Question

Graph A: A horizontal line goes from (1, 0.5) to (2, 0.5). Another horizontal line goes form (2, 0.2) to (7, (0, 2). Graph B: A curve starts at (0, 0), curves up to (1, 1), and then curves down to (2, 0).
Which graph represents a density curve, and why?


graph A only, because the curve is above the horizontal axis, and the area under the curve from 2 to 7 is 1


graph B only, because the curve is above the horizontal axis, and the area under the curve is equal to 1.57


both graph A and graph B, because both curves are above the horizontal axis, and both areas are positive


neither graph A nor graph B, because, even though both curves are above the horizontal axis, neither graph has an area of 1


Pictures posted below

Answer 1

The statement - "neither graph A nor graph B, because, even though both curves are above the horizontal axis, neither graph has an area of 1" is true.

A few fundamental principles apply to density curves:

• A density curve's area beneath it represents probability.

• A density curve's area under it equals one.

• Base x height in a uniform density curve equals one.

• The likelihood that x = a will never occur.

• The likelihood that x < a is the same as that of x ≤ a.

Neither curve of Graph A nor of Graph B has the area under the curve summed up as 1, though the curve is above the horizontal axis.

Hence, because neither graph has an area of 1, even if both curves are above the horizontal axis, the statement "neither graph A nor graph B, because, even though both curves are above the horizontal axis, neither graph has an area of 1" is true.

Learn more about density curves here-

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Answer 2

Answer: neither graph A nor graph B, because, even though both curves are above the horizontal axis, neither graph has an area of 1

Step-by-step explanation:

Areas under the graphs:

Graph A

$(1)(0.5)+(7-2)(0.2)=1.5$

Graph B

$\frac{\pi}{2}(1^{2})=\frac{\pi}{2}$

As neither of these graphs have an area of 1, neither of them are density curves.