The chapter's hinge, and a genuine change of machinery. A continuous random variable can take any value in an interval rather than isolated whole numbers, and its distribution is described by a probability density function whose defining property is that the area between it and the horizontal axis equals a probability. Since the largest probability is one, the largest area is one. Three consequences follow immediately: probabilities belong to intervals rather than to single values, the probability of any one exact value is zero because a vertical line has no width and therefore no area, and strict and non-strict inequalities give the same answer because they enclose the same region. The cumulative distribution function gives the area to the left of a value, and the area to the right is one minus it.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 5 — Continuous Random Variables
Continuous Probability Functions
Objectives
Five outcomes. The first is the whole section, and the other four are its consequences.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, pp. 292-294 — the section these objectives are drawn from
Warm-up
Chapter 4's variables took isolated whole numbers, and each carried its own probability.
Discussion prompt
A number is picked at random from between 0 and 20, with every value equally likely. What is the probability it is exactly 15? And what is the probability it lies between 4 and 15?
Hint: Try to list the possible values the way you would for a discrete variable.
Answer:
The list cannot be made. Between 0 and 20 there is no next value after 15, and no finite or even countable collection of values to spread probability across — so the chapter 4 method of assigning a probability to each value and adding them has nothing to work on.
The second question is easier than the first, which is a reversal. Between 4 and 15 is eleven twentieths of the range, so the answer ought to be 0.55, and that reasoning needs no list at all — it compares the size of a region with the size of the whole.
That is exactly the move this section makes. Probability becomes a measure of a region rather than a total over values, and once probability is an area the first question answers itself: a single value occupies no width, so it encloses no area, so its probability is zero.
Concept
A continuous probability density function is written f(x), and it is defined so that the area between it and the horizontal axis is equal to a probability. Since the maximum probability is one, the maximum area is also one.
probability density function — A function f(x) whose graph encloses area equal to probability. The total area between it and the horizontal axis is exactly one, and the area above any interval is the probability of landing in that interval.
\[ \text{PROBABILITY} = \text{AREA} \]
It is worth being clear that f(x) itself is not a probability. In Example 5.1 the height is one twentieth, which is not the probability of anything in particular; multiply it by a base and you get a probability. Densities can even exceed one in height — a uniform on an interval of length one half has height two — without anything going wrong, because it is the area and not the height that is capped at one.
Figure (svg): A horizontal line segment at one twentieth from zero to twenty, with the whole rectangle beneath it shaded to show a total area of one
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 292
Section
Section 1
Concept
For the density in Example 5.1 the region is a rectangle, so its area is base times height. The probability that x lies between two values is the area of the rectangle standing on that interval.
P(c < x < d) — The probability that X is in the interval between c and d. It is the area under the curve, above the horizontal axis, to the right of c and to the left of d.
\[ P(c < x < d) = (d - c) \cdot f(x) \quad \text{for a flat density} \]
The book's reminder at this point is a single line — area of a rectangle equals base times height — and its plainness is deliberate. Nothing about continuous probability is harder than that while the density is flat. The difficulty of the general case is entirely in finding areas under curved shapes, which is what the next idea's list of four methods addresses.
Figure (svg): The same rectangle with the strip from four to fifteen shaded, representing a probability of nought point five five
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 293 — Example 5.1 and the reminder about rectangles
Picture it
Base 20, height one twentieth.
Figure (svg): A horizontal line segment at one twentieth from zero to twenty, with the whole rectangle beneath it shaded to show a total area of one
The total area being one is not a coincidence of this example but a requirement on every density. It is the continuous counterpart of section 4.1's condition that a discrete distribution's probabilities must sum to one, and it plays exactly the same role: a check that nothing has been left out and nothing double-counted.
Worked example
Example 5.1's first question.
\[ f(x) = \tfrac{1}{20} \text{ on } [0, 20]; \text{ find } P(0 < x < 2) \]
Find the base
Why: From zero to two.
\[ \text{base } 2 \]
Find the height
Why: The value of f.
