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Math Help - need help with these Q's please

  1. #1
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    need help with these Q's please

    Players select one chip froma bag with 8 zerosand 4 ones and one chip froma bag with 4 zeros and 8 ones. Their score is the larger of the two numbers on the chips.

    Present the probability distribution of the scores that could be achieved by a player in one game and compute its mean. [mu = 0.78]

    __________________________________________________ ______________

    In 1979,a random sample of 200 families in a large city was found to spend an average of $85.44 per week on food, with a standard deviation of $8.12.

    a) Construct a 90% confidence interval to estimate the mean weekly expenditure on food in the population. [84.50 to 86.38]

    b) Construct a 99% confidence interval to estimate the mean weekly expenditure on food in the population. [83.96 to 86.92]
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by aptiva
    Players select one chip froma bag with 8 zerosand 4 ones and one chip froma bag with 4 zeros and 8 ones. Their score is the larger of the two numbers on the chips.

    Present the probability distribution of the scores that could be achieved by a player in one game and compute its mean. [mu = 0.78]
    <br />
p(0)=prob(\mbox{0 from bag 1})prob(\mbox{0 from bag 2}) =\frac{8}{12} \frac{4}{12}\approx 0.2222<br />

    <br />
p(1)=1-p(0)<br />


    Mean score:

    <br />
\mu=0\times 0.2222+1 \times 0.7778<br />
.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by aptiva

    In 1979,a random sample of 200 families in a large city was found to spend an average of $85.44 per week on food, with a standard deviation of $8.12.

    a) Construct a 90% confidence interval to estimate the mean weekly expenditure on food in the population. [84.50 to 86.38]
    The critical z-scores for a symmetric interval containing 90% of a normal
    distribution are \pm 1.645.

    The mean of a sample of 200 drawn from a population with mean \$ 85.44
    is \$ 85.44. The standard deviation of the mean of a sample of 200 drawn
    from a population with standard deviation of \$ 8.12 is \$ 8.12/\sqrt{200}=\$ 0.57.

    Therefore the z-scores of the end points of the required interval are:

    <br />
z_{lo}=(x_{lo}-85.44)/0.57=-1.645<br />

    and:

    <br />
z_{hi}=(x_{hi}-85.44)/0.57=+1.645<br />

    So the required interval is (x_{lo},x_{hi})=(84.50,86.38)


    b) Construct a 99% confidence interval to estimate the mean weekly expenditure on food in the population. [83.96 to 86.92]
    This part is exactly the same except you will be working with the critical
    z-scores for a 99% region.

    RonL
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