4.5 Continuous Probability Functions

Learning Objectives

By the end of this section, you will be able to:

Chapter 4 dealt with discrete random variables, where you could list the values one at a time — 0 babies, 1 baby, 2 babies — and give each one its own probability. This chapter takes on quantities that do not come in separate pieces. Waiting time at a bus stop, the weight of a bag of flour, the length of a phone call: between any two values you name there is another value, and another one after that. You cannot list them, so you cannot give each one its own row in a table.

That forces a change in how we measure probability. Instead of a table, we use a function \(f(x)\), and instead of reading a number off a row we measure an area. The rule for the whole chapter fits on one line: for a continuous random variable, probability is area. The area between the curve \(f(x)\) and the x-axis, between two values you pick, is the probability the variable lands between those two values.

You have seen function notation before, in algebra. What is new here is what the function is for. We build \(f(x)\) specifically so that areas under it come out as probabilities, which puts one hard constraint on it right away: the largest probability there is, is 1, so the largest area there can be is also 1.

4.5.1 Probability Is Area

The probability that \(X\) lands between two values \(a\) and \(b\) is the area under \(f(x)\) between \(x = a\) and \(x = b\):

The first property is the continuous version of "no negative probabilities" — a curve that dipped below the axis would hand you a negative area, and there is no such thing as a negative probability. The second is the continuous version of "the column sums to one." Every outcome the variable can produce is somewhere under that curve, so the whole area has to account for the whole of the probability.

Notice what the definition does not say. It never claims \(f(x)\) is itself a probability. The height of the curve at \(x = 3\) is not "the probability of 3" — it is a density, a rate of probability per unit of \(x\), and it only becomes a probability once you multiply it by a width. That distinction is the single most common place students get stuck in this chapter, and the exercises below are built to make you use it rather than memorize it.

That is worth taking seriously, because it tells you what kind of work these problems actually are. A flat density is called a uniform density, and for a uniform density every probability question in this section reduces to the same two measurements: how wide is the stretch of x-values I care about, and how tall is the line above it. Multiply them. If the answer comes out bigger than 1 you have made an arithmetic mistake, because no region under the curve can be larger than the whole region, and the whole region has area 1. If the answer comes out negative you have subtracted your endpoints in the wrong order. Those two sanity checks catch nearly every error people make here, and both take about a second to run. It is worth building the habit now, while the densities are flat and the arithmetic is one multiplication, because the same two checks still work in the next two sections when the curve is no longer a straight line and the area is no longer something you can measure with a ruler. The shape of the region changes; the question you ask about your answer does not. Try the check for yourself on the exercise below before you look at how it works out.

Definition 4.5.1 — The whole area under the density is 1, and a slice of it is a probability.

Example 4.5.1: Areas Under a Flat Density

Consider the function \(f(x) = \frac{1}{20}\) for \(0 \leq x \leq 20\), where \(x\) is a real number. The graph of \(f(x) = \frac{1}{20}\) is a horizontal line. However, since \(0 \leq x \leq 20\), \(f(x)\) is restricted to the portion between \(x = 0\) and \(x = 20\), inclusive.

Figure 4.5.1 — The density f(x) = 1/20 drawn as a horizontal segment at height 1/20 from x = 0 to x = 20, closed off at both ends and left unshaded.

$$ f(x) = \frac{1}{20} \text{ for } 0 \leq x \leq 20 $$

a. Show that the total area under \(f(x)\) is one.

b. Find \(P(0 < x < 2)\).

c. Find \(P(4 < x < 15)\).

d. Find \(P(x = 15)\).

Solution

a. Total area. The graph of \(f(x) = \frac{1}{20}\) is a horizontal line segment when \(0 \leq x \leq 20\). The area between \(f(x) = \frac{1}{20}\) and the x-axis is the area of a rectangle with base 20 and height \(\frac{1}{20}\):

$$ \text{AREA} = 20\left(\frac{1}{20}\right) = 1 $$

So this is a legitimate density, and every probability we read off it will be a piece of that single unit of area.

b. The area from 0 to 2. Suppose we want the area between \(f(x) = \frac{1}{20}\) and the x-axis where \(0 < x < 2\). That region is again a rectangle — the base runs from 0 to 2 and the height is still \(\frac{1}{20}\).

