# 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.)

(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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• harrowed2

#### #RoadTo31Atar

##### Well-Known 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.

• shashysha and mikikieko12

#### 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:

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

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#### #RoadTo31Atar

##### Well-Known 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

#### 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

##### Active 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.

• B1andB2 and 大きい男

#### tywebb

##### dangerman
Have a look at Example 4 too. That is also not a pdf.

I would of course like to replace this with 2 questions

Question 1.

Question 2 (a little more challenging).

What can you do to fix this question to make it a probability density function?

• harrowed2

Moderator

#### idkkdi

##### Well-Known Member
Have a look at Example 4 too. That is also not a pdf.

I would of course like to replace this with 2 questions

Question 1.

Question 2 (a little more challenging).

What can you do to fix this question to make it a probability density function?
integrate for q2, then stretch vertically by the factor required to make the area under the function 1.

#### idkkdi

##### Well-Known Member
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.)

(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.)
hint 2, subbing in the extreme values into the new equation yields 0 for f(x) at the extreme points.

However, how would you actually realise what to switch the equation to?

#### CM_Tutor

##### Moderator
Moderator
For to be a PDF, we require

Assuming that it is otherwise suitable, and thus that , we can form a suitable PDF and find by requiring

and then checking that . In this case, the latter condition will be fine so long as

• idkkdi