Density curves: area, mean and median
With thousands of values, the bars of a histogram blur into one smooth outline. How do you read proportions, a median and a mean off a curve that has no bars at all?
Where you are 8 in Distributions
Why this exists
A relative frequency histogram answers “what share of the values lies in this bar?”. It answers it only for whole bars. Ask for the share of waits between 2.5 and 3.5 minutes when the bars run from 2 to 4, and the histogram has nothing to say, because a bar does not show how its values spread inside it. The picture also depends on a choice that has nothing to do with the data: move the bin edges or change their width and the bars come out different.
Now imagine a great many values, sorted into narrower and narrower bins. The jagged staircase of bar tops settles down into one smooth outline. That outline no longer depends on any choice of bins, and it can be cut anywhere: between 2.5 and 3.5 as easily as between 2 and 4. A density curve is that outline, made exact. It is a model of a distribution, an idealized picture of how the values of a numerical variable are spread, in which every proportion is an area.
It is also the shape every later lesson on the normal distribution is built on. A normal curve is one particular density curve, and every question asked of it is a question about area.
Definition
A density curve is the graph of a function that
- never goes below the horizontal axis, and
- has exactly 1 unit of area between it and the horizontal axis.
It models a numerical variable: the proportion of values between any two numbers a and b is the area under the curve from a to b. Writing X for the value of one item picked at random, that area is written P(a \le X \le b), and the same number is the probability that the item’s value lands between a and b.
The median of a density curve is the value with half the area to its left and half to its right.
The mean of a density curve, written \mu, is its balance point: the place where the region under the curve would balance on a knife edge if it were cut out of stiff card of even thickness.
The height of the curve is called its density. It is not itself a proportion. It is a proportion per unit of the variable, such as “share of waits per minute”, and only height times width gives a share.
Intuition
Bars on a density scale. In a relative frequency histogram the height of a bar is its share. To make the area of a bar its share instead, divide each height by the bar’s width:
Then each bar’s area is \displaystyle \text{width} \times \frac{\text{share}}{\text{width}} = \text{share}, and the areas of all the bars add to 1 because the shares do. With bars of equal width this changes only the numbers up the side, never the shape. Its point is that bars of any width now speak the same language: a bar half as wide holding half the share has the same height.
Narrow the bars. With a lot of data, cut each bar in two, and in two again. Every bar’s area is still its share, so the total stays 1, but the tops follow the pattern more and more closely and settle on a smooth curve.
The curve keeps the two facts the bars had: it never dips below the axis, since a share cannot be negative, and the total area under it is 1. That is the definition. The curve is drawn for a model rather than for one data set, so its areas are idealized shares, close to what a large data set would show rather than exactly equal to it.
Center, two ways. The two centers of data from Mean, median and skew: choosing the center carry straight over. The median of data sits in the middle of the sorted values, so half the bar area lies on each side of it; for a curve, the median splits the area in half. The mean of data is the balance point of its histogram (Mean as balance point); for a curve, the mean is the balance point of the region under it. On a symmetric curve both sit at the center of symmetry. When one tail is long, a little area far out has a lot of leverage on the balance point but cannot move the halfway point much, so the mean is pulled out toward the tail, exactly as with skewed data.
Results
Area is proportion. For a density curve and any numbers a < b, the share of values between a and b is the area under the curve from a to b.
The whole is 1, so the rest is 1 minus the part. The share above a is 1 - P(X \le a), and the share outside an interval is 1 minus the share inside it.
A single value has area 0. The region above one point has no width, so P(X = a) = 0, and whether an end is included or not makes no difference: P(a \le X \le b) and P(a < X < b) are equal.
Flat density. A density that is flat from c to d and zero elsewhere has height \displaystyle \frac{1}{d - c}, so
This is the uniform shape as a curve. Its median and mean are both the midpoint \displaystyle \frac{c + d}{2}.
Median and quartiles. The median m has \displaystyle P(X \le m) = \frac{1}{2}. In the same way the first quartile has area \displaystyle \frac{1}{4} to its left and the third quartile area \displaystyle \frac{3}{4}, and the value at any percentile has that share of the area to its left (percentiles).