Figure 4.5.2 — The same flat density with only the strip from x = 0 to x = 2 shaded, a narrow sliver at the left end of the rectangle.

Reminder: the area of a rectangle is (base)(height). Here the base is \((2 - 0) = 2\), so

$$ \text{AREA} = (2 - 0)\left(\frac{1}{20}\right) = 0.1 $$

The area corresponds to a probability. The probability that \(x\) is between zero and two is 0.1, which is written \(P(0 < x < 2) = 0.1\).

c. The area from 4 to 15. Same move, wider base. The base is \((15 - 4) = 11\):

Figure 4.5.3 — The flat density with the strip from x = 4 to x = 15 shaded, covering just over half the rectangle.

$$ \text{AREA} = (15 - 4)\left(\frac{1}{20}\right) = 0.55 $$

So \(P(4 < x < 15) = 0.55\).

d. The area at the single point 15. On an x-y graph, \(x = 15\) is a vertical line. A vertical line has no width — its base is zero. So

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Figure 4.5.4 — The flat density with a single vertical line drawn at x = 15, a segment with no width and therefore no area.

$$ P(x = 15) = (\text{base})(\text{height}) = (0)\left(\frac{1}{20}\right) = 0 $$

Answer: the total area is 1; \(P(0 < x < 2) = 0.1\); \(P(4 < x < 15) = 0.55\); and \(P(x = 15) = 0\).

Part d is not a quirk of this particular function. For any continuous random variable the probability of landing on one exact value is zero, because a single point has no width to give it area. That is why \(P(x < 15)\) and \(P(x \leq 15)\) are the same number here — the endpoint contributes nothing.

4.5.2 The Cumulative Distribution Function

Because a single point contributes no area, \(P(X \leq x)\) and \(P(X < x)\) are the same number for a continuous distribution. The area to the right is whatever is left over:

Notice the "less than or equal to" symbol in the definition. The CDF gives "area to the left," and the complement rule turns that into "area to the right" whenever you need it — you never have to measure the right-hand region directly.

Figure 4.5.5 — A flat density split at an interior point: the region to the left of the split is shaded and the region to the right is left white, showing the area-to-the-left and area-to-the-right split.

This picture is the whole idea. The shaded block is \(P(X < x)\) and the white block is \(P(X > x)\), and together they are the entire rectangle, whose area is 1. So the two numbers always add to 1, which is exactly what the complement rule says.

There is one habit worth building before you start calculating, and it costs nothing: draw the picture first, then label it. Label the vertical axis \(f(x)\) and the horizontal axis \(x\). Scale each axis with the largest value it needs to show — for \(f(x) = \frac{1}{20}\) on \(0 \leq x \leq 20\), that means the x-axis runs to 20 and the height is \(\frac{1}{20}\). Then shade the region the question asks about and read the base and the height straight off the drawing. Students who skip the sketch tend to subtract the endpoints backwards, or forget which of two numbers was the height; students who draw it almost never do, because a shaded rectangle makes both measurements visible at once. The exercise below is a good place to practise the habit while the arithmetic is still easy.

Example 4.5.2: A Probability Between Two Interior Values

Return to \(f(x) = \frac{1}{20}\) on \(0 \leq x \leq 20\). Label the graph with \(f(x)\) and \(x\), and scale the axes with the maximum \(x\) and \(y\) values. Then shade the region between \(x = 2.3\) and \(x = 12.7\) and calculate the shaded area to find \(P(2.3 < x < 12.7)\).

Figure 4.5.6 — The density f(x) = 1/20 with the strip between x = 2.3 and x = 12.7 shaded, covering roughly half the rectangle.

