Probaility density function Determine A

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Discussion Overview

The discussion revolves around determining the constant A in a given probability density function (PDF) for a random variable X, as well as calculating the probability P(2.5 <= X <= 7.5). The scope includes mathematical reasoning and integration techniques related to probability theory.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • Post 1 presents the PDF and proposes that A = 1/25, questioning if this is correct.
  • Post 2 agrees with the calculation of A and outlines an approach for finding P(2.5 <= X <= 7.5), but raises a question about the need for squares in the integrals.
  • Post 3 reiterates the question about the squares and suggests a method of doubling one of the integrals due to the symmetry of the distribution around 5.
  • Post 4 confirms that integrating Ax would yield an x² term, implying a potential oversight in the previous calculations.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the integration steps and whether squares should be included in the calculations. There is no consensus on the correct approach to the integrals or the value of A.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in the integration process, and there are differing opinions on the necessity of including squares in the calculations.

TomJerry
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Problem :
A random variable X has probability density function
[tex] f(n) = <br /> \begin{cases} <br /> Ax, & \mbox{0 lessthan equal x less than 5 } \\<br /> A(10-x), & \mbox{5 less than equal x less than 10 }\\ <br /> 0, & \mbox{otherwise } <br /> \end{cases}<br /> [/tex]

i)Determine A
ii)Find P(2.5 <= X <= 7.5)

Solutions
i) let P(A) = Ax and P(b)= A(10-x)

P(A) + P(B) = 1

Therefore on calculating i get A = 1/25 [IS THIS CORRECT]

ii)
integrate from [0-5] for Ax and substract it by [0-2.5] A(x) --> I will get [2.5-5] A(x)
then
integrate from [5-10] for A(10-x) and substract it by [7.5-10] A(10-x) --> I will get [5-7.5] A(10-x)

Add them both to get answer for ii)

IS THIS CORRECT
 
Last edited:
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Hi TomJerry! :smile:
TomJerry said:
Therefore on calculating i get A = 1/25 [IS THIS CORRECT]

Yes! :biggrin:
integrate from [0-5] for Ax and substract it by [0-2.5] A(x) --> I will get [2.5-5] A(x)
then
integrate from [5-10] for A(10-x) and substract it by [7.5-10] A(10-x) --> I will get [5-7.5] A(10-x)

Add them both to get answer for ii)

IS THIS CORRECT

shouldn't there be squares here? :confused:

(and i'd just do one of the integrals, and double it … the distribution is obviously syymmetric about 5 :wink:)
 
tiny-tim said:
shouldn't there be squares here? :confused:

(and i'd just do one of the integrals, and double it … the distribution is obviously syymmetric about 5 :wink:)

I didnt get it... what squares
 
If you're integrating Ax, won't you get an x2 ?
 

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