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  1. #1
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    X is uniformly distributed over

    Question : X is uniformly distributed over (0,10) calculate the probability that
    i) X < 3
    ii) X > 7
    iii) 1 < X < 6
    -------------------------------------------------------------------
    Solution :

    P(x<3) = \frac{1}{3}

    P(X>7) = \frac{1}{3}

    P(1<X<6) = \frac{1}{5}

    Are these answer correct
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  2. #2
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  3. #3
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    Quote Originally Posted by zorro View Post
    Question : X is uniformly distributed over (0,10) calculate the probability that
    i) X < 3
    ii) X > 7
    iii) 1 < X < 6
    -------------------------------------------------------------------
    Solution :

    P(x<3) = \frac{1}{3}

    P(X>7) = \frac{1}{3}

    P(1<X<6) = \frac{1}{5}

    Are these answer correct
    Unfortunately, none of them are correct.

    Draw the distribution (note, and I am assuming that X is continuous, that f(x) = 1/10 for 0 \leq x \leq 10 and zero otherwise) and shade the area represented by each question. The questions boil down to finding areas of rectangles.
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    My work

    Quote Originally Posted by mr fantastic View Post
    Unfortunately, none of them are correct.

    Draw the distribution (note, and I am assuming that X is continuous, that f(x) = 1/10 for 0 \leq x \leq 10 and zero otherwise) and shade the area represented by each question. The questions boil down to finding areas of rectangles.

    -----------------------------My work---------------------------------

    Pdf of uniform distribution

    Pr(x;a,b) = \begin{cases} \frac{1}{b-a} & a \le x < b \\ 0 & otherwise \end{cases}

    Pr(X< 3) = Pr(0 \le X < 3) = \frac{1}{3-0 } = \frac{1}{3}

    Pr(X>7 ) = Pr(7 < X < 10) = \frac{1}{10 - 7} = \frac{1}{3}

    Pr(1 < X < 6) = \frac{1}{6 -1} = \frac{1}{5}

    Please let be know what i have done wrong
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  5. #5
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    Quote Originally Posted by zorro View Post
    -----------------------------My work---------------------------------

    Pdf of uniform distribution

    Pr(x;a,b) = \begin{cases} \frac{1}{b-a} & a \le x < b \\ 0 & otherwise \end{cases}

    Pr(X< 3) = Pr(0 \le X < 3) = \frac{1}{3-0 } = \frac{1}{3}

    Pr(X>7 ) = Pr(7 < X < 10) = \frac{1}{10 - 7} = \frac{1}{3}

    Pr(1 < X < 6) = \frac{1}{6 -1} = \frac{1}{5}

    Please let be know what i have done wrong
    You need to review - very, very thoroughly - what the definitions of a and b are in that formula for the pdf.

    b = 10 and a = 0.
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  6. #6
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    i know b=10 and a=0 but what should i do with that there is no question asking for the interval between 0-10
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  7. #7
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    Quote Originally Posted by zorro View Post
    i know b=10 and a=0 but what should i do with that there is no question asking for the interval between 0-10
    I gave the pdf in my first post (post #3). And you should know how to calculate probability from a pdf.
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  8. #8
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    Is this correct

    Quote Originally Posted by mr fantastic View Post
    I gave the pdf in my first post (post #3). And you should know how to calculate probability from a pdf.

    Pr(X<3)= \int_{0}^{3} \frac{1}{10} dx = \frac{3}{10}

    Pr(X>7) = \int_{7}^{10} \frac{1}{10} dx = \frac{3}{10}

    Pr(1<X<6) = \int_{1}^{6} \frac{1}{10} dx = \frac{1}{2}

    Are these correct ????
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  9. #9
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    Yes they are correct.
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  10. #10
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    Thanks mite for replyiing
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