Solution

Step 1 — set up the picture. The vertical axis is \(f(x)\), scaled to the height \(\frac{1}{20}\); the horizontal axis is \(x\), scaled from 0 to 20. Shade the region between \(x = 2.3\) and \(x = 12.7\).

Step 2 — the shaded region is a rectangle. Its height is \(\frac{1}{20}\) and its base is \(12.7 - 2.3 = 10.4\).

Step 3 — multiply.

$$ P(2.3 < x < 12.7) = (\text{base})(\text{height}) = (12.7 - 2.3)\left(\frac{1}{20}\right) = 0.52 $$

Check: the shaded strip is a little over half of a rectangle whose total area is 1, and 0.52 is a little over half. The answer is consistent with the picture.

Answer: \(P(2.3 < x < 12.7) = 0.52\).

Figure 4.5.7 — Reading area to the left as a probability in rāSHio: Distributions → Normal.

Problem Set 4.5

Problem 1. Which type of distribution does the graph illustrate?

Exercise Figure 4.5.1 — A density drawn as a flat horizontal segment at a constant height from x = 3 to x = 8, on an axis running from 0 to 10 and zero elsewhere.

Solution

Step 1 — Read the shape of the graph: The curve is a flat horizontal segment sitting at one constant height above the x-axis, running from \(x = 0\) to \(x = 9\) and stopping there.

Step 2 — Match the shape to a family: A density with the same height everywhere on its interval gives equal probability to every stretch of equal width. That is the defining behavior of a uniform density.

Answer: Uniform distribution.

Problem 2. Which type of distribution does the graph illustrate?

Exercise Figure 4.5.2 — A density that starts high against the y-axis and falls away steadily toward the x-axis, still above it at x = 10.

Solution

Step 1 — Read the shape of the graph: The curve starts high on the left, near \(f(x) = 1.0\) at \(x = 0\), and drops away steadily, flattening out toward the x-axis without ever touching it — about 0.03 by \(x = 10\).

Step 2 — Match the shape to a family: A density that decays from a peak at the left edge and trails off to the right is the exponential shape.

Answer: Exponential distribution.

Problem 3. Which type of distribution does the graph illustrate?

Exercise Figure 4.5.3 — A symmetric mound-shaped density centered at x = 0, with a vertical line up to the peak and tails tapering toward the axis at x = -3 and x = 3.

Solution

Step 1 — Read the shape of the graph: The curve is a single mound, highest at \(x = 0\) and falling away symmetrically in both directions, down to about 0.01 at \(x = -3\) and \(x = 3\).

Step 2 — Match the shape to a family: A symmetric, bell-shaped mound with tails on both sides is the normal shape.

Answer: Normal distribution.

Problem 4. What does the shaded area represent? \(P(\_\_\_\_ < x < \_\_\_\_)\)

Exercise Figure 4.5.4 — A flat density from x = 1 to x = 8 with the strip between x = 2 and x = 5 shaded.

Solution

Step 1 — Find where the shading starts and stops: The density is flat from \(x = 1\) to \(x = 8\), and the shaded block runs from \(x = 2\) on the left to \(x = 5\) on the right.

Step 2 — Translate the picture into notation: The shaded area is the probability that \(x\) lands between those two endpoints.

Answer: \(P(2 < x < 5)\).

Problem 5. What does the shaded area represent? \(P(\_\_\_\_ < x < \_\_\_\_)\)

Exercise Figure 4.5.5 — A density falling steadily from left to right, with a narrow strip between x = 6 and x = 7 shaded.

Solution

Step 1 — Find where the shading starts and stops: The density falls steadily from left to right, and the narrow shaded sliver sits between \(x = 6\) and \(x = 7\).

Step 2 — Translate the picture into notation: The shaded area is the probability that \(x\) lands between those two endpoints.

Answer: \(P(6 < x < 7)\).

Problem 6. For a continuous probability distribution, \(0 \leq x \leq 15\). What is \(P(x > 15)\)?

