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Math Help - binomial

  1. #1
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    binomial

    3. Suppose thirty-five percent of adults did not visit their physicians' offices last year. Let x be the number of adults in a random sample of 40 adults who did not visit their physicians' offices last year. For the probability distribution of x:
    a] The mean is: _______ because: ______________________
    b] The standard deviation is: ________ because: ____________________
    NOTE: HINT: use the t statistic to answer problem 4 (you are working with a small sample, n=12)
    4.Construct a 99% confidence interval for the mean of the population of all statistics students from which the sample in problem 1 was drawn.
    5.According to the American Federation of Teachers, the average salary of 1998 college graduates with a degree in accounting was $33,702 (USA Today, June 18, 1999). Assume that the distribution of current salaries of recent graduates with a degree in accounting is skewed to the right with a mean of $33,702 and a standard deviation of $3,900. Find the probability that the mean current salary of a random sample of 200 recent college graduates with a degree in accounting is within $400 of the population mean.

    thank you for any help and suggestion in advance.
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  2. #2
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    Quote Originally Posted by mathdummy View Post
    3. Suppose thirty-five percent of adults did not visit their physicians' offices last year. Let x be the number of adults in a random sample of 40 adults who did not visit their physicians' offices last year. For the probability distribution of x:
    a] The mean is: _______ because: ______________________
    b] The standard deviation is: ________ because: ____________________
    This is a question on the application of the binomial distribution.

    The number in the sample who did not visit the doctor last year is ~B(n;40,0.35).

    The mean of X~B(n,N,p) is np, and the standard deviation is sqrt(Np(1-p)).

    RonL
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  3. #3
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    Quote Originally Posted by mathdummy View Post
    NOTE: HINT: use the t statistic to answer problem 4 (you are working with a small sample, n=12)
    4.Construct a 99% confidence interval for the mean of the population of all statistics students from which the sample in problem 1 was drawn.
    Can't be done without the data from question 1.

    RonL
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  4. #4
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    Quote Originally Posted by mathdummy View Post
    5.According to the American Federation of Teachers, the average salary of 1998 college graduates with a degree in accounting was $33,702 (USA Today, June 18, 1999). Assume that the distribution of current salaries of recent graduates with a degree in accounting is skewed to the right with a mean of $33,702 and a standard deviation of $3,900. Find the probability that the mean current salary of a random sample of 200 recent college graduates with a degree in accounting is within $400 of the population mean.
    You are expected to use the result that a sample mean is approximatly normally distributed with mean equal to the population mean, and sd = population sd/sqrt(N), N the sample size, for large enough samples. (this is the Central Limit Theorem)

    So assume the sample mean m~N(33900, (3900^2)/200), so now you need to find the probability p(-400<m-33900<400), or dividing through by 3900/sqrt(N) that:

    p(-1.450475<z<1.450475), where z=[m-33900]/[3900/sqrt(200)], has a standard normal distribution.

    RonL
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  5. #5
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    data

    1. The test grades for twelve students in a certain statistics class are:
    87, 60, 72, 98, 99, 59, 67, 94, 92, 89, 93, 100
    a] The mode is Not Available because there are no repetitive pattern within the number set.
    b] The mean is 84.167 because this is the calculated result of the mean of this number set.
    c] The std. dev. is 15.3435 (show the formula you used)
    d] The median is 90.5 because this is the mid range of the number within this set.
    e] The range is 41 because this is the highest value and least value gap within the set.
    f] The variance is 235.424 because it is the square of the standard deviation.
    g] The 35th percentile is _____ because___________________________________________ _
    this is the data set for number 4 what information should i use to solve the problem?
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  6. #6
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    many thanks

    wow thank you that was so helpful. really appreciate the pointers. well its a lot more than just pointers thanks
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