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Thread: Help with stats concepts. Help. Statistics overwhelm me.

  1. #1
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    Help with stats concepts. Help. Statistics overwhelm me.

    1. A corporation has four investment possibilities: A, B, and C with respective gains of $10, $20, and $50 million and investment D with a loss of $30 million. If one investment will be made and the probabilities of choosing A, B, C, or D are .1, .4, .2, and .3, respectively, find the expected gain for the corporation.


    2.. A marketing research survey shows that approximately 80% of the car owners surveyed indicated that their next car purchase will be either a compact or economy car. Assume the 80% figure is correct and five prospective buyers are interviewed. a. Find the probability that all five indicate that their next car purchase will be either a compact or economy car. b. Find the probability that at most one indicates that her next purchase will be either a compact or economy car.
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    Quote Originally Posted by bombo31
    1. A corporation has four investment possibilities: A, B, and C with respective gains of $10, $20, and $50 million and investment D with a loss of $30 million. If one investment will be made and the probabilities of choosing A, B, C, or D are .1, .4, .2, and .3, respectively, find the expected gain for the corporation.
    My intution telle me you need to find,
    $\displaystyle (.1)10+(.4)20+(.2)50-.3(30)=1+8+10-9=10\mbox{ million}$
    Cuz since 10% probability means you are expected to get 10% of A which means 10% of 10 million, similarly the other 3. Then you need to add them up to see what they give, while remember to subtract D
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    Quote Originally Posted by bambi31
    2.. A marketing research survey shows that approximately 80% of the car owners surveyed indicated that their next car purchase will be either a compact or economy car. Assume the 80% figure is correct and five prospective buyers are interviewed. a. Find the probability that all five indicate that their next car purchase will be either a compact or economy car.
    The probability of the first person saying yes is .8
    The probability of two people is $\displaystyle .8\times .8$
    The probability of three $\displaystyle .8\times .8\times.8$
    The that all five is $\displaystyle (.8)^5\approx .33$

    Quote Originally Posted by bambi31
    b. Find the probability that at most one indicates that her next purchase will be either a compact or economy car.
    Find the probability that NONE will says yes. That is,
    $\displaystyle (.2)^5$ because given that probability is .8 for saying yes then the probability of saying no is 1-.8=.2 since there are five people you rasie to the fifth az in the first question. But you need at most 1 which is the same as saying what is the probability of this (no one saying yes) will not happen which is,
    $\displaystyle 1-(.2)^5=.99968$
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  4. #4
    MHF Contributor Quick's Avatar
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    Question 2


    Find the probability that NONE will says yes. That is,
    (.2)^5 because given that probability is .8 for saying yes then the probability of saying no is 1-.8=.2 since there are five people you rasie to the fifth az in the first question. But you need at most 1 which is the same as saying what is the probability of this (no one saying yes) will not happen which is,
    1-(.2)^5=.99968
    The question was not what are the chances that at least one will say yes but AT MOST one will say an yes. Heres what I get
    $\displaystyle \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}=\frac{5}{3125} = 0.0016 = 0.16$%
    Last edited by Quick; Jun 3rd 2006 at 07:04 PM.
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  5. #5
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    Quote Originally Posted by Quick
    The question was not what are the chances that at least one will say yes but AT MOST one will say an yes. Heres what I get
    $\displaystyle \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}=\frac{5}{3125} = 0.0016 = 0.16$%
    You want:

    $\displaystyle P((n_y=0) \lor (n_y=1))=P(n_y=0)+P(n_y=1)$

    Now:

    $\displaystyle P(n_y=0)=0.2^5\approx 0.00032$,

    and:

    $\displaystyle P(n_y=1)={5 \choose 1}\ 0.2^4\ 0.8=5\ 0.2^4\ 0.8\approx 0.0064$

    Therefore:

    $\displaystyle P((n_y=0) \lor (n_y=1))\approx 0.00672$,

    or $\displaystyle \approx 0.67 \%$

    RonL
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    MHF Contributor Quick's Avatar
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    Why

    Well, it appears I forgot to multiply by five.
    Last edited by Quick; Jun 4th 2006 at 03:55 AM.
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    Quote Originally Posted by Quick
    Well, it appears I forgot to multiply by five.
    Because you only considered one case, namely,
    YNNNN

    But there are 4 more:
    NYNNN
    NNYNN
    NNNYN
    NNNNY

    This is why you need to multiply by five.
    ----
    Note, in may probability exacmple you need to use binomial coeificcients because of this reason.
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  8. #8
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    Thank you

    Sorry about the double posts. I did not realize, and thanks for your help.
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