Solution

Step 1 — Note where the density lives: The distribution is defined only on \(0 \leq x \leq 15\). Outside that interval the density is zero, so there is no curve above the axis past \(x = 15\).

Step 2 — Measure the area to the right of 15: No curve means no region, and no region means no area.

$$ P(x > 15) = 0 $$

Answer: \(P(x > 15) = 0\).

Problem 7. What is the area under \(f(x)\) if the function is a continuous probability density function?

Solution

Step 1 — Recall the second property of a density: By Definition 4.5.1, the total area between \(f(x)\) and the x-axis must account for all of the probability.

Step 2 — State the total: All of the probability is 1.

Answer: The area under \(f(x)\) is one.

Problem 8. For a continuous probability distribution, \(0 \leq x \leq 10\). What is \(P(x = 7)\)?

Solution

Step 1 — Picture what \(x = 7\) looks like on the graph: A single value is a vertical line, and a vertical line has no width.

Step 2 — Compute the area: Area is (base)(height), and the base here is zero.

$$ P(x = 7) = (0)(\text{height}) = 0 $$

Answer: \(P(x = 7) = 0\). For any continuous random variable the probability of one exact value is zero.

Problem 9. A continuous probability function is restricted to the portion between \(x = 0\) and \(x = 7\). What is \(P(x = 10)\)?

Solution

Step 1 — Check whether 10 is even in range: The function is restricted to the portion between \(x = 0\) and \(x = 7\), so \(f(10) = 0\) — there is no curve above \(x = 10\) at all.

Step 2 — Add the single-point rule: Even if 10 had been inside the interval, a single value is a vertical line of zero width and therefore zero area.

Answer: \(P(x = 10) = 0\).

Problem 10. \(f(x)\) for a continuous probability function is \(\frac{1}{5}\), and the function is restricted to \(0 \leq x \leq 5\). What is \(P(x < 0)\)?

Solution

Step 1 — Note where the density lives: \(f(x) = \frac{1}{5}\) only on \(0 \leq x \leq 5\); to the left of \(x = 0\) the density is zero.

Step 2 — Measure the area to the left of 0: There is no region above the axis there, so there is no area.

$$ P(x < 0) = 0 $$

Answer: \(P(x < 0) = 0\).

Problem 11. \(f(x)\), a continuous probability function, is equal to \(\frac{1}{12}\), and the function is restricted to \(0 \leq x \leq 12\). What is \(P(0 < x < 12)\)?

Solution

Step 1 — Identify the base and the height: The interval \(0 < x < 12\) is the whole domain of the function, so the base is \(12 - 0 = 12\) and the height is \(\frac{1}{12}\).

Step 2 — Multiply:

$$ P(0 < x < 12) = (12)\left(\frac{1}{12}\right) = 1 $$

Step 3 — Sanity-check: Asking for the probability across the entire domain is asking for the total area, which Definition 4.5.1 says must be 1. The answer agrees.

Answer: \(P(0 < x < 12) = 1\).

Problem 12. Find the probability that \(x\) falls in the shaded area.

Exercise Figure 4.5.6 — A flat density at height 1/9 from x = 0 to x = 9, with the strip between x = 6 and x = 8 shaded.

Solution

Step 1 — Read the height: The graph labels the flat line at \(f(x) = \frac{1}{9}\), and the density runs from \(x = 0\) to \(x = 9\).

Step 2 — Read the base: The shaded block runs from \(x = 6\) to \(x = 8\), so its base is \(8 - 6 = 2\).

Step 3 — Multiply:

$$ \text{AREA} = (8 - 6)\left(\frac{1}{9}\right) = \frac{2}{9} \approx 0.2222 $$

Answer: \(P(6 < x < 8) = \frac{2}{9} \approx 0.2222\).

Problem 13. Find the probability that \(x\) falls in the shaded area.

Exercise Figure 4.5.7 — A flat density at height 1/8 from x = 0 to x = 8, with the strip between x = 0 and x = 5 shaded.

