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Integration involving continuous random variable

  1. Aug 21, 2014 #1
    1. The problem statement, all variables and given/known data
    please refer to the question, i cant figure out which part i did wrongly. i 'd been looking at this repeatedly , yet i cant find my mistake. thanks for the help! the correct ans is below the question. where the c= 283/5700 , q = 179/5700


    2. Relevant equations



    3. The attempt at a solution
     

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  2. jcsd
  3. Aug 21, 2014 #2

    Simon Bridge

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    OK = your strategy was to put $$\int_3^5k(v)\; dv = 33/50 \qquad \int_0^6 k(v)\;dv=1$$ and solve the simultaneous equations.

    This is the correct approach - either the model answer is wrong or you made a mistake in your arithmetic.
     
  4. Aug 21, 2014 #3
    can you please check my working? i looked at it continuously but still couldnt trace my mistake.
     
  5. Aug 21, 2014 #4

    Simon Bridge

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    You have: $$k(v)=\begin{cases} cv+q &: 0\leq v<5 \\ qv+c &: 5\leq v< 6\end{cases}$$ $$p(3\leq v < 5)=\frac{23}{50}$$ You need to find q and c.

    $$p(3\leq v < 5)=\int_3^5 (cv+q)\;dv = \frac{25}{2}c +5q - \frac{9}{2}c-3q =8c+2q =\frac{23}{50}\\ \qquad \implies 400c + 100q = 23 \qquad \text{... (1)}$$

    $$p(0\leq v <6)=\int_0^5 (cv+q)\;dv + \int_5^6 (qv+c)\;dv\\
    \qquad = \frac{25}{2}c + 5q + \left[\frac{36}{2}q + 6c - \frac{25}{2}q - 5c \right] \\
    \qquad = \frac{25+12-10}{2}c+\frac{10+36-25}{2}q = \frac{27}{2}c+\frac{21}{2}q =1\\
    \qquad \implies 27c+21q=2 \qquad \text{... (2)}$$

    ... I'm getting the same as you up to here.
    From here I get the same as the model answers ... so take a closer look at the simultaneous equations.
    What is (23/50)x10.5 =?
     
    Last edited: Aug 21, 2014
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