Multiply
Why: Base times height.
\[ \frac{2}{20} \]
Simplify
Why: One tenth.
\[ 0.1 \]
Figure (svg): The solution to Worked example a probability of 0.1 shown as a ladder of expressions, one row per legal move
\[ P(0 < x < 2) = (2)\left(\tfrac{1}{20}\right) = 0.1 \]
Verify: confirm why P(0 < x < 2) and P(x < 2) are the same
Why: The density is zero outside the interval from 0 to 20, so there is no area to the left of zero to add. Writing P(x < 2) therefore describes the same region as P(0 < x < 2). This is worth noticing because it does not hold in general: for a density defined on the whole line, P(x < 2) would include everything below zero as well.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 293
Faded example
f(x) = 1/20 on [0, 20]. Find P(2.3 < x < 12.7).
Fill in the blanks
P(2.3 < x < 12.7) = (10.4)\left(\tfrac0.52___\right) = ___
Why: The base is 12.7 minus 2.3, which is 10.4, and ten point four twentieths is 0.52. The proportional check agrees: the interval is a little over half the range, and the answer is a little over one half.
Worked example
Example 5.1's second question, with the region in the middle.
\[ \text{find } P(4 < x < 15) \]
Find the base
Why: Fifteen minus four.
\[ \text{base } 11 \]
Find the height
Why: Unchanged, since f is flat.
Multiply
Why: Eleven twentieths.
\[ \frac{11}{20} \]
Convert
Why: As a decimal.
\[ 0.55 \]
Figure (svg): The solution to Worked example a probability of 0.55 shown as a ladder of expressions, one row per legal move
\[ P(4 < x < 15) = (11)\left(\tfrac{1}{20}\right) = 0.55 \]
Verify: confirm the answer is plausible before trusting the arithmetic
Why: The interval from 4 to 15 is a bit more than half of the range from 0 to 20, so the answer should be a bit more than 0.5 — and 0.55 is. That check works because the density is flat, which makes probability proportional to length. It is the fastest sanity test available in this section, and it catches the commonest slip, which is subtracting the endpoints in the wrong order.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, pp. 293-294
Trap
\[ P(4 < x < 15) = (4 + 15)\left(\tfrac{1}{20}\right) = 0.95 \]
Take the base to be the two endpoints combined
Why: Both numbers describe the interval, so both belong in the base.
\[ 0.95 \text{ for an interval covering just over half the range} \]
The base of a rectangle is the distance between its edges, which is a difference and never a sum.
\[ P(4 < x < 15) = (15 - 4)\left(\tfrac{1}{20}\right) = 0.55 \]
Subtract the smaller endpoint from the larger to get the width
Why: Base means width, and width is a difference.
The proportional check catches it at once: an interval of length 11 out of 20 cannot carry 95 percent of the probability when the density is flat. Whenever a flat density is involved, compare the interval's share of the range against the answer before moving on — they must agree.
Estimation
f(x) = 1/8 on [0, 8].
Predict first
Roughly what is P(2.5 < x < 7.5)?
Correct: About 0.63.
Why: The interval has width 5 out of a range of 8, so the probability is five eighths, which is 0.625. Because the density is flat, probability is simply the interval's share of the range — no arithmetic beyond a fraction is needed. The option 1.25 is impossible on its face, since no probability exceeds one.
Two truths and a lie
All three concern area and probability.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. It is the AREA that is capped at one, not the height: a uniform density on an interval of length one half has height two, and its total area is still one. Confusing the height of f(x) with a probability is the commonest misreading of these pictures.
Prediction
Commit before reasoning.
Predict first
A uniform density on [0, 20] is replaced by one on [0, 5]. What happens to the height?
Correct: It rises to one fifth.
Why: The total area must remain one, so a narrower base forces a taller height: five times one fifth is one. This trade-off is worth holding on to, because it is what makes the uniform's density formula one over b minus a, and it explains why squeezing a distribution into a shorter interval makes it taller.
Section
Section 2
Concept
On an x-y graph a single value such as x = 15 is a vertical line, and a vertical line has no width. Its area is base times height with a base of zero, so the area is zero — and since probability is area, the probability is zero.
P(x = c) = 0 — The probability that a continuous random variable takes any one particular value is exactly zero. Not small, not negligible: zero, because the region enclosed has no width.
\[ P(x = 15) = (\text{base})(\text{height}) = (0)\left(\tfrac{1}{20}\right) = 0 \]
This is genuinely strange on first meeting, because x does take some value every time the experiment is run, and whatever value it takes had probability zero beforehand. The resolution is that zero probability does not mean impossible for continuous variables — it means the event occupies no width in a continuum. Impossibility and probability zero come apart here, and they did not in chapter 4.