Mean of pieces. If the region under a curve is split into pieces with areas A_1, A_2, \dots and balance points x_1, x_2, \dots, then
a weighted mean with the areas as weights (they already add to 1). A rectangle balances at its middle; a triangle balances at its Centroid, whose horizontal position is the mean of its three corners’ horizontal positions.
Shape and center. On a symmetric density curve the mean equals the median. On a curve with one clearly longer tail to the right the mean is greater than the median, and with the tail to the left it is less.
Why it is true
Total area 1. Take a density-scale histogram of any data. Its bars’ areas are the shares in the bars, and every value is in exactly one bar, so the areas add to 1. Narrowing the bars never changes that total, so the curve the bars settle on also has area 1.
One value has no area. The area from a to a + w is at most w times the greatest height of the curve there. As w shrinks to 0 this goes to 0, so the single point a carries no area at all. In the model, the share of values exactly equal to a is 0, even though values near a occur.
Flat density. A rectangle of width d - c must have area 1, so its height is \displaystyle \frac{1}{d - c}, and the piece from a to b is a rectangle of area \displaystyle (b - a) \cdot \frac{1}{d - c}. By the mirror symmetry of the rectangle, half its area lies on each side of the midpoint and it balances there.
Mean of pieces. A data set made of two groups has mean equal to the groups’ means weighted by their shares of the data (weighted mean). Each piece of the region is such a group: its share is its area and its own balance point is its mean. So the whole balances at the area-weighted mean of the pieces’ balance points.
Symmetric curves. Reflect a symmetric curve in its line of symmetry. The reflection is the same curve, so its median and mean must land on themselves, and the only points a reflection leaves in place are on the line of symmetry. So both are there.
The tail pulls the mean. Take a curve and drag some area that lies to the right of the median further to the right. The area on each side of the median has not changed, so the median stays put. The balance point does change: weight moved further from it on one side tips it that way. A long right tail is exactly area dragged out to the right, so it sets the mean to the right of the median.
Worked examples
Waiting for a shuttle bus
A shuttle bus leaves a parking lot every 10 minutes, and a traveler reaches the stop at a random moment, so a model for the wait is spread evenly from 0 to 10 minutes. Find the height of the density curve, the share of waits between 2 and 6 minutes, the share longer than 7.5 minutes, and the share of exactly 5 minutes.
SolutionThe density is a rectangle of width 10. Its area must be 1, so its height is \displaystyle \frac{1}{10} per minute.
The share between 2 and 6 minutes is the rectangle under the curve from 2 to 6: width 4 times height \displaystyle \frac{1}{10}, so P(2 \le X \le 6) = 0.4. Four tenths of the waits fall there.
Longer than 7.5 minutes is the rectangle from 7.5 to 10, of width 2.5: \displaystyle P(X > 7.5) = 2.5 \times \frac{1}{10} = 0.25.
A wait of exactly 5 minutes is a region of width 0, so its share is 0. In the model, “exactly 5” means 5.000\ldots to every decimal place; what a stopwatch reports as 5 minutes is really an interval, such as 4.5 to 5.5, with share \displaystyle 1 \times \frac{1}{10} = 0.1.
Which of these are density curves?
Decide which of these are density curves, each being zero outside the stated range.
(a) Flat at height \displaystyle \frac{1}{4} from 0 to 3. (b) A triangle on the axis from 0 to 8 with its peak \displaystyle \frac{1}{4} above x = 2. (c) Flat at height 2 from 0 to \displaystyle \frac{1}{2}.
SolutionAll three stay on or above the axis, so the question is the area.
(a) A rectangle of area \displaystyle 3 \times \frac{1}{4} = \frac{3}{4}. Not a density curve: a quarter of the values would be missing.
(b) A triangle has area \displaystyle \frac{1}{2} \times \text{base} \times \text{height} (Area of triangles), here \displaystyle \frac{1}{2} \times 8 \times \frac{1}{4} = 1. A density curve, whatever the position of the peak.
