Can you see what is wrong with this question? (1 Viewer)

tywebb

dangerman
There is something wrong with this example from New Senior Mathematics Advanced on pages 561-562. Can you see what it is?

(Hint 1: there is nothing wrong with the calculation.)

$\bg_white \text{Example 2}$

$\bg_white \text{A particular continuous random variable has the following probability density function:}$

$\bg_white f(x) = \begin{cases}-3(x-1)^2+2,&0\le x\le1 \\0,&\text{otherwise}\end{cases}$

$\bg_white \text{Find }P(X)\le0.7.$

$\bg_white \text{Solution}$

$\bg_white \text{Identify the bounds of integration to use: the lower bound is 0 and the upper bound is 0.7.}$

$\bg_white \text{Expand the quadratic expression:}$

\bg_white \begin{aligned}-3(x-1)^2+2&=-3(x^2-2x+1)+2\\&=-3x^2+6x-3+2\\&=-3x^2+6x-1\end{aligned}

$\bg_white \text{Find the required integral:}$

\bg_white \begin{aligned}\int_0^{0.7}(-3x^2+6x-1)\ dx&=[-x^3+3x^2-x]_0^{0.7}\\&=(-(0.7)^3+3\times(0.7)^2-0.7)-0\\&=0.427\end{aligned}

(Hint 2: you can fix it by replacing -3(x-1)2+2 with -6(x-0.5)2+1.5 in which case the answer is 0.784 - but can you see why?)

(Hint 3: without changing the function find P(X ≤ 0.1) instead.)

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Active Member
Idk if I'm really dumb but I can't understand what you wrote, I have no clue what you're asking or what your hints are talking about. I don't see anything wrong in the working out.

tywebb

dangerman
Idk if I'm really dumb but I can't understand what you wrote, I have no clue what you're asking or what your hints are talking about. I don't see anything wrong in the working out.
OK. Here is hint 4:

$\bg_white \text{For a function }f(x)\text{ to be a probability density function:}$

$\bg_white \bullet f(x)\ge0\text{ for all values of }x$

$\bg_white \bullet\int_{-\infty}^\infty f(x)\ dx=1,\text{ i.e., the area enclosed by the graph }y=f(x)\text{ and the }x\text{-axis is equal to 1.}$

Do you think these conditions are satisfied by the example above?

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Active Member
OK. Here is hint 4:

View attachment 29166

Do you think these conditions are satisfied by the example above?
Yes I found that f(x) was less than 0 from hint 3 but I don't think it mattered for the question since it asked for 0.7 and it's just an example question.

tywebb

dangerman
Yes I found that f(x) was less than 0 from hint 3 but I don't think it mattered for the question since it asked for 0.7 and it's just an example question.
That means that the function is NOT a probability density function.

If you find P(X ≤ 0.1) then you get a negative answer. So it is NOT a probability.

You can turn it into a probability density function so that it satisfies both conditions by replacing -3(x-1)2+2 with -6(x-0.5)2+1.5.

大きい男

Member
That means that the function is NOT a probability density function.

If you find P(X ≤ 0.1) then you get a negative answer. So it is NOT a probability.

You can turn it into a probability density function so that it satisfies both conditions by replacing -3(x-1)2+2 with -6(x-0.5)2+1.5.
Do you think it would be necessary, in an exam, to verify that f(x) is a pdf or do you just assume that it is?

tywebb

dangerman
Do you think it would be necessary, in an exam, to verify that f(x) is a pdf or do you just assume that it is?
If this was an exam question it would be completely invalid.

An exam might ask you to verify that a function qualifies as a probability density function, in which case you show that it satisfies both conditions.

The authors of this textbook thought that it was an example of a probability density function. But it is actually an example of a function which is NOT a probability density function.