AP Statistics Help?

Question

A density curve looks ike an inverted letter "v". The first segment goes point (0,.06) to the point (.05,1.4). The second segment goes from (.5,1.4) to (1,.6)
1. Verify that the area under the curve is 1, so that is a valid density curve.
2. Determine the median. Approximate the locations of Q1 and Q3.
3. What percent of the observations lie below .3?
4. What percent of the observations lie between .3 and .7?

I figured out number one. .5bh+bh=1

Number 2, the median is .5 and Q1=.25 and Q3=.75

Answer 1

3. To find the percent of the observations that lie below .3, you need to find the area under the curve between 0 and .3

Let's look at the left half of the graph, the graph from 0 to .5. This half should look like a right triangle (with height .8 and width .5) on top of a rectangle (with height .6 and width .5)

Shade the area between 0 and .3.

You will have shaded a smaller right triangle, on top of a rectangle with height .6 and width .3.

The area of the rectangle is .6 * .3 = .18

To find the area of the triangle, we need to find its height. Fortunately, it's similar to the bigger right triangle, so its sides are in proportion, that is
h/b (bigger triangle) = h/b (smaller triangle)
or
(.8)/(.5) = h/(.3)
So h = (.3)(.8)/(.5) = .48

So now we can find the area of the triangle:
1/2*b*h = 1/2*.3*.48 = .072

Thus the total area is: .072 + .18 = .252, or 25.2%

4. Since the density curve is symmetrical, the area above .7 is also .252!
Since we are interested in the observations which are not below .3 and not above .7, the area between .3 and .7 is the total area minus these two areas, or: 1 - 2*(.252)

Space People Unite! ;)

Answer 2

This is the crazy talk! Space people stay off online!