(c) A rectangle of area \displaystyle \frac{1}{2} \times 2 = 1. A density curve, even though its height is 2. A height is a share per unit, not a share, so it can be larger than 1; here the values are crowded into an interval half a unit wide, so the share per unit is high.
Ribbon from a cutting machine
A machine cuts lengths of ribbon spread evenly from 40 to 65 cm. What share of pieces are from 45 to 52 cm long? What share are longer than 60 cm? Below what length do 30\% of the pieces fall?
SolutionThe rectangle has width 65 - 40 = 25, so its height is \displaystyle \frac{1}{25} = 0.04 per centimeter.
From 45 to 52 is width 7: the share is 7 \times 0.04 = 0.28.
Longer than 60 is the piece from 60 to 65, width 5: 5 \times 0.04 = 0.2.
For the 30th percentile p, the area from 40 to p must be 0.3:
So 30\% of the pieces are shorter than 47.5 cm.
How far seeds land
A model for the distance, in meters, that a plant’s seeds land from it is the straight line \displaystyle f(x) = \frac{4 - x}{8} for 0 \le x \le 4, and zero elsewhere. Check that it is a density curve, then find the share of seeds landing within 1 meter and the share landing beyond 3 meters.
SolutionThe line runs from height \displaystyle \frac{4}{8} = \frac{1}{2} at x = 0 down to 0 at x = 4, never below the axis. The region under it is a triangle with base 4 and height \displaystyle \frac{1}{2}, area \displaystyle \frac{1}{2} \times 4 \times \frac{1}{2} = 1. So it is a density curve.
Beyond 3 meters is the small triangle from 3 to 4: base 1, height \displaystyle f(3) = \frac{1}{8}, area \displaystyle \frac{1}{2} \times 1 \times \frac{1}{8} = \frac{1}{16}.
Within 1 meter is the region from 0 to 1, a trapezium with parallel sides \displaystyle f(0) = \frac{1}{2} and \displaystyle f(1) = \frac{3}{8} and width 1. It is quicker to subtract the triangle from 1 to 4, with base 3 and height \displaystyle \frac{3}{8}, from the whole:
So \displaystyle \frac{7}{16} = 0.4375 of the seeds land within a meter, and only \displaystyle \frac{1}{16} beyond three meters.
The median and mean of the seed distances
Find the median and the mean of the seed density in the last example, and say why they differ.
SolutionMedian. Work from the right, where the region is a single triangle. Beyond a distance m the triangle has base 4 - m and height \displaystyle f(m) = \frac{4 - m}{8}, so its area is \displaystyle \frac{(4 - m)^2}{16}. The median has area \displaystyle \frac{1}{2} beyond it:
The positive square root is taken because 4 - m is a length. So \displaystyle m = 4 - 2\sqrt{2} \approx 1.17 meters.
Mean. The region is a triangle with corners at x = 0, x = 0 (the top of the vertical side) and x = 4. It balances at its centroid, whose horizontal position is the mean of the corners’ positions:
The mean is larger. Most seeds land near the plant, but the curve trails off in a long tail to the right, and the few far seeds pull the balance point out toward them without moving the halfway point as much: a right-skewed density, with mean greater than median.
Ride times on a density scale
The times of 400 cyclists in a 10 km time trial are shown as a density histogram in bars 5 minutes wide, from 20 to 45 minutes, with heights 0.02, 0.06, 0.07, 0.04 and 0.01. (a) Check that the total area is 1. (b) How many cyclists finished in under 30 minutes? (c) Estimate the median and the mean.
(a) Each area is 5 times the height, so the shares are 0.10, 0.30, 0.35, 0.20 and 0.05. They add to 1.
(b) Under 30 minutes is the first two bars: 0.10 + 0.30 = 0.40 of the cyclists, and 0.40 \times 400 = 160 cyclists.
(c) Median. The first two bars hold 0.40, so the halfway point is in the 30 to 35 bar, which still has to supply 0.10. Treat the bar as the model, so its values are spread evenly across it: a strip of width w at height 0.07 has area 0.07w, and
So the median is about 30 + 1.43 \approx 31.4 minutes.