Figure (svg): A rectangle with a single vertical red line drawn at x equals fifteen, showing that a line of zero width encloses zero area
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 294 — the vertical line at x = 15
Picture it
The same rectangle, with x = 15 drawn on it.
Figure (svg): A rectangle with a single vertical red line drawn at x equals fifteen, showing that a line of zero width encloses zero area
Notice how short the argument is: a line has no width, so it encloses no area, so its probability is zero. Nothing about limits or infinitesimals is needed. Every strange consequence in this lesson comes from the same single line of reasoning, which is why it is worth being able to reproduce it.
Worked example
Example 5.1's third question.
\[ f(x) = \tfrac{1}{20} \text{ on } [0, 20]; \text{ find } P(x = 15) \]
Draw the region
Why: x = 15 is a vertical line.
Find its base
Why: A line has no width.
\[ \text{base } 0 \]
Find its height
Why: The density there.
Multiply
Why: Zero times anything.
\[ 0 \]
Figure (svg): The solution to Worked example the probability of exactly 15 shown as a ladder of expressions, one row per legal move
\[ P(x = 15) = (0)\left(\tfrac{1}{20}\right) = 0 \]
Verify: confirm this does not make every outcome impossible
Why: Every run of the experiment produces some value, and each of those had probability zero in advance — so probability zero cannot mean impossible here. What it means is that the event has no width in a continuum. Ask instead for the probability of landing within a small interval around 15, say from 14.9 to 15.1, and you get 0.2 times one twentieth, which is 0.01: small, positive and answerable.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 294
Prediction
Commit before reasoning.
Predict first
For a continuous X on [0, 20], what is P(x = 7.4)?
Correct: 0.
Why: Any single value is a vertical line with no width, so it encloses no area whatever the density looks like. The last option is tempting but wrong: the answer is zero for every continuous density, flat or not, which is why it can be stated as a general rule rather than computed each time.
Worked example
Watching the probability approach zero as the width does.
\[ \text{intervals of width } 2, \; 0.2, \; 0.02 \text{ around } 15 \]
Width 2
Why: From 14 to 16.
\[ 0.10 \]
Width 0.2
Why: From 14.9 to 15.1.
\[ 0.01 \]
Width 0.02
Why: From 14.99 to 15.01.
\[ 0.001 \]
Width 0
Why: The single value itself.
\[ 0 \]
Figure (svg): The solution to Worked example shrinking an interval shown as a ladder of expressions, one row per legal move
\[ P \text{ scales with the width; width } 0 \Rightarrow P = 0 \]
Verify: confirm the pattern is proportionality rather than a limit argument
Why: Each row's width is a tenth of the one above and each probability is a tenth of the one above, because the density is flat and probability is base times height. The last row is not a limit being taken but the same multiplication with a base of zero. The pattern is reassuring, but the one-line argument from the previous example is the actual proof.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, pp. 293-294
Trap
\[ P(x = 15) = 0 \;\Rightarrow\; x \text{ can never equal } 15 \]
Carry chapter 4's meaning of probability zero across
Why: There, an outcome with probability zero was one that could not happen.
\[ \text{but } x \text{ takes SOME value every time} \]
Whatever value it takes had probability zero in advance, so if zero meant impossible then nothing could happen at all.
\[ P(x = 15) = 0 \text{ means the event encloses no area} \]
Read zero probability as zero width, not as impossibility
Why: For continuous variables the two ideas come apart.
The practical upshot is that questions about exact values are the wrong questions to ask of a continuous variable. Ask for an interval instead — within a tenth of 15, or above 15 — and you get an answer with content. Every worked problem in the rest of this book asks about intervals for exactly this reason.
Two truths and a lie
All three concern single values.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false, and the word exactly matters. The base is zero, not merely tiny, so the product is exactly zero rather than a small positive number. Treating it as small-but-positive leads directly to the error of thinking probabilities of single values could be added up.
Sorting
Ask whether the region has width.
Sort into buckets
Sort each question about a continuous X on [0, 20].
Item (d) is worth a second look: x = 0 is an endpoint of the interval, which makes it feel different from an interior value, but the argument is unchanged. A line at the edge has no width either.
Explain it
A classmate objects that if every value has probability zero, then nothing can ever happen.
Discussion prompt
In three sentences or fewer, resolve it.
Hint: Ask what they would say the probability of landing in a small interval is.
