The columns become narrower, their heights become more accurate but possibly more variable. The chart contains more of the underlying detailed information.
Class interval width increases.
central tendency become more obvious
central tendancy gets more obvious
number of classes
highest value-lowest value/number of classes
you have to find the class size by: (max-min)/number of classes Then use that class size to setup the class ranges Then use the class ranges to determine the frequency a sample occurs in each class. make a chart using the class ranges and the sample frequencies to display the histogram
central tendency become more obvious
central tendancy gets more obvious
number of classes
highest value-lowest value/number of classes
you have to find the class size by: (max-min)/number of classes Then use that class size to setup the class ranges Then use the class ranges to determine the frequency a sample occurs in each class. make a chart using the class ranges and the sample frequencies to display the histogram
4
try sqrt(N) where N represents the number of observations you have...
5
Increasing the temperature the number of particles remain constant and the pressure increase.
i think you divide the histogram in two, so there are two equal halves. The number in the middle is the median,
All that histogram equalization does is remap histogram components on the intensity scale. To obtain a uniform (­at) histogram would require in general that pixel intensities be actually redistributed so that there are L groups of n=L pixels with the same intensity, where L is the number of allowed discrete intensity levels and n is the total number of pixels in the input image. The histogram equalization method has no provisions for this type of (arti®cial) redistribution process.
They increase