Calculate frequency density and class counts from supplied numerical data for unequal-width histogram classes. Estimate partial-class counts using an even-spread assumption.
You used class boundaries on a cumulative frequency curve. Wren Lane's next 10k has 80 finishers: 8 in 30 ≤ t < 40, 12 in 40 ≤ t < 45, 15 in 45 ≤ t < 50, 25 in 50 ≤ t < 60 and 20 in 60 ≤ t < 80. Here t is time in minutes. Each class includes its lower boundary and excludes its upper boundary. A time of exactly 45 belongs to 45 ≤ t < 50.
The 60 to 80 class has 20 runners over 20 minutes: an average of 1 per minute. The 40 to 45 class has 12 over 5 minutes: 2.4 per minute. It has fewer runners but a higher average rate. A histogram draws numerical widths, so these bars must be 20 and 5 minutes wide. If height were frequency, area would be frequency x width. Wider classes would receive more area per runner. Density heights make areas equal counts.
Class width = upper boundary - lower boundary. For a density histogram, height = frequency / width, the average frequency per unit. Area = density x width = frequency. If density is positive, width = frequency / density, and a wider class at that density has more values. Equal-width frequency heights preserve the shape. Density equals frequency numerically when width is 1; an empty class has both zero at any positive width.
To draw the histogram, label the horizontal axis time in minutes and the vertical axis frequency density. Draw each rectangle from its lower to its upper boundary. The five widths are 10, 5, 5, 10 and 20. Their heights are 0.8, 2.4, 3, 2.5 and 1. For 45 to 50, width 5 x height 3 gives frequency 15. The boundary line below shows the numerical spacing; it is not the histogram.
For part of a class, estimate with density x the requested width. From 50 to 55: 2.5 x 5 = 12.5, or about 13 runners. From 42 to 48, split at 45: 2.4 x 3 + 3 x 3 = 16.2, or about 16. Split at every boundary crossed and add the contributions. A complete class contributes its known frequency. Partial classes require an even-spread assumption. Keep decimals during calculation and round at the end if requested.
A categorical count chart uses heights for counts; its widths do not measure category sizes. A density histogram uses numerical widths and areas for counts. Adjacent nonempty classes meet at their shared boundary. An empty class has density zero and may leave a blank interval. Emptiness alone proves neither possibility nor impossibility. Frequency = density x width. Classes count each value once, so areas sum to the total. Find a missing count by subtracting known frequencies from that total.
To merge adjacent classes, add their frequencies and widths separately. Divide the combined frequency by the combined width. Widths 4 and 6 with densities 2 and 3 give frequencies 8 and 18. The merged density is (8 + 18) / (4 + 6) = 2.6. Adding densities or taking their unweighted mean does not generally work. The merged bar preserves total frequency but loses the density change at the old boundary.
Splitting a class into equal-width pieces under the even-spread assumption keeps the estimated density unchanged. Each piece receives an equal share of the class frequency. The grouped count alone does not tell you the actual count in each piece, or where individual observations lie.
For a density histogram, frequency = density x width = bar area. Density = frequency / width. Estimate partial classes using an even spread, splitting at boundaries and adding the contributions.
Lesson and practice text written for this dataset. Histogram conventions, average density, class areas and grouped-data estimates are explained with original examples.
The four readings and nine extra questions were written for this dataset. Video references identify Khan Academy resources; their content licenses have not been verified.
A class contains 27 measurements across a width of 9. Its frequency density is 27 / 9 = 3. A density histogram uses this number for the rectangle's height. Multiplying height by width returns the count: 3 x 9 = 27.
The grouped table does not reveal the count in each smaller interval. Some one-unit intervals might contain more than three measurements and others fewer. Density 3 describes the class average.
Check the vertical axis. Equal-width histograms can use frequency heights. Their shape is proportional to the density version. Width 1 guarantees equal numerical values for density and frequency. An empty class also has density 0 and frequency 0 at any positive width.
To estimate within part of a class, use that class's density and the width of the requested part. This treats the observations as evenly spread inside the class.
Density 1.8 across a requested width of 4 gives an estimated frequency of 7.2. A decimal estimate is allowed even when the observations are people. If a whole-person answer is requested, round after finishing the calculation.
An interval crossing class boundaries needs several contributions. Complete classes contribute their known counts. Partial classes contribute estimates. Add these counts, not the density heights.
For specified class boundaries, width is upper boundary minus lower boundary. The class 7.25 ≤ x < 11.75 has width 11.75 - 7.25 = 4.5.
This class includes 7.25 and excludes 11.75. If the next class begins at 11.75, an observation at that boundary belongs to the next class. Using the signs prevents double counting.
An inclusive label such as 20 to 29 may describe integer observations rather than the boundaries used to draw the bar. Read the question's stated boundaries or convention before calculating a width.
Two classes each contain 18 values. Their widths are 3 and 9. If both are drawn at height 18 on a numerical scale, the areas become 54 and 162. Equal counts have been given unequal areas.
Density heights are 18 / 3 = 6 and 18 / 9 = 2. Now both areas are 18: 3 x 6 = 18 and 9 x 2 = 18.
A categorical chart can compare the two counts by height without using numerical widths. The error is combining frequency heights with numerical histogram widths while expecting areas to represent counts.
You used class boundaries on a cumulative frequency curve. The next Wren Lane 10k has 80 finishers in the lesson's five bands. Sam says 20 runners in 60 to 80 minutes means a greater average rate than 12 in 40 to 45.
Which calculation checks Sam's claim?
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