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Math Help - Help would be appreciated..........

  1. #1
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    Help would be appreciated..........

    (i)Suppose 10% of the population of a city said they agreed with a new speed limit. In a poll of 400 randomly chosen people from that city, what is the probability that fewer than 7.5% of them will say they agree with the new speed limit?



    (ii)Suppose that X is a normal random variable with unknown mean and variance of 4, if Pr[X>20] = .1056, what is the mean?
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  2. #2
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    Quote Originally Posted by rebeccajm
    (i)Suppose 10% of the population of a city said they agreed with a new speed limit. In a poll of 400 randomly chosen people from that city, what is the probability that fewer than 7.5% of them will say they agree with the new speed limit?
    You could restate it as "what is the probability that <30 say they agree with the new spped limit?".

    .075(400)=30

    {\sigma}=\sqrt{npq}\;\ and\;\ {\mu}=np

    Use z=\frac{x-{\mu}}{\sigma}

    Don't forget the continuity correction for x. In a binomial distribution, the possible midpoint values for "fewer than 30" are ....27, 28, 29.
    Add 0.5 to the right hand boundary to get x=29.5.

    Look up the z-score in the table.
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