Answer:
First, plot points A & B on a graph.
Collinear just means 3 or more points in a straight line (because just 2 points are always collinear, since a straight line can always be drawn through two points.
The instructions don't state a specific area in which points C & D have to be in, so you can put them anywhere, as long as they are collinear with each other, but not any other points,
- i.e. putting three units up and two units left of points A & B
So let's make up some points for C & D that are on a straight line.
- Remember, this line does <em>not</em> have to be horizontal! As long as it's a straight line, any direction will do.
Here are some points that you can choose from:
- C(-1, 1); D(-1, -1)
- C(4, 5); D(4, -5)
- C(3, 4); D(3, 5)
- Anything that doesn't fall on x=2 or y=±3.
For "F" just pick a set of coordinates off to the side and label it
You can even use half values if you want:
- (0.5, 3.2)
- (1.2, -4.1)
- (-9.1, -0.2)
As long as your plotted points meet the criteria:
- C & D are <em>Collinear</em>
- A, B, C, D, & F must not land on the same straight line.
So we see our equation, clear:
y=8^x
Let's just plug in values of x from 1-4.
y= 8^1=8; So, it would be (1,8).
y= 8^2=64; So, it would be (2, 64).
y = 8^3= 512; So, it would be (3,512).
y= 8^4 = 4096; So, it would be (4, 4096).
I assumed that 3 households are sampled.
Now, is the number in each of these households equal to 1, 3 and 8 people?
Next, the question says, samples of n=2 are randomly selected (with replacement) from these 3.
So are sample space is
11
13
31
33
18
81
38
88
83
88
This is done 9 times.
Is that how you understand the problem?
When they select the household with only 1 person, the variance is 0
When they select the household with 8 people, they variance will be from a discrete uniform distribution
Similarly for household with 3 individuals.