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Math Help - Questions I'm struggling with Part 1

  1. #1
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    Question Questions I'm struggling with Part 1

    1.)

    The table below gives the number of people who attended the U.S. Open tennis tournament in Flushing Meadows for the past 7 years. However, due to a typing error the attendance for 2005 was entered incorrectly and should have been 631,870. Which measure of variability remains unchanged after correcting the data set?

    Year Attendance
    2007 715, 587
    2006 640,000
    2005 659,538
    2004 531,870
    2003 615,456
    2002 628,738
    2001 639,343

    A.) Variance B.) Standard deviation C.) No information provided D.) Range

    2.) A study on the IQ's of 300 six year-old children yielded the following frequency distribution. Find the estimated variance of the IQ's for the six-year-olds.

    IQ Frequency
    150-159 5
    140-149 9
    130-139 19
    120-129 37
    110-119 79
    100-109 69
    90-99 65
    80-89 17 the frequencys are the numbers that are not devided with a dash. example 5, 9, 19, 37 etc

    3.) The fifth percentile for wait times at emergency rooms across the 50 states in the United States of America was 135 minutes, the fiftieth percentile was 175 minutes, and the ninety-fifth percentile was 215 minutes. Determine the shape of the distribution of wait times in an emergency room.

    A.) Symmetric B.) Positively skewed C.) Negatively Skewed

    D.) Unknown
    Last edited by helpmeplease2323; October 25th 2009 at 11:39 AM.
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  2. #2
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    1. Answering this problem requires you to know how to solve for each of those parameters. The variance is dependent on the mean, and the mean is dependent on your attendance, so thats not going to remain the same. The standard deviation is just the square root of the variance, so again - wont remain the same. The range is the difference between the lowest and the highest value - as the incorrect entry is neither the highest or the lowest, the range will not change.

    2. Do you know how to determine the mean from a frequency distribution?

    \overline{x}=\frac{\Sigma(f*x)}{\Sigma(f)};

    Where x is the midpoint of your classes (150-159 would be 154.5), and f is the frequency of each class - so (154.5)(5).

    3. Do you know how percentile works? I can tell you the data is symetric. They tell you that that 5th Percentile is 135 minutes, and the 50th Percentile (aka, the mean) is 175, and the 95th Percentile is 215 minutes. If we were constructing a graph, with a scale from 0 to 100, 5 and 95 would be an equal distance from the left and right edges yes? So for that graph to be symetric, an equal number of "stuff" would have to be between 5 and 50, as there is between 95 and 50. Is there?
    Last edited by ANDS!; October 26th 2009 at 08:20 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ANDS! View Post
    A. Answering this problem requires you to know how to solve for each of those parameters. The variance is dependent on the mean, and the mean is dependent on your attendance, so thats not going to remain the same. The standard deviation is just the square of the variance, so again - wont remain the same. The range is the difference between the lowest and the highest value - as the incorrect entry is neither the highest or the lowest, the range will not change.
    Typo:

    The standard deviation is just the square root of the variance

    CB
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ANDS! View Post
    A. Answering this problem requires you to know how to solve for each of those parameters. The variance is dependent on the mean, and the mean is dependent on your attendance, so thats not going to remain the same. The standard deviation is just the square of the variance, so again - wont remain the same. The range is the difference between the lowest and the highest value - as the incorrect entry is neither the highest or the lowest, the range will not change.

    B. Do you know how to determine the mean from a frequency distribution?

    \overline{x}=\frac{\Sigma(f*x)}{\Sigma(f)};

    Where x is the midpoint of your classes (150-159 would be 154.5), and f is the frequency of each class - so (154.5)(5).

    C. Do you know how percentile works? I can tell you the data is symetric. They tell you that that 5th Percentile is 135 minutes, and the 50th Percentile (aka, the mean) is 175, and the 95th Percentile is 215 minutes. If we were constructing a graph, with a scale from 0 to 100, 5 and 95 would be an equal distance from the left and right edges yes? So for that graph to be symetric, an equal number of "stuff" would have to be between 5 and 50, as there is between 95 and 50. Is there?
    Please quote the post you are answering, and cut out the irrelevant questions.

    In particular tell us which question you are answering

    CB
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