Mean. Each bar is a rectangle balancing at its midpoint, so the mean is the area-weighted mean of the midpoints:
The mean is barely above the median. The bars rise to a peak in the middle and fall away on the two sides at much the same rate, so the shape is close to symmetric, and on a roughly symmetric shape the two centers nearly agree. Both are estimates from the bars; the cyclists’ actual times would give slightly different values.
Calls to a help line
A model for the length of calls to a help line, in minutes, is flat at a height k from 0 to 1, then falls in a straight line to 0 at 5, and is zero elsewhere. Find k, the share of calls longer than 3 minutes, the median and the mean.
Height. The region is a rectangle from 0 to 1 of area k, and a triangle from 1 to 5 of base 4 and height k, area \displaystyle \frac{1}{2} \times 4 \times k = 2k. The total must be 1: k + 2k = 3k = 1, so \displaystyle k = \frac{1}{3}.
The falling line goes from height \displaystyle \frac{1}{3} at x = 1 to 0 at x = 5; its height at x is \displaystyle \frac{5 - x}{12}, which gives \displaystyle \frac{4}{12} = \frac{1}{3} at x = 1 as it must.
Longer than 3 minutes. The triangle from 3 to 5 has base 2 and height \displaystyle \frac{5 - 3}{12} = \frac{1}{6}, so its area is \displaystyle \frac{1}{2} \times 2 \times \frac{1}{6} = \frac{1}{6}.
Median. The rectangle holds only \displaystyle \frac{1}{3}, less than half, so the median is past 1. Work from the right again: beyond m is a triangle of base 5 - m and height \displaystyle \frac{5 - m}{12}, area \displaystyle \frac{(5 - m)^2}{24}. Setting this to \displaystyle \frac{1}{2}:
Mean. Split into the two pieces. The rectangle has area \displaystyle \frac{1}{3} and balances at \displaystyle \frac{1}{2}. The triangle has area \displaystyle \frac{2}{3} and corners at x = 1, 1 and 5, so it balances at \displaystyle \frac{1 + 1 + 5}{3} = \frac{7}{3}. Weighting by area:
So half the calls are over in about 1.54 minutes, the mean call lasts about 1.72 minutes, and the long right tail of long calls is what puts the mean above the median.
Boundaries
A density curve is a model, not the data. Areas under it are idealized shares. A real data set of a few hundred values matches them only roughly, and how closely is the question the later lessons on inference take up.
The variable has to be measured on a continuous scale. Lengths, times and masses can take any value in a range. A count, such as the number of children in a family, takes only whole-number values with gaps between them; its distribution is a table of separate shares, each value with a share of its own, and the rule “a single value has area 0” would throw all of it away.
Both conditions are needed. A curve with total area other than 1 gives shares that do not add to the whole, and a curve dipping below the axis would give a negative share. Neither is a density curve.
The height alone is never a share. Only an area is. A density can be larger than 1.
A median can fail to be a single number. If the curve is zero along a stretch exactly where the area to the left reaches \displaystyle \frac{1}{2}, every point of that stretch splits the area in half. The usual choice then is the middle of the stretch.
A mean can fail to exist. Every curve in this lesson stops somewhere. A curve whose tail runs on for ever can have so much area so far out that no balance point exists. That does not happen for the curves met at this level, the normal curve included.
“Tail pulls the mean” is about clear one-tailed shapes. For a curve with one long tail and the bulk on the other side, the rule holds. For a lumpy or two-peaked curve the mean and median can sit in either order, and the shape has to be looked at directly.
Where this goes next
The most important density curve by far is the bell-shaped normal curve, next in this unit. Its areas cannot be found with rectangles and triangles, so z-scores and the empirical rule and the standard normal curve turn every normal curve into one standard one, and normal calculations read its areas from a table or a calculator. Everything there is the idea of this lesson: a share is an area, the whole is 1, and a percentile is a cut with a given area to its left.
The link between a value picked at random and an area is the start of continuous Random variables. And the question “what is the area under a curve that is not made of straight pieces?” is answered by calculus: thin rectangles first (Riemann sums, the same narrowing of bars as in the first figure) and then the The definite integral.
