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  • one year ago

What is the mean of the given distribution, and which type of skew does it exhibit? {4.5, 3, 1, 2, 4, 3, 6, 4.5, 4, 5, 2, 1, 3, 4, 3, 2}

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  1. Ciarán95
    • one year ago
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    To find the mean of the set of values, you can calculate: Mean = (Sum of all the values)/Number of values When we refer to a set of data being 'skewed' , we would typically be saying that its distribution about this mean value is not symmetric. We can identify this by drawing a histogram of the data, but we can also numerically identify it through the median. |dw:145674332872:dw| When the distribution is symmetric, the middle value (the median) will be approximately equal to the mean. However, if the data set is skewed, then it contains outliers or unusual values which do not follow the natural pattern of the majority of the values present (An example of such a set would be 2, 5, 6, 3, 8, 6, 4, 18, 31 - the last two values are outliers). These unusual values will tend to 'pull' the mean towards them - the bigger they are compared to the rest of the data, the more the mean gets shifted towards larger values. The same would happen for unusually small values being present, which would shift the mean to a lower value compared to the median. The reason why the mean is affected by these outliers so much compared to the median is that we use the specific value when calculating it. We don't do this when dealing with the median, which works by identifying the 'middle' value in the data. If we have 'n' values, where n is an odd number we order them from smallest to largest and take the (n+1)/2 value in the ordering (i.e. in a set of 5 ordered numbers, we would take the third value to be the median). In a set of 'n' ordered numbers, where n is even, we look at the n/2 and (n+2)/2 value. So, for the set of 16 values in the question, we would identify the median by taking the 8th and 9th values in the ordered set and getting the mean of these. If: Mean > Median, then the distribution of the data is right-skewed. Mean < Median, then the distribution of the data is left-skewed. Hope that helps! :)

  2. Ciarán95
    • one year ago
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