Answer:
Agree that every individual value has probability zero, and that some value nevertheless occurs — so zero cannot mean impossible in this setting.
Point out that probability here measures width, and a single point has no width, while any interval around it does, however small.
So the meaningful questions are about intervals: the chance of landing within a tenth of 15 is 0.01, positive and perfectly ordinary, and it shrinks toward zero only as the interval does.
Section
Section 3
Concept
Because the probability of each endpoint is zero, including or excluding the endpoints changes nothing. The book states it directly: P(c < x < d) is the same as P(c at most x at most d), because probability is equal to area.
inclusive and exclusive endpoints — For a continuous variable these give identical probabilities, since each endpoint contributes zero area. For a discrete variable they differ by the probability of the endpoint itself, which is generally not zero.
\[ P(c < x < d) = P(c \le x \le d) \]
This is a genuine convenience rather than a technicality, and it removes a whole category of chapter 4 anxiety. There, whether at least five meant from five upward or from six upward changed the answer, and getting it wrong was the commonest error in the binomial section. Here the distinction simply does not arise, and it will not arise again for any continuous variable in the rest of the book.
Figure (svg): Two columns contrasting a discrete random variable, whose values carry probabilities, with a continuous one, where only intervals do
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 292 — the chapter's opening list of properties
Picture it
Six differences, of which the last is the one met first.
Figure (svg): Two columns contrasting a discrete random variable, whose values carry probabilities, with a continuous one, where only intervals do
The right-hand column is not a set of exceptions to memorise but six consequences of one definition. Whenever a continuous fact seems arbitrary, the thing to do is trace it back to probability equals area, and it usually takes one step.
Worked example
All giving the same answer for the density on [0, 20].
\[ P(4 < x < 15), \; P(4 \le x \le 15), \; P(4 \le x < 15), \; P(4 < x \le 15) \]
Compute the first
Why: Base 11, height one twentieth.
\[ 0.55 \]
Add the endpoint at 4
Why: A line of zero width.
\[ \text{adds } 0 \]
Add the endpoint at 15
Why: Also zero width.
\[ \text{adds } 0 \]
Compare all four
Why: No difference anywhere.
\[ \text{all } 0.55 \]
Figure (svg): The solution to Worked example four ways to write one probability shown as a ladder of expressions, one row per legal move
\[ P(4 < x < 15) = P(4 \le x \le 15) = 0.55 \]
Verify: confirm the same four expressions would differ for a discrete variable
Why: Take the binomial of section 4.3 with values 4 and 15 both possible. Including 4 adds P(x = 4), which is a positive number, so the four expressions there give up to four different answers — and keeping them straight was the whole point of that section's habit of listing which values are in. Here nothing needs listing, because the endpoints contribute nothing.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, pp. 292-293
Matching
Match each statement to the kind of variable it holds for.
Match the pairs
Why: The four rows are two pairs, and each pair is one difference stated twice. Deciding which kind of variable you have is therefore the first step in any problem, because it settles the method as well as the endpoint conventions.
Worked example
The same question asked of a discrete and a continuous variable.
\[ \text{is } P(x < 12) = P(x \le 12)? \]
Continuous, on [0, 20]
Why: The endpoint has zero area.
Discrete, a die roll
Why: P(x = 12) may be positive.
Say what makes the difference
Why: Whether the endpoint carries probability.
State the rule
Why: Continuous variables only.
Figure (svg): The solution to Worked example a boundary that would have mattered shown as a ladder of expressions, one row per legal move
\[ \text{continuous: } P(x < 12) = P(x \le 12) = 0.60 \]
Verify: confirm the rule is about the variable rather than the question
Why: Nothing in the phrasing of the question tells you which case you are in — the words less than mean the same thing in both. It is the variable that decides, so the first thing to establish in any probability problem is whether X is discrete or continuous. That decision governs the whole method, not just the endpoints.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, pp. 292-294
Error analysis
The density is flat at one twentieth. Which claims hold?
Annotate
On: \( \begin{aligned} &(1)\; P(x \le 12) = P(x < 12) + P(x = 12) = 0.60 + 0.05 \\ &(2)\; P(x = 12) \text{ is small but positive} \\ &(3)\; P(4 < x < 15) \ne P(4 \le x \le 15) \\ &(4)\; P(x \le 12) = P(x < 12) = 0.60 \end{aligned} \)
Claim (1) is the instructive one, because it makes two errors that partly hide each other: it treats a single value as carrying probability, and it uses the density's height as that probability. Height is not probability — only area is.