Solution

Step 1 — Read the height: The flat line is labelled \(f(x) = \frac{1}{8}\), and the density runs from \(x = 0\) to \(x = 8\).

Step 2 — Read the base: The shaded block runs from \(x = 0\) to \(x = 5\), so its base is \(5 - 0 = 5\).

Step 3 — Multiply:

$$ \text{AREA} = (5 - 0)\left(\frac{1}{8}\right) = \frac{5}{8} = 0.625 $$

Answer: \(P(0 < x < 5) = 0.625\).

Problem 14. Find the probability that \(x\) falls in the shaded area.

Exercise Figure 4.5.8 — A flat density at height 1/10 from x = 0 to x = 10, with the strip between x = 2.5 and x = 5.5 shaded.

Solution

Step 1 — Read the height: The flat line is labelled \(f(x) = \frac{1}{10}\), and the density runs from \(x = 0\) to \(x = 10\).

Step 2 — Read the base: The shaded block runs from \(x = 2.5\) to \(x = 5.5\), so its base is \(5.5 - 2.5 = 3\).

Step 3 — Multiply:

$$ \text{AREA} = (5.5 - 2.5)\left(\frac{1}{10}\right) = \frac{3}{10} = 0.3 $$

Answer: \(P(2.5 < x < 5.5) = 0.3\).

Problem 15. \(f(x)\), a continuous probability function, is equal to \(\frac{1}{3}\) and the function is restricted to \(1 \leq x \leq 4\). Describe \(P\left(x > \frac{3}{2}\right)\).

Solution

Step 1 — Describe the region in words: \(P\left(x > \frac{3}{2}\right)\) is the area from \(x = \frac{3}{2}\) across to the right-hand end of the domain, \(x = 4\), above the x-axis and up to the line \(f(x) = \frac{1}{3}\).

Step 2 — Measure it: That region is a rectangle with base \(4 - \frac{3}{2} = \frac{5}{2}\) and height \(\frac{1}{3}\).

$$ P\left(x > \frac{3}{2}\right) = \left(\frac{5}{2}\right)\left(\frac{1}{3}\right) = \frac{5}{6} \approx 0.8333 $$

Answer: It is the area from \(x = \frac{3}{2}\) to \(x = 4\) above the x-axis and up to \(f(x) = \frac{1}{3}\), which works out to \(\frac{5}{6} \approx 0.8333\).

For the following exercises, draw the picture.

Problem 16. Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine the percent of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer "yes" or "no." You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor.

a) What part of the experiment will yield discrete data?

b) What part of the experiment will yield continuous data?

Problem 17. When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why?

Solutions 16–17

Problem 16

Step 1 — Separate what is counted from what is measured: Two different quantities come out of this experiment. One is the tally of "yes" answers; the other is the percentage you report to your supervisor.

a. The discrete part. Asking each nurse and recording "yes" or "no" produces a count — the number of nurses in your sample of 100 who hold an R.N. degree. That count can only be a whole number: 0, 1, 2, up to 100, with nothing in between. Countable and separate means discrete.

b. The continuous part. The percentage you calculate and hand to your supervisor is a proportion rather than a tally. A percent is a number on a scale, and treated as a measurement of "what share of nurses hold the degree" it can sit anywhere between 0 and 100. That is continuous data.

Answer: a. counting the nurses who answer "yes" yields discrete data; b. the percentage of nurses with an R.N. degree yields continuous data.

Problem 17

Step 1 — Ask what the variable actually is: Age is how long someone has been alive. Time runs without gaps — a person is 19.4 years old on the way from 19 to 20 — so age is a measurement, not a count.

Step 2 — Ask what rounding changes: Rounding to the nearest year changes how the value is recorded, not what the value is. The underlying quantity still varies continuously; you have just chosen to report it coarsely.

Answer: The data stay continuous. Age is a measurement regardless of the accuracy used to report it; rounding is a reporting convention, not a change in the variable.

Key Terms