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Assume your data are the number of car accidents and

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Unformatted text preview: cel follow the route Data ‐> Data Analysis ‐> Histogram • Select the data as “Input Range” • Select a free cell for “Output Range” • Mark “Pareto (sorted histogram)” and “Chart Output” and click “OK” You will obtain a pareto chart and the underlying ordered frequency distribution as follows: Histogram Frequency Bin Frequency 184,6667 37 110 2 More 1 259,3333 0 334 0 408,6667 0 483,3333 0 1,5 1 0,5 0 Frequency Bin 02 March 2011 You can quickly observe that a pareto chart is simply an ordered histogram in which you can see most frequently occurring data first. Assume your data are the number of car accidents and the speed of the cars during accident. In such a case, bins can represent the speed ranges and the bars of the diagram represent the number of accidents. Then in a pareto chart we can quickly see which speed ranges are more critical in the occurrences of accidents. To construct a relative frequency histogram: We have N = 40 data in our sample. From the frequencies that we have found in drawing the histogram, we can calculate the relative frequencies with the Excel formula: Frequency * 100 / N and list tem under new column Relative Frequency Distribution. Then we can obtain relative frequency histogram as follows: • • • • • In Excel follow the route Insert ‐> Column ‐> 2‐D Column‐> Clustered Column A new tab “Design” will automatically open To identify the input data click “Select Data” Choose the data under column “Relative Frequency Distribution” as “Chart data range” Click “OK” Bin Frequency 110 2 184,6667 37 259,3333 0 334 0 408,6667 0 483,3333 0 More 1 Relative Frequency Distribution 5,00% 92,50% 0,00% 0,00% 0,00% 0,00% 2,50% 100,00% 80,00% 60,00% Series1 40,00% 20,00% 0,00% 1 2 3 4 5 6 7 (c) What feature in this data set is responsible for the substantial difference between mean and median? Mean and median are two different location measures of a sample. Mean estimates the location of the sample by considering all data, however median takes into account only the data points that are in the middle of the ordered sequence. Observe that there is an outlier in the given data, namely 558,...
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