the relative frequencies (equal to the probability) are:
[tex]P(0)=k/n=547/14,770=0.037\\\\P(1)=5,012/14,770=0.339\\\\P(2)=6,110/14,770=0.414\\\\P(3)=2,544/14,770=0.172\\\\P(4)=557/14,770=0.038[/tex]
Answer:
a)
P(0) = 547/14860 = 0.0368
P(1) = 5012/14860 = 0.3373
P(2) = 6100/14860 = 0.4105
P(3) = 2644/14860 = 0.1779
P(4) = 557/14860 = 0.0375
b) Expected Value = 1.842
Variance = 0.7874
c)
P(0) = 23 / 16869 = 0.0001
P(1) = 541 / 16869 = 0.0321
P(2) = 3832 / 16869 = 0.2272
P(3) = 8690 / 16869 = 0.5151
P(4) = 3783 / 16869 = 0.2243
d) Expected Value = 2.929
Variance = 0.5763
e) Owner-occupied houses have more bedrooms in general than renter-occupied houses.
Step-by-step explanation:
Incomplete Question.
See the complete question: a. Define a random variable x number of bedrooms in renter-occupied houses and develop a probability distribution for the random variable. (Let x 4 represent 4 or more bedrooms.) b. Compute the expected value and variance for the number of bedrooms in renteroccupied houses. c. Define a random variable y number of bedrooms in owner-occupied houses and develop a probability distribution for the random variable. (Let y 4 represent 4 or more bedrooms.) d. Compute the expected value and variance for the number of bedrooms in owneroccupied houses. e. What observations can you make from a comparison of the number of bedrooms in renter-occupied versus owner-occupied homes?
Bedrooms Number ofRenter-Occupied Houses Owner-Occupied (1000)
0 547 23
1 5012 541
2 6100 3832
3 2644 8690
4 or more 557 3783
a) Total number of rental
= 547 + 5012 + 6100 + 2644 + 557 = 14860
P(0) = 547/14860 = 0.0368
P(1) = 5012/14860 = 0.3373
P(2) = 6100/14860 = 0.4105
P(3) = 2644/14860 = 0.1779
P(4) = 557/14860 = 0.0375
b) The expected value:
[tex]E(X) = 0 * 0.0368 + 1 * 0.3373 + 2 * 0.4105 + 3 * 0.1779 + 4 * 0.0375\\= 0 + 0.3373 + 0.821 + 0.5337 + 0.15\\= 1.842[/tex]
The variance:
[tex]Var(X) = (0-1.842)^{2} * 0.0368 + (1 - 1.842)^{2} * 0.3373 + (2 - 1.842)^2 * 0.4105 + (3-1.842)^2 * 0.1779 + (4 - 1.842)^2 * 0.0375\\= 0.7874[/tex]
c) Probabilities for renter-occupied
total number of rental-occupied
= 23 + 541 + 3832 + 8690 + 3783 = 16869
P(0) = 23 / 16869 = 0.0001
P(1) = 541 / 16869 = 0.0321
P(2) = 3832 / 16869 = 0.2272
P(3) = 8690 / 16869 = 0.5151
P(4) = 3783 / 16869 = 0.2243
d) The Expected value:
[tex]E(X) = 0 * 0.0001 + 1 * 0.0321 + 2 * 0.2272 + 3 * 0.5151 + 4 * 0.2243\\= 0 + 0.0321 + 0.4544 + 1.5453 + 0.8972\\= 2.929[/tex]
The Variance is:
[tex]Var(X) = (0-2.929)^{2} * 0.0001 + (1 - 2.929)^{2} * 0.0321 + (2 - 2.929)^2 * 0.2272 + (3-2.929)^2 * 0.5151 + (4 - 2.929)^2 * 0.2243\\= 0.5763[/tex]