Fill the middle
The reason strict and non-strict inequalities coincide.
Fill in the blanks
P(c < x < d) = P(c \le x \le d) \textarea ___
Why: Zero area, because a vertical line has no width. Since probability is area, the endpoints contribute nothing and the two expressions describe the same region.
Two truths and a lie
All three concern endpoints.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false because it drops the condition that does all the work. The interchangeability is a property of continuous variables specifically, and applying it to a binomial would produce wrong answers on every boundary question.
Discrimination
Decide by the kind of variable, not by the wording.
Sort into buckets
Sort each situation by whether P(x < c) and P(x at most c) agree.
Section
Section 4
Concept
P(X at most x), which can also be written P(X < x) for continuous distributions, is called the cumulative distribution function, or CDF. It gives area to the left. The area to the right is found by subtracting from one.
cumulative distribution function — The function giving P(X at most x): the area under the density to the left of x. Its complement, P(X > x), is one minus it, because the two regions together make up the whole area of one.
\[ P(X > x) = 1 - P(X < x) \]
Almost every table and calculator command in the rest of the book reports a CDF, which is why it is worth naming here rather than in passing. The normal table of chapter 6 gives area to the left; so do the calculator's normalcdf and tcdf. Any question about a right tail or a middle region is therefore answered by combining left-tail values, and the two moves needed are subtraction from one, and subtraction of one left tail from another.
Figure (svg): Two rectangles side by side, the left shaded up to twelve and the right shaded from twelve onward, showing that the two areas total one
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 294 — the CDF and the complement
Picture it
The area to the left of 12, and the area to its right.
Figure (svg): Two rectangles side by side, the left shaded up to twelve and the right shaded from twelve onward, showing that the two areas total one
The complement rule here is the same one section 3.2 stated for events, restated in areas. That is a pattern worth watching for across the rest of the book: the probability rules do not change when the variable becomes continuous, only the way the probabilities are computed does.
Worked example
The density on [0, 20], with the cut at 12.
\[ \text{find } P(X > 12) \text{ from the CDF} \]
Compute the left area
Why: Base 12, height one twentieth.
\[ 0.60 \]
Recall the total
Why: The whole density.
\[ \text{area } 1 \]
Subtract
Why: One minus the left area.
\[ 0.40 \]
Check by direct area
Why: Base 8, height one twentieth.
\[ 0.40\text{ again} \]
Figure (svg): The solution to Worked example a right tail from a left one shown as a ladder of expressions, one row per legal move
\[ P(X > 12) = 1 - 0.60 = 0.40 \]
Verify: confirm both routes must agree, and why one is usually preferred
Why: The direct route works here only because the region to the right is another rectangle. For the normal density of chapter 6 there is no formula for the area of the right-hand region, and the table gives only left tails — so the complement is not a convenience there but the only available route. Practising it on a case where both work is the point of doing it this way.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 294
Faded example
For the density on [0, 20], the left tail at 18 is 0.90 and at 6 is 0.30.
Fill in the blanks
P(6 < X < 18) = 0.90 - 0.30 = 0.60
Why: The upper tail value comes first, and removing the lower one leaves the strip between. The direct check agrees: a base of 12 at height one twentieth is 0.60.
Worked example
The move that answers every between question later in the book.
\[ \text{find } P(4 < X < 15) \text{ using CDF values} \]
Left tail at 15
Why: Base 15.
\[ 0.75 \]
Left tail at 4
Why: Base 4.
\[ 0.20 \]
Subtract
Why: The larger minus the smaller.
\[ 0.55 \]
Compare with the direct area
Why: Base 11.
\[ 0.55,\text{ as before} \]
Figure (svg): The solution to Worked example a middle region from two left tails shown as a ladder of expressions, one row per legal move
\[ P(4 < X < 15) = 0.75 - 0.20 = 0.55 \]
Verify: confirm the subtraction removes exactly the right region
Why: The area to the left of 15 includes everything below 4, and subtracting the area to the left of 4 removes precisely that surplus, leaving the strip between them. Getting the order backwards gives negative 0.55, which is impossible and therefore self-announcing — always subtract the smaller left tail from the larger.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 294
Trap
\[ P(4 < X < 15) = 0.20 - 0.75 = -0.55 \]
Subtract the larger left tail from the smaller
Why: The interval starts at 4, so 4 comes first.
\[ -0.55 \quad \text{(a negative probability)} \]
The order of the endpoints in the question has nothing to do with the order of the subtraction.
\[ P(4 < X < 15) = 0.75 - 0.20 = 0.55 \]
Take the left tail at the UPPER endpoint and remove the left tail at the lower
Why: The bigger region minus the surplus it contains.
