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Math Help - Probability and proportions?

  1. #1
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    Probability and proportions?

    I can't work out how to do this question. It doesn't look like anything we've done before, but I don't know if I'm just misreading it. I'm assuming it has something to do with binomial distributions, but I'm not sure. Any help is appreciated!


    A road is constructed so that the right-turn lane at an intersection has a capacity of 3 cars. Suppose that 30% of cars approaching the intersection want to turn right, and they do so independently.

    a) If 15 cars approach the intersection, what is the probability that the right-turn lane will not hold all the cars wanting to turn right?

    b) Over a week, 1428 cars used the intersection. What is the distribution of the sample proportion of cars turning right at this intersection, in random samples of this size?
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  2. #2
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    Quote Originally Posted by mightysparks View Post
    I can't work out how to do this question. It doesn't look like anything we've done before, but I don't know if I'm just misreading it. I'm assuming it has something to do with binomial distributions, but I'm not sure. Any help is appreciated!


    A road is constructed so that the right-turn lane at an intersection has a capacity of 3 cars. Suppose that 30% of cars approaching the intersection want to turn right, and they do so independently.

    a) If 15 cars approach the intersection, what is the probability that the right-turn lane will not hold all the cars wanting to turn right?

    b) Over a week, 1428 cars used the intersection. What is the distribution of the sample proportion of cars turning right at this intersection, in random samples of this size?
    a) Binomial(n=15, p=3/10). You want P(X>3).
    b) Binomial(n=1428, p=3/10)
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  3. #3
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    So a) is just P(X>3) = .2969?

    and for b) is B(1428, 3/10) all I have to write? These answers seem too simple, which is why I questioned it in the first place.

    Thanks.
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    Or is it 1-P(X<3) = 1 -.2969 = .7031?
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  5. #5
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    it is the 0.7031

    b) i) Binomial(n=1428, p=3/10)
    that is not the case, it is not a binomial distribution it is a normal distribution
    ii) you need to work out the SD=square root of variance; the mean=np
    as well as the Z score

    i just have a quick question for you....
    b) i) how do you expect the sample mean brain weight to be distributed for random sample of size 20 from this population?
    (i think that i need to use the central limit theorem for this answer, i just do not know how to word it)

    ii) in such samples, how unusual would it be to find a mean brain weight in excess of 1470g?

    can you help?
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