Frequency Density and Histogram Calculator

What is frequency density?

Frequency density is the frequency of a class divided by its class width: frequency density = frequency ÷ class width. It goes on the vertical axis of a histogram, and that makes the area of each bar equal to the frequency of its class, because frequency density × class width = frequency. Area, not height, is what a histogram asks you to compare, which is how it can show classes of different widths fairly.

Its unit is the frequency per unit of whatever was measured. The calculator opens on the journey times of 80 commuters, so its frequency densities are in commuters per minute: the 11 commuters who took between 10 and 15 minutes are spread over 5 minutes, which is 11 ÷ 5 = 2.2 commuters per minute. The textbook Statistics by Freedman, Pisani and Purves calls this vertical axis the density scale.

Using the calculator

Each row of the class table is one class. Type the lower boundary of the first class, then the upper boundary and the frequency of each class in turn. A class starts where the one before it ends, because the bars of a histogram meet with no gaps between them. Add a class or remove the last one with the buttons under the rows, and type what was measured, with its unit, to label the horizontal axis.

The histogram redraws as you type. The number over each bar is its frequency density, the darker bar is the modal class, the dashed line marks the estimated median and the dotted line the estimated mean. Type a pair of values into the last two boxes and the shaded area is the estimate of how many items lie between them. Under the histogram, the table gives each class’s width, frequency density, midpoint, f × x and cumulative frequency, and the working sets out every sum with your numbers in it.

Worked example: 80 commuters’ journey times

The opening table has five classes of unequal width: 0 to 10 minutes with 6 commuters, 10 to 15 with 11, 15 to 20 with 19, 20 to 30 with 20 and 30 to 60 with 24.

• Frequency densities. 6 ÷ 10 = 0.6, 11 ÷ 5 = 2.2, 19 ÷ 5 = 3.8, 20 ÷ 10 = 2 and 24 ÷ 30 = 0.8. These are the bar heights.
• Modal class. The tallest bar is 15 to 20 minutes, at 3.8. The 30 to 60 class holds more commuters, 24, but they are spread over 30 minutes, so its bar is one of the lowest.
• Mean. The midpoints are 5, 12.5, 17.5, 25 and 45, so Σfx = 6 × 5 + 11 × 12.5 + 19 × 17.5 + 20 × 25 + 24 × 45 = 2080 and the estimated mean is 2080 ÷ 80 = 26 minutes.
• Median. Half of 80 is 40. The cumulative frequency is 36 at 20 minutes and 56 at 30, so the median lies in the 20 to 30 class: 20 + (40 − 36) ÷ 20 × 10 = 22 minutes.
• Frequency between 25 and 40 minutes. The area between the two values is 2 × 5 = 10 from the 20 to 30 bar and 0.8 × 10 = 8 from the 30 to 60 bar, so about 10 + 8 = 18 commuters took between 25 and 40 minutes.

The mean is 4 minutes above the median because the long last class pulls it upwards, the usual sign of data skewed to the right. The readouts also give an interquartile range of 35 − 15.79 = 19.21 minutes and an estimated standard deviation of 13.56 minutes.

Why unequal class widths need frequency density

Draw the same table with frequency on the vertical axis and the 30 to 60 bar becomes the tallest. Being 30 minutes wide as well, it covers more of the chart than the other four bars put together, and a reader comparing areas would conclude that most commuters take over half an hour, when 24 of 80 do. Dividing by the width removes the distortion: the bar keeps exactly the area it should, 0.8 × 30 = 24, by being low and wide.

Area is also how a histogram is read. The number of items in any range is the area over that range, and the total area is the total frequency, 80 here. It is the same idea as approximating the area under a curve with rectangles in the Riemann sum explorer. Some exam questions scale the axis so that one square of the grid stands for a number of items, which makes area proportional to frequency rather than equal to it, so find what one square is worth first.

Estimating from grouped data

Once data are grouped the individual values are gone, so the mean, median and mode can only be estimated. The mean counts every item at its class midpoint. The median and the quartiles assume the items in a class are spread evenly across it, which is linear interpolation: median = L + (n ÷ 2 − F) ÷ f × w, where L is the lower boundary of the median class, F the cumulative frequency before it, f its frequency and w its width. The quartiles use n ÷ 4 and 3n ÷ 4 in the same way. The standard deviation is estimated as √(Σfx² ÷ n − mean²), dividing by n as school courses do for grouped data. When you have the raw values rather than a table, the standard deviation calculator gives the exact mean, median and spread instead.

What this model leaves out

• Where values sit inside a class. Every estimate assumes the items are spread evenly through their class, or sit at its midpoint. Real data rarely are, so the estimated mean can be out by up to half the widest class width.
• Open-ended classes. A class such as “60 or more” has no width, so it has no frequency density and no bar. Choose a sensible upper boundary and say that you did.
• Class limits. Whole numbers grouped as 10 to 19 and 20 to 29 leave gaps between the printed limits. Enter the class boundaries, 9.5, 19.5 and 29.5, so each class is 10 wide and the bars meet.
• Two variables at once. A histogram describes one measured quantity. To see how one quantity changes with another, the linear regression calculator fits a line of best fit, and the error bars and worst-fit line calculator finds the uncertainty in its gradient.

Common mistakes

• Plotting frequency instead of frequency density. With unequal widths the wide classes then look far bigger than they are.
• Taking the class with the highest frequency as the modal class. The modal class has the highest frequency density: it is the tallest bar.
• Reading a frequency off the vertical axis. The height is a density. Multiply it by the width of the bar, or of the part of the bar you need.
• Working out widths from class limits. A class of 10 to 19 whole marks runs from 9.5 to 19.5, so it is 10 wide, not 9.
• Dividing Σfx by the number of classes. Divide by the total frequency, 80 here, not by 5.
• Giving the median class as the median. The 20 to 30 class says where the median is; interpolation gives the estimate, 22 minutes.