A negative answer is impossible, so this error announces itself immediately — which makes it much less dangerous than the endpoint errors of chapter 4, where wrong answers stayed plausible. Reading the subtraction as bigger region minus the part you do not want, rather than as a rule about which number comes first, prevents it entirely.
Two truths and a lie
All three concern the CDF.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and confuses the CDF with the density. The exact value has probability zero; the CDF accumulates everything up to and including x. The word cumulative is the reminder — it is a running total of area, not a value read off the curve.
Prediction
Commit before reasoning.
Predict first
As x increases, what does the CDF do?
Correct: It increases toward one.
Why: Moving right adds more area to the left and never removes any, so the CDF never decreases, and it approaches the total area of one. That shape — rising from zero to one — is the same for every continuous distribution in the book, which makes it a useful check on any CDF value you compute or look up.
Estimation
For some continuous X, the CDF at 30 is 0.82.
Predict first
What is P(X > 30)?
Correct: 0.18.
Why: One minus 0.82. The two regions must total one, so a large left tail forces a small right one. Note that this needed no knowledge of the density at all — the complement rule works for any continuous distribution, which is why it survives into every later chapter.
Section
Section 5
Concept
The book finds areas by geometry, formulas, technology or probability tables. It notes that in general calculus is needed to find the area under the curve for many probability density functions, and that where formulas are used in this textbook they were themselves obtained by integral calculus.
the four routes to an area — Geometry when the region is a rectangle or triangle; a formula derived elsewhere; a calculator or software command; or a printed probability table. Which one applies depends on the density, not on the question.
\[ \text{rectangle} \to \text{geometry}; \quad \text{normal} \to \text{table or technology} \]
It is worth being clear about what is and is not being given up. Not using calculus limits which densities can be integrated by hand; it does not limit which densities can be used, nor which questions can be answered. The exponential's formula in section 5.3 and the normal's table in chapter 6 both come from integrals someone else performed, and they are exact rather than approximate.
Figure (svg): Four ways of finding the area under a density, listed as a numbered procedure
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 292 — the note about calculus and the four methods
Picture it
Each finds the same area; they differ in what the density allows.
Figure (svg): Four ways of finding the area under a density, listed as a numbered procedure
Chapter 5's two distributions are chosen partly because they are tractable: the uniform's region is a rectangle and the exponential has a short closed form. The normal, which needs a table, is held back to chapter 6 for exactly that reason.
Worked example
Three densities, three methods.
\[ \text{a uniform; an exponential; a normal} \]
The uniform
Why: The region is a rectangle.
The exponential
Why: A short closed form exists.
The normal
Why: No elementary formula for the area.
Say what decides
Why: The shape of the density.
Figure (svg): The solution to Worked example choosing the route shown as a ladder of expressions, one row per legal move
\[ \text{uniform} \to \text{geometry}, \quad \text{exponential} \to \text{formula}, \quad \text{normal} \to \text{table} \]
Verify: confirm that all three give exact answers
Why: Geometry gives an exact area for a rectangle, and the exponential's formula is exact. The normal's table is rounded to four decimal places, which is a limit of the printed table rather than of the method — a calculator gives as many digits as wanted. So none of these routes is an approximation in the way that, say, the Poisson approximation to a binomial was.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, p. 292
Sorting
Ask what shape the region is.
Sort into buckets
Sort each density by how its areas are found in this book.
Everything in the left column belongs to chapter 5 and everything in the right belongs to chapter 6 and beyond. That is not an accident of ordering: the book teaches the idea on shapes you can measure before moving to ones you must look up.
Worked example
Geometry still works when the shape is simple.
\[ f(x) = \tfrac{x}{8} \text{ on } [0, 4]; \text{ total area} \]
Identify the shape
Why: A line from the origin.
