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Math Help - Probability Distribution for a continuous random variable

  1. #1
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    Probability Distribution for a continuous random variable

    The length of time required by students to complete a exam is a random variable with a density function given by:

    f(x) = cx^3 + x if 0 <= x <= 1
    0 otherwise

    1) find c
    2) Find the distribution function F(y).
    3) Find the probability that a randomly selected student will finish the exam in less than half an hour
    4) Given that a particular student needs at least 0.25 hours to complete the exam. What is the probability Heather will require at least 0.5 hours to complete the exam?

    This is what I did:
    1) I find the integral from 1 to 0 of x^3 + x
    c[(x^4)/4 + (x^2)/2] = 1
    c[(1/4) + (1/2) = 1
    c(3/4) = 1
    c is 4/3.

    2)
    4/3[(x^4)/4 + (x^2)/2]
    [(x^4)/3] + [(2x^2)/3]

    Is that correct? What about the rest of the problem? The book doesn't have any example so I am stuck.
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  2. #2
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    Quote Originally Posted by dkmathguy View Post
    The length of time required by students to complete a exam is a random variable with a density function given by:

    f(x) = cx^3 + x if 0 <= x <= 1
    0 otherwise

    1) find c
    2) Find the distribution function F(y).
    3) Find the probability that a randomly selected student will finish the exam in less than half an hour
    4) Given that a particular student needs at least 0.25 hours to complete the exam. What is the probability Heather will require at least 0.5 hours to complete the exam?

    This is what I did:
    1) I find the integral from 1 to 0 of x^3 + x
    c[(x^4)/4 + (x^2)/2] = 1
    c[(1/4) + (1/2) = 1
    c(3/4) = 1
    c is 4/3.

    2)
    4/3[(x^4)/4 + (x^2)/2]
    [(x^4)/3] + [(2x^2)/3]

    Is that correct? What about the rest of the problem? The book doesn't have any example so I am stuck.
    Your answer for (1) is incorrect: c only multiplies the first term (you have corrected this in post #3). Your answer for (2) is incomplete (a common error made by students) and therefore wrong. The correct answer is:

    F(x) = 0 for x < 0
    F(x) = \frac{x^4}{2} + \frac{x^2}{2} for 0 \leq x \leq 1 (from post #3).
    F(x) = 1 for x > 1.

    You probably think that I'm being pedantic here, but I'm not.


    (3) By definition: Pr(X < 1/2) = F(1/2) = ....

    (4) Conditional probability is required: \Pr(X > 0.5 | X > 0.25) = \frac{\Pr(X > 0.5)}{\Pr(X > 0.25)} (from Bayes Theorem).
    Last edited by mr fantastic; April 12th 2010 at 01:05 AM. Reason: Corrected an error - misread the pdf.
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  3. #3
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    Wait, I think I mess up on 1. The correct question is f(x) = cx^3 + x, not f(x) = c(x^3 + x)

    So is this correct?

    c((x^4)/4) + (x^2)/2 = 1
    c((1/4)) + (1/2) = 1
    c(1/4) = 1/2
    c is 2

    For 2, I have to right all 3 of those for the answer?
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  4. #4
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    Quote Originally Posted by dkmathguy View Post
    Wait, I think I mess up on 1. The correct question is f(x) = cx^3 + x, not f(x) = c(x^3 + x)

    So is this correct?

    c((x^4)/4) + (x^2)/2 = 1
    c((1/4)) + (1/2) = 1
    c(1/4) = 1/2
    c is 2

    [snip]
    Yes.
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  5. #5
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    (3) By definition: Pr(X < 1/2) = F(1/2) = ....

    ((.5)^4)/2 + ((.5)^2)/2 = .15625

    (4) Conditional probability is required: (from Bayes Theorem).

    ((.5)^4)/2 + ((.5)^2)/2 = 0.15625
    ((.25)^4)/2 + ((.25)^2)/2 = 0.03320

    (1 - 0.15625)/(1 - 0.03320) = .8727.

    Is that correct? I am so tired, its 3 a.m over here. Good night, thank Mr. F, I learn a lot today. Thank you very much.
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