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Stem and leaf display statistics

  1. Jul 21, 2008 #1
    I am working on statistics assignment and since statistics is not my primary field of studies I have difficulties with it. can someone please help me out?? I did stem and leaf display which I think looks OK but am not sure.

    1. The problem statement, all variables and given/known data

    The following data represent the weights (in kilograms) of a sample of 25 horses:

    164 148 137 157 173 156 177 172 169 165
    145 168 163 162 174 152 156 168 154 151
    174 146 134 140 171

    Use these data to answer the following questions:
    a. Construct a stem-and-leaf display for the weights.
    b. Find the median weight.
    c. Compute the sample mean weight.
    d. Compute the sample variance and sample standard deviation.
    e. Construct a frequency distribution for the data, using five class intervals and the value 130 as the lower limit of the first class.



    2. Relevant equations

    it possible for the standard deviation of a data set to be larger than its variance? Explain.


    3. The attempt at a solution
     
  2. jcsd
  3. Jul 21, 2008 #2

    HallsofIvy

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    Re: Statistics

    What have you done with these? Don't you have any work to show?

    As for your final question, what are the definitions of "variance" and "standard deviation"? Under what conditions can one number be larger than the other?
     
  4. Jul 21, 2008 #3
    Re: Statistics

    I first sorted numbers from smallest to the largest. Based on that I believe median weight is 162kg because is the 13th in the row. My stem and leaf looks something like this;

    STEM LEAF
    13 4,7
    14 0,5,6,8
    15 1,2,4,6,6,7
    16 2,3,4,5,8,8,9
    17 1,2,3,4,4,7

    Mean weight I calculated by adding all numbers together and then deviding by 25 which give me a sum of 159.04kg.
    This is where I got stuck. Any suggestions??
     
  5. Jul 21, 2008 #4

    HallsofIvy

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    Re: Statistics

    Okay, good so far! For the last one, "Construct a frequency distribution for the data, using five class intervals and the value 130 as the lower limit of the first class." I would note that all the data lies between 130 and 180 and that 180- 130= 50 so that "five class intervals" would divide that into 10 pound intervals. Basically, you can just look at your stem and leaf chart to see how many data points lie in each interval.

    The only things left then are the variance and standard distribution. Again, what are their definitions?
     
  6. Jul 21, 2008 #5
    Re: Statistics

    HallsofIvy,

    Thanks for your help so far. At least I got something right. My understanding is that standard deviation is a measure of the set values. I think I should deduct mean average, which in this case is 159, from each horse to get deviation of each number from the mean.
    Each deviation should then be squared to get any negative values positive.

    Next step (I think) is finding mean of those squared deviations by adding all deviations and deviding by 25. Square root of that would be standard deviation. Does this makes any sense??
     
  7. Jul 22, 2008 #6

    HallsofIvy

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    Re: Statistics

    This is meaningless- every statistic is "a measure of the set values"!

    The variation is defined as the sum of the squares of differences from the mean divided, in this case, by 25. The standard deviation is the square root of the variation.
     
    Last edited: Jul 23, 2008
  8. Jul 23, 2008 #7
    Re: Statistics

    Hi HallsofIvy,

    Thanks again for your valuable input. This is just the begining of my strugle and there is worst to come. Luckilly I have only one unit of statistics to study which is bad enough.

    Thanks again and good luck.
     
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