In statistics, a histogram is a graphical representation of tabular frequencies, shown as adjacent rectangles. Each rectangle is stretched over interval, with an area equal to the frequency of the interval. The attitude of a rectangle is equal to the frequency density of the interval, i.e. the frequency is made to divide by the width of the interval. The sum of the area of histogram is equal to the total number of data.
Histogram standard deviation:
Histogram is nothing but the plotting of the data given in an order. If the x-axis has the value 1 then the histograms are said to be relative frequency plotting. To construct the histogram, the rectangles or blocks are made on the x-axis without leaving any space between the blocks are rectangles. The usage of the histograms are found were the values given are in discontinues manner. In other words the histogram can be named as frequency.
I like to share this Histogram Analysis with you all through my article.
The standard deviation explains you how the number of examples is clustered around the mean of the set of data. The computation of the value of the standard deviation is little complicated. In other words the square root of variance is named as standard deviation and the historical volatility. The standard deviation is used as the gauge for the amount of expected volatility.
Example problems for Histogram standard deviation:
Standard deviation from histograms in a workload management configuration
Example 1 : Count1 count2 count3 count4 count5 count6 count7 count8
20 30 80 10 5 3 2 0
To compute the average lifetime, count8 must be empty.count8 only exists to makes you to know the upper range of the data. For this reason, you must specify the upper bound for the range.
Average = (20 x 1 + 30 x 3 + 80 x 6 + 10 x 12 + 5 x 24 + 3 x 48 + 2 x 96) / 150
= (20 + 90 + 480 + 120 + 120 + 144 + 192) / 150
= 1166 / 150
= 7.77 seconds
Standard deviation = [(20 x (1 - 7.77)2 + 30 x (3 - 7.77)2 +.) / 150]1/2
Example 2:
Data base partition count1 count2 count3 count4 count5 count6 count7 count8
1 20 30 80 10 5 3 2 0
2 1 5 20 20 4 0 0 0
Since the count sizes are the same across all database partitions, the overall histogram is easy to compute:
Count1 count2 count3 count4 count5 count6 count7 count8
21 35 100 30 9 3 2 0
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The combined histogram, you can calculate the overall lifetime average and standard deviation.
Average lifetime = (21 x 1 + 35 x 3 + 100 x 6 + 30 x 12 + 9 x 24 + 3 x 48 + 2 x 96) / 200
= (21 + 105 + 600 + 360 + 216 + 144 + 192) / 200
= 1638 / 200
= 8.19 seconds
Standard deviation = [(21 x (1 – 8.19)2 + 35 x (3 - 7.77)2 + ) / 200]1/2
histogram
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