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Thread: Sampling distribution of the sample proportion.

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    Sampling distribution of the sample proportion.




    A random sample of 20 students is selected. What is the probability that the sample
    proportion exceeds 75%, i.e., more than 15 students (out of 20 selected) will agree that
    that the course is interesting? You may leave your answer as, for example, 1 − 0.88. (As a
    standard error use 0.8. Also, in your calculations instead of 1/8 = 0.125 you may use 0.13).
    Solution: given by professor

    1. Sampling distribution of the sample proportion.
    P(p >b 0.75) = P (Z >( 0.75 − 0.7 )/0.8)= P (Z > 0.25) = 1 − PZ ≤ 0.25= 1 − 0.5987 = 0.4013.

    2. Normal approximation to a Binomial distribution.
    Let X be the number of students in the sample who find the course interesting, then
    the mean of X is 14.
    P(X ≥ 15) = P ((Z ≥ 14.5 − 14 )/0.8) = P (Z ≥ 0.65 )= 1 − P( Z < 0.65) = 1 − 0.7422 = 0.2578

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    Re: Sampling distribution of the sample proportion.


    Hi! We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.
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