Find the base
Why: From 0 to 4.
\[ \text{base } 4 \]
Find the height at x = 4
Why: Four eighths.
\[ \text{height } 0.5 \]
Take half the product
Why: Area of a triangle.
\[ 1 \]
Figure (svg): The solution to Worked example a region that is not a rectangle shown as a ladder of expressions, one row per legal move
\[ \tfrac{1}{2}(4)(0.5) = 1 \]
Verify: confirm the check that any candidate density must pass
Why: Total area one is the requirement, and it is checkable by geometry whenever the shape is simple. Had the total come to 2, the function would not be a density and could be repaired by halving it. This is the continuous counterpart of section 4.1's requirement that a discrete distribution's probabilities sum to one, and it is used the same way — as the first test of whether a proposed distribution is legitimate at all.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, pp. 292-293
Trap
\[ \text{the area needs an integral, so this cannot be done} \]
Stop at the first curved density
Why: Areas under curves are a calculus topic.
\[ \text{but the normal's areas are tabulated to four decimals} \]
Someone else performed the integration, and the result is printed in the back of the book.
\[ \text{geometry, formula, technology, or table} \]
Match the route to the density's shape
Why: Only the first of the four needs the region to be simple.
The distinction worth holding is between finding an area and computing one. Calculus is how these areas were found, once, by someone; using them needs only a table lookup or a calculator command. The book is explicit about this so that a student without calculus does not conclude that the subject is closed to them.
Two truths and a lie
All three concern finding areas.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Calculus is how the areas were originally found, not what is needed to use them. Every technique in chapters 6 through 13 relies on tabulated or computed areas, and none of them asks the reader to integrate anything.
Faded example
f(x) = x/8 on [0, 4], a straight line from the origin.
Fill in the blanks
\text0.5 = \tfrac1___(4)(___) = ___
Why: The height at the right end is four eighths, or 0.5, and half of base times height is 1. Total area one is what makes it a legitimate density, and geometry settles it without any calculus.
Explain it
A classmate says they cannot do this chapter because they have not taken calculus.
Discussion prompt
In three sentences or fewer, reassure them accurately — without pretending calculus is irrelevant.
Hint: Distinguish finding an area once from using it many times.
Answer:
Agree that calculus is genuinely what produces these areas, and that the book says so rather than hiding it.
But the integration was done once, by someone else, and its results are printed as tables or built into calculator commands — so using them needs arithmetic, not integration.
In chapter 5 it is simpler still: the uniform's region is a rectangle, so its area is base times height, and no calculus is involved at any point.
Comparison
Fill the blanks. Every row is a consequence of one definition.
Comparison matrix
| Question | Discrete (ch. 4) | Continuous (ch. 5) |
|---|---|---|
| How is probability found? | by adding values' probabilities | as an area under f(x) |
| What is P(x = c)? | generally positive | exactly zero |
| Do strict and non-strict agree? | no, they can differ | yes, always |
| What must total one? | the sum of the probabilities | the total area |
The last row is the one that carries across unchanged, and it is the most useful. Whether you are adding stems or measuring area, the total must be one, and checking it is the fastest way to catch a distribution that has been set up wrongly.
Pattern
Five steps, and the first is the one that decides the method.
For a right tail or a middle region, work from left tails: subtract from one, or subtract the smaller left tail from the larger.
OpenStax Introductory Business Statistics 2e, §5.1 Properties of Continuous Probability Density Functions §5.1 Properties of Continuous Probability Density Functions
Check
The defining property.
Check your understanding
For a continuous probability density function, what must the total area between the curve and the horizontal axis equal?
Answer: A
Why: Since the maximum probability is one, the maximum area is also one, and the total area is exactly one for every density.
Check
An exact value.
Check your understanding
For a continuous random variable X, what is P(x = 7)?
Answer: A
Why: A single value is a vertical line with no width, so it encloses no area, so its probability is exactly zero.
Check
Working from the CDF.
Check your understanding
The CDF of a continuous X at 40 is 0.73. What is P(X > 40)?
Answer: A
Why: The area to the right is one minus the area to the left, and one minus 0.73 is 0.27.
Real world
A quality engineer records the diameter of machined pins, specified as 12.00 millimetres. A colleague reports that in 5,000 measurements not a single pin came out at exactly 12.00 millimetres, and concludes that the machine is incapable of hitting its target.
Discussion prompt
Assess the conclusion, and say what question the engineer should have asked instead.
Hint: Ask what the probability of any exact value is for a continuous measurement.