Exercises
Flat densities
- A computer picks a number X spread evenly from 0 to 5. Find the height of its density curve and P(1 \le X \le 4).
- For the number in Exercise 1, find P(X > 3.5) and P(X = 2).
- A tram reaches a stop at a moment spread evenly between 10 and 30 minutes past the hour. On what share of days does it arrive between 12 and 21 minutes past?
- For the tram in Exercise 3, by how many minutes past the hour has it arrived on 80\% of days?
- A flat density runs from 2 to 6 and is zero elsewhere. Find its height, its first and third quartiles, its median and its mean.
Is it a density curve?
- A curve is flat at height 0.3 from 0 to 3 and zero elsewhere. Is it a density curve? If not, what flat height on the same interval would make it one?
- A triangle stands on the axis from 0 to 10 with its peak 0.2 above x = 3, and the curve is zero elsewhere. Is it a density curve?
- A triangular density stands on the axis from 0 to 6. How high must its peak be?
Triangles and pieces
- A density is \displaystyle f(x) = \frac{x}{2} for 0 \le x \le 2 and zero elsewhere. Check that the area is 1, then find P(X \le 1) and P(X > 1.2).
- For the density in Exercise 9, find the median and the mean. Which is larger, and what about the shape makes it so?
- A symmetric triangular density stands on the axis from 0 to 4 with its peak above x = 2. Find the height of the peak, P(1 \le X \le 3), the median and the mean.
- For the seed density \displaystyle f(x) = \frac{4 - x}{8} on 0 \le x \le 4 of the worked examples, find the first quartile.
- A density is \displaystyle \frac{1}{8} from 0 to 4 and \displaystyle \frac{1}{4} from 4 to 6, and zero elsewhere. Check that the area is 1, then find the median and the mean.
Histograms on a density scale
- A relative frequency histogram has bins 0 to 2, 2 to 4, 4 to 6 and 6 to 8 with shares 0.1, 0.3, 0.4 and 0.2. Find the heights of the same bars on a density scale.
- A density histogram of 300 values has bars 0 to 10, 10 to 20, 20 to 30 and 30 to 40 with heights 0.01, 0.04, 0.03 and 0.02. Check that the area is 1, and find how many of the values lie between 10 and 30.
- For the histogram in Exercise 15, estimate the median and the mean, treating each bar’s values as spread evenly across it.
- Three bars each 0.5 wide hold shares 0.15, 0.6 and 0.25. Find their heights on a density scale.
Explaining
- A density curve has median 12 and mean 15. On which side is its long tail? Explain.
- Explain why the total area under a density curve has to be exactly 1.
- A variable modeled by a density curve can take the value 3. Explain why P(X = 3) = 0 anyway, and why it follows that P(X \le 3) = P(X < 3).
- Explain why the mean and the median of a symmetric density curve are equal.
- The same data are drawn as two relative frequency histograms, one with bins 1 wide and one with bins 2 wide. The bars of the second are roughly twice as tall. Explain why, and why on a density scale the two would have bars of roughly the same height.
Answers
- Height \displaystyle \frac{1}{5} = 0.2. P(1 \le X \le 4) = 3 \times 0.2 = 0.6.
- P(X > 3.5) = 1.5 \times 0.2 = 0.3, the rectangle from 3.5 to 5. P(X = 2) = 0: a single value has no width.
- 0.45. The height is \displaystyle \frac{1}{20}, and the interval has width 9, so \displaystyle 9 \times \frac{1}{20} = 0.45.
- 26 minutes past. The area from 10 to t is \displaystyle \frac{t - 10}{20} = 0.8, so t - 10 = 16.
- Height \displaystyle \frac{1}{4}. First quartile 3 and third quartile 5, since the area to their left is \displaystyle 1 \times \frac{1}{4} and \displaystyle 3 \times \frac{1}{4}. Median 4 and mean 4, the midpoint, by symmetry.
- No: its area is 3 \times 0.3 = 0.9. A height of \displaystyle \frac{1}{3} gives area 1.
