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Math Help - Empirical/Chebyshev

  1. #1
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    Empirical/Chebyshev

    A sample of 200 people were given a test. The distribution of the test scores was mound-shaped and symmetrical with a mean of 100. One person, whose test score was 125, was found to be at the 84th percentile.

    a)Approximately how many people obtained scores between 65 and 85?


    Using Chebyshev’s theorem, a score of 65 corresponds to a standard deviation of 1.4 and a score of 85 corresponds to a standard deviation of 0.6 but I don’t know where to go from there … I believe my prof told me the answer to this question was 38 people

    b)Suppose the distribution was not mound-shaped, but rather skewed. At least how many people would be expected to obtain scores between 65 and 135?

    This is how I did the problem but my answer (98.97) is a little off from the original answer (98) so I don’t know where I am going wrong with this:
    Z1 = 65-100/25 = -1.4 and Z2 = 135-100/25 = 1.4 so value of k = 1.4
    Using Chebyshev’s theorem, 1 – (1/1.4^2) = 1-0.510 = 0.48979 x 100 = 48.97 + 50 = 98.97


    2)Charles Darwin discovered a strange species of shrubs on a remote island and recorded their heights in inches as follows: “I measures 800 plants. The distribution of their heights is mound-shaped and symmetrical and has a variance of 9. One plant, whose height was 4 inches, falls at the 33rd percentile of the distribution.”

    a)Approx how many of the 800 plants in this sample are between 8.5 and 10.5 inches tall?


    I realize that this question is similar to the one above, but how can I use Chebyshev’s theorem without knowing the mean? The answer to this question is suppose to be 72 plants

    b)Approx what percentage of the sample of 800 plants would be expected to have a Z-score of 1.28? (Assuming plant heights can be positive or negative)
    Answer is 87.78%
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  2. #2
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    Quote Originally Posted by aptiva
    A sample of 200 people were given a test. The distribution of the test scores was mound-shaped and symmetrical with a mean of 100. One person, whose test score was 125, was found to be at the 84th percentile.

    a)Approximately how many people obtained scores between 65 and 85?


    Using Chebyshev’s theorem, a score of 65 corresponds to a standard deviation of 1.4 and a score of 85 corresponds to a standard deviation of 0.6 but I don’t know where to go from there … I believe my prof told me the answer to this question was 38 people
    How have you obtained an (gu)estimate of the standard deviation?
    Are you sure you are not assuming normality?
    State which of Chebyshev's theorems you are using, it is not Chebyshev's
    inequality.


    b)Suppose the distribution was not mound-shaped, but rather skewed. At least how many people would be expected to obtain scores between 65 and 135?

    This is how I did the problem but my answer (98.97) is a little off from the original answer (98) so I don’t know where I am going wrong with this:
    Z1 = 65-100/25 = -1.4 and Z2 = 135-100/25 = 1.4 so value of k = 1.4
    Using Chebyshev’s theorem, 1 – (1/1.4^2) = 1-0.510 = 0.48979 x 100 = 48.97 + 50 = 98.97
    1) the 84th percentile is 1.0364 SD for a normal distribution not 1 SD, so
    the SD~=25/1.0364 (but this is now an invalid assumption).
    2) Chebyshev’s inequality tells us that P(|z|>n*SD)<1/n^2, so the
    probability P(|z|<n*SD)>1/n^2, so a lower bound on the expected number
    with scores between 65 and 135 is 200/(35/(25/1.0364))^2 ~= .96 by my
    reckoning (but using an invalid assumption).


    2)Charles Darwin discovered a strange species of shrubs on a remote island and recorded their heights in inches as follows: “I measures 800 plants. The distribution of their heights is mound-shaped and symmetrical and has a variance of 9. One plant, whose height was 4 inches, falls at the 33rd percentile of the distribution.”

    a)Approx how many of the 800 plants in this sample are between 8.5 and 10.5 inches tall?


    I realize that this question is similar to the one above, but how can I use Chebyshev’s theorem without knowing the mean? The answer to this question is suppose to be 72 plants
    You are told that the variance is 9, and you are told a percentile, you
    are expected to use this information (and a normality assumption) to
    calculate the mean.


    I must say that I'm not very happy answering this problem. There seems
    to be context information missing.

    RonL
    Last edited by CaptainBlack; May 9th 2006 at 01:18 AM.
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  3. #3
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    A sample of 200 people were given a test. The distribution of the test scores was mound-shaped and symmetrical with a mean of 100. One person, whose test score was 125, was found to be at the 84th percentile.

    a)Approximately how many people obtained scores between 65 and 85?
    ANS: 38 people

    In the question it says that the mean is 100 and that one person got a score of 125 which was found to be at the 84th percentile.

    So by the empirical rule (sorry not Chebyshev as i originally stated), 68% of the distribution is between -1s and +1s around the mean so half of that (34%) is betweent he mean and 1s above the mean so 50% + 34% = 84% which would mean that the std. dev. is 25

    and so using the z-score, I find that 65-100/25 = 1.4 so a score of 65 corresponds to a standard deviation of 1.4 and a score of 85 corresponds to a standard deviation of 0.6 using the same formula

    after this i kind of get stuck..
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  4. #4
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    Quote Originally Posted by aptiva
    A sample of 200 people were given a test. The distribution of the test scores was mound-shaped and symmetrical with a mean of 100. One person, whose test score was 125, was found to be at the 84th percentile.

    a)Approximately how many people obtained scores between 65 and 85?
    ANS: 38 people

    In the question it says that the mean is 100 and that one person got a score of 125 which was found to be at the 84th percentile.

    So by the empirical rule (sorry not Chebyshev as i originally stated), 68% of the distribution is between -1s and +1s around the mean so half of that (34%) is betweent he mean and 1s above the mean so 50% + 34% = 84% which would mean that the std. dev. is 25

    and so using the z-score, I find that 65-100/25 = 1.4 so a score of 65 corresponds to a standard deviation of 1.4 and a score of 85 corresponds to a standard deviation of 0.6 using the same formula

    after this i kind of get stuck..
    Try using this normal distribution calculator to see the next step here. Plug in mean 100, SD 25, select "Between" and enter the bounds 65 and 85. The calculator gives the probability as .1934, which multiplied by 200 yields 38.68.

    You can also use mean 0, SD 1 and plug in your Z-scores of -1.4 = (65-100)/25 and -.6 = (85-100)/25 with the same result.
    Last edited by JakeD; May 9th 2006 at 03:01 PM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by aptiva
    A sample of 200 people were given a test. The distribution of the test scores was mound-shaped and symmetrical with a mean of 100. One person, whose test score was 125, was found to be at the 84th percentile.

    a)Approximately how many people obtained scores between 65 and 85?
    ANS: 38 people

    In the question it says that the mean is 100 and that one person got a score of 125 which was found to be at the 84th percentile.

    So by the empirical rule (sorry not Chebyshev as i originally stated), 68% of the distribution is between -1s and +1s around the mean so half of that (34%) is betweent he mean and 1s above the mean so 50% + 34% = 84% which would mean that the std. dev. is 25

    and so using the z-score, I find that 65-100/25 = 1.4 so a score of 65 corresponds to a standard deviation of 1.4 and a score of 85 corresponds to a standard deviation of 0.6 using the same formula

    after this i kind of get stuck..
    As I said before the 84-th percentile corresponds to 1.0364 ~ 1.04 standard
    deviations for a normal distribution -not that that is important given the
    probable size of the error in estimating the SD this way.

    RonL
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