Answer:
The conclusion does not follow, and the observation was inevitable. Diameter is a continuous measurement, so the probability of any exact value — including 12.00 — is zero. Five thousand pins missing the target exactly is precisely what the model predicts, and it would remain true of a perfect machine.
The reasoning also mistakes the recording for the measurement. A gauge reporting two decimal places is rounding: a pin logged as 12.00 is really anywhere in the interval from 11.995 to 12.005. So pins recorded as 12.00 do exist, and what the colleague has actually observed is either a gauge with more digits, or a genuine offset in the machine.
\[ P(X = 12.00) = 0 \quad \text{but} \quad P(11.995 < X < 12.005) > 0 \]
The right question is about an interval. Engineering specifications are always written as a target plus a tolerance — 12.00 give or take 0.05, say — and the useful quantity is the probability of falling inside that band, which is a genuine area and can be estimated from the data. That number has content: it is the expected proportion of conforming parts.
Two further points are worth making. The proportion outside tolerance is the complement of that area, computed by subtracting from one, exactly as in this lesson. And the colleague's observation could still be evidence of a problem if the measurements cluster somewhere other than 12.00 — but the evidence would be the location of the cluster, not the absence of exact hits, which carries no information at all.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why is P(x = c) equal to zero for a continuous random variable?
Correct: Because a vertical line has no width and therefore no area.
\[ P(x = c) = (\text{base})(\text{height}) = (0) \cdot f(c) = 0 \]
Why: Probability equals area, area equals base times height, and the base of a single value is zero. The result is exact rather than approximate, and it has nothing to do with measurement precision — it would hold for a perfect instrument. Confusing the two is the commonest way this fact gets misremembered.
Explain it
They wrote P(x at most 12) = P(x < 12) + P(x = 12) and added the density's height of 0.05.
Discussion prompt
In three sentences or fewer, correct both errors.
Hint: Ask them what P(x = 12) is, and separately what 0.05 actually measures.
Answer:
Start with the second term: P(x = 12) is zero for a continuous variable, because a single value is a line with no width and encloses no area.
Then point at the 0.05 they used for it — that is the HEIGHT of the density, and a height is not a probability; only a height multiplied by a base is.
So the equation is right in form but both extra pieces vanish, leaving P(x at most 12) equal to P(x < 12), which is 0.60.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, remember that the total area is one because the largest probability is one. For the second, subtract the endpoints to get the base and multiply. For the third, the base of a line is zero, so the product is zero. For the fourth, subtract from one for a right tail, and subtract the smaller left tail from the larger for a middle region. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write PROBABILITY EQUALS AREA and under it the reason the total area is one. Below that, draw the density f(x) = 1/20 on the interval from 0 to 20 three times, side by side. Shade the first from 0 to 2, the second from 4 to 15, and on the third draw a single vertical line at x = 15. Under each, write the base, the height and the product, and check that the three answers are 0.1, 0.55 and 0. Beside the third, write one sentence explaining why zero does not mean impossible here. In the middle of the page, make a two-column table headed discrete and continuous with four rows: how probability is found, what P(x = c) is, whether strict and non-strict inequalities agree, and what must total one. At the bottom, draw the rectangle twice more, shading one to the left of 12 and the other to the right, write the two areas, and check they sum to one. Finish by writing the four routes to an area and naming which one chapter 5 will use and which one chapter 6 will need.
Check your work by adding the two areas in the bottom row: 0.60 and 0.40 must give exactly 1. And check the middle table by asking, for each row, whether the continuous entry follows from probability equals area — all four of them do, which is the point of the lesson.
Recap
Five things, all consequences of one sentence.
| If you see | Then |
|---|---|
| A continuous random variable | Probability is an area, not a sum |
| P(x = c) asked of a continuous X | Zero, exactly |
| Endpoints included or excluded | No difference: same region, same area |
| A flat density | Base times height, and check the proportion |
| A right tail wanted | One minus the left tail |
| A middle region wanted | The larger left tail minus the smaller |
| A curved density | A table or technology, not geometry |
Section 5.2 takes the flat density of this lesson and names it. The uniform distribution describes events that are equally likely across an interval, and because its region is always a rectangle every question about it — probabilities, percentiles, even conditional probabilities — reduces to arithmetic on base and height.
OpenStax Introductory Statistics 2e, §5.1 Continuous Probability Functions §5.1, pp. 292-294 — everything on these slides traces back here
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