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

tywebb

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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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#RoadTo31Atar

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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

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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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tywebb

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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

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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

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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.
 

tywebb

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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?
 

idkkdi

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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

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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

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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

 

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