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Math Help - Continuous random variable

  1. #1
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    Continuous random variable

    The amount of vegetables eaten by a family in a week is a random variable W kg. The probability density function is given by:

     f(w) = \begin{cases} \frac {20}{5^5}w^3(5-w), &\text {$0 \leq w \leq 5$,} \\ 0, & \text {otherwise.} \end{cases}

    (a) Find the cumulative distribution function of W.
    (b) Find the probability that the family eats between 2kg & 4kg of vegetables in one week.
    (c) Verify that the amount, m, of vegetables such that the family is equally likely to eat more of less than m in any week is about 3.431 kg.

    I have done part (a) & (b), and the answer of (b) is 0.650. But how could I verify (c)?
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  2. #2
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    Quote Originally Posted by geton View Post
    The amount of vegetables eaten by a family in a week is a random variable W kg. The probability density function is given by:

     f(w) = \begin{cases} \frac {20}{5^5}w^3(5-w), &\text {$0 \leq w \leq 5$,} \\ 0, & \text {otherwise.} \end{cases}

    (a) Find the cumulative distribution function of W.
    (b) Find the probability that the family eats between 2kg & 4kg of vegetables in one week.
    (c) Verify that the amount, m, of vegetables such that the family is equally likely to eat more of less than m in any week is about 3.431 kg.

    I have done part (a) & (b), and the answer of (b) is 0.650. But how could I verify (c)?
    I agree with your answer to (b). As for (c), you are asked to check that P(W\leq m)\simeq P(W\geq m) for m=3.421\,{\rm kg}. Since the two probabilities sum to 1 (justify), this amounts to checking that P(W\leq m)\simeq\frac{1}{2} (I wrote \simeq since m is "about" 3.421 kg). You have to compute an integral, like in (b).
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