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Math Help - Random variables

  1. #1
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    Random variables

    1.
    X is a random variable which can take the following values: (This is a table, c/2 means c over 2)
    x.............0....1.....2...3....4
    P(X=x)....c/2...c....2c..c...2/c

    State the conditions for P(X = x) to be a probability distribution, then find the value of c for which this will be the case.

    2.
    The random variable X represents the number of visitors a hospital patient has in a visiting period. A maximum of four visitors is permitted. C is found to have probability distribution:
    x...........0.......1......2......3.......4
    P(X=x)..0.1....0.2....0.4....0.2....0.1

    Show that E[X] = 2 and E [X] = 5.2, then write down the probability distribution of the random variable Y = (X-2) and hence, or otherwise, find E[(X-2)].
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Kiwigirl
    1.
    X is a random variable which can take the following values: (This is a table, c/2 means c over 2)
    x.............0....1.....2...3....4
    P(X=x)....c/2...c....2c..c...2/c

    State the conditions for P(X = x) to be a probability distribution, then find the value of c for which this will be the case.
    For P(X=x) to be a probability distribution we need:

    <br />
\sum_{x} P(X=x)=1<br />

    Which in this case is:

    <br />
\sum_0^4 P(X=x)=c/2+c+2c+c+2/c=\frac{9c^2+4}{2c}=1<br />

    So 9c^2-2c+4=0, but this has no roots so no value of c
    will make P(X=x) a probability distribution.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Kiwigirl
    2.
    The random variable X represents the number of visitors a hospital patient has in a visiting period. A maximum of four visitors is permitted. C is found to have probability distribution:
    x...........0.......1......2......3.......4
    P(X=x)..0.1....0.2....0.4....0.2....0.1

    Show that E[X] = 2 and E [X] = 5.2, then write down the probability distribution of the random variable Y = (X-2) and hence, or otherwise, find E[(X-2)].
    By definition the expectation of a discrete RV R is:

    <br />
E(R)=\sum r_i\ P(R=r_i)<br />

    So:

    <br />
E(X)=0 \times 0.1+1\times 0.2+2 \times 0.4+3 \times 0.2+4\times 0.1=2<br />

    and:

    <br />
E(X^2)=0 \times 0.1+1\times 0.2+4 \times 0.4+9 \times 0.2+16\times 0.1=5.2<br />

    The probability distribution of Y=(X-2)^2 is:

    <br />
\begin{array}{cccc}y&0&1&4\\p(Y=y)&0.4&0.4&0.2\end  {array}<br />

    So:

    <br />
E(Y)=E((X-2)^2)=0\times 0.4+1 \times 0.4 +4\times 0.2=1.2<br />

    RonL
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