- Yes. It never dips below the axis, and its area is \displaystyle \frac{1}{2} \times 10 \times 0.2 = 1.
- \displaystyle \frac{1}{3}. The area \displaystyle \frac{1}{2} \times 6 \times h = 3h must be 1.
- The region is a triangle of base 2 and height f(2) = 1, area 1. \displaystyle P(X \le 1) = \frac{1}{2} \times 1 \times \frac{1}{2} = \frac{1}{4}. Up to 1.2 the area is \displaystyle \frac{1}{2} \times 1.2 \times 0.6 = 0.36, so P(X > 1.2) = 1 - 0.36 = 0.64.
- Median \displaystyle \sqrt{2} \approx 1.41: the area to the left of m is \displaystyle \frac{m^2}{4} = \frac{1}{2}, so m^2 = 2. Mean \displaystyle \frac{0 + 2 + 2}{3} = \frac{4}{3} \approx 1.33, the centroid of the triangle with corners at x = 0, 2 and 2. The mean is smaller: the curve has its long tail on the left, toward small values, a left-skewed shape.
- Peak \displaystyle \frac{1}{2}, since \displaystyle \frac{1}{2} \times 4 \times h = 1. Left of 1 is a triangle of base 1 and height \displaystyle \frac{1}{4}, area \displaystyle \frac{1}{8}, and by symmetry the same lies right of 3, so \displaystyle P(1 \le X \le 3) = 1 - \frac{2}{8} = \frac{3}{4}. Median and mean are both 2.
- \displaystyle q = 4 - 2\sqrt{3} \approx 0.54 meters. The area beyond q must be \displaystyle \frac{3}{4}: \displaystyle \frac{(4 - q)^2}{16} = \frac{3}{4}, so (4 - q)^2 = 12 and \displaystyle 4 - q = 2\sqrt{3}.
- Area \displaystyle 4 \times \frac{1}{8} + 2 \times \frac{1}{4} = \frac{1}{2} + \frac{1}{2} = 1. Median 4, since the first rectangle holds exactly half. Mean \displaystyle \frac{1}{2} \times 2 + \frac{1}{2} \times 5 = 3.5, weighting the rectangles’ midpoints by their areas. The mean is below the median: the long, low part of the curve is on the left.
- Divide each share by the width 2: 0.05, 0.15, 0.20 and 0.10. Check: 2 \times (0.05 + 0.15 + 0.20 + 0.10) = 1.
- Area 10 \times (0.01 + 0.04 + 0.03 + 0.02) = 1. Between 10 and 30 the share is 0.4 + 0.3 = 0.7, which is 0.7 \times 300 = 210 values.
- Median 20: the first two bars hold 0.1 + 0.4 = 0.5 exactly. Mean 5(0.1) + 15(0.4) + 25(0.3) + 35(0.2) = 21, the bar midpoints weighted by the shares. The mean is above the median, a mild right skew.
- Divide by 0.5: 0.3, 1.2 and 0.5. The middle bar is taller than 1, which is allowed, because a height is a share per unit and the bars are narrow.
- On the right. The mean is pulled out toward a long tail while the median stays near the bulk, and here the mean is the larger of the two.
- The areas are the shares of the values in each range, and every value lies somewhere, so the shares of all the ranges together make up the whole: 100\%, which is 1.
- The region above the single value 3 has width 0, so its area is 0: the area from 3 to 3 + w is at most w times the greatest height there, and that goes to 0 as w does. So including or leaving out the end point 3 adds nothing: P(X \le 3) is P(X < 3) + P(X = 3), which is P(X < 3) + 0.
- Reflecting the curve in its line of symmetry gives the same curve. The point with half the area on each side, and the balance point, must therefore each be their own mirror image, and only points on the line of symmetry are. So both lie on it, and they are equal.
- A bin twice as wide holds about twice the share of values, so its relative frequency is about twice as large. On a density scale each share is divided by its bin width, 2 instead of 1, which cancels the doubling, so the heights come out about the same. A density measures share per unit, which does not depend on how wide the bins are.