Case Study 2 — How Long Until the Next Call? FTC Behind the CDF

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Case Study 2 — How Long Until the Next Call? FTC Behind the CDF

Field: Probability and statistics (operations / reliability analysis) FTC tools used: Accumulation function (§14.2), FTC Part 1 linking density and CDF (§14.3), FTC Part 2 evaluation (§14.4), net change as probability (§14.7, §14.10)

A Staffing Question with a Hidden Integral

Marcus Bell runs the overnight support desk for a mid-size internet provider. His staffing model lives or dies on one question: how long until the next call? If he understood the distribution of the gap between incoming calls, he could size his crew, set service-level targets, and predict how often two calls collide before the first is resolved. Marcus has months of timestamped call logs, and from them a single number jumps out: on the overnight shift, calls arrive at an average rate of one every four minutes.

That average is the seed of a model. When events arrive independently at a constant average rate, the waiting time $T$ between consecutive events follows the exponential distribution, whose probability density function is $$p(t) = \lambda e^{-\lambda t}, \qquad t \ge 0,$$ where $\lambda$ is the arrival rate. An average gap of four minutes means $\lambda = \tfrac{1}{4} = 0.25$ calls per minute. The density $p(t)$ is not itself a probability — it is a rate of probability per minute, exactly the way velocity is a rate of distance per second. To turn a rate into an amount, Marcus needs to integrate. This is where FTC enters, and it enters as the very definition of a cumulative distribution function.

The CDF Is an Accumulation Function

The probability that the next call arrives within $x$ minutes is the cumulative distribution function (CDF): $$F(x) = P(T \le x) = \int_0^{x} p(t)\,dt = \int_0^x \lambda e^{-\lambda t}\,dt.$$ Read that integral against §14.2. The lower limit is frozen at $0$; the upper limit $x$ slides to the right; $F(x)$ is the running total of probability swept out from $0$ up to $x$. The CDF is the accumulation function of the density. And FTC Part 1 (§14.3) tells Marcus the relationship the other direction: $F'(x) = p(x)$ — the density is the derivative of the CDF. Density and CDF are an FTC pair, the probabilistic twins of velocity and displacement.

Marcus evaluates the accumulation in closed form using FTC Part 2. An antiderivative of $\lambda e^{-\lambda t}$ is $-e^{-\lambda t}$ (check: $\tfrac{d}{dt}(-e^{-\lambda t}) = \lambda e^{-\lambda t}$ ✓), so $$F(x) = \int_0^x \lambda e^{-\lambda t}\,dt = \Big[-e^{-\lambda t}\Big]_0^x = -e^{-\lambda x} - (-e^{0}) = 1 - e^{-\lambda x}.$$ With $\lambda = 0.25$, $$\boxed{F(x) = 1 - e^{-0.25 x}.}$$ Two sanity checks confirm the formula behaves like a probability. At $x = 0$, $F(0) = 1 - e^0 = 0$: there is zero chance the call has already arrived before any time passes. As $x \to \infty$, $e^{-0.25x} \to 0$, so $F \to 1$: the call eventually arrives with certainty. The accumulation runs from $0$ to $1$, exactly as a probability must.

Question 1 — Will a Call Arrive Within Four Minutes?

Marcus wants to know how often a call lands inside the average gap itself. That is just $F$ evaluated at $x = 4$: $$P(T \le 4) = F(4) = 1 - e^{-0.25 \cdot 4} = 1 - e^{-1} = 1 - 0.3679 = 0.6321.$$ About 63% of inter-call gaps are shorter than the four-minute mean. This surprises Marcus — he expected the average to sit near the middle, with calls "usually" coming after four minutes. But the exponential distribution is right-skewed: most gaps are short, a few are very long, and the long tail drags the mean above the typical gap. FTC gave him the number; the number corrected his intuition.

Question 2 — The Long-Gap Risk

For shift planning, Marcus cares about quiet stretches — gaps long enough that an agent might step away. What is the chance of a gap longer than ten minutes? He can read this straight off the complement of the CDF: $$P(T > 10) = 1 - F(10) = 1 - \big(1 - e^{-0.25 \cdot 10}\big) = e^{-2.5} = 0.0821.$$ So roughly 8% of gaps exceed ten minutes. Equivalently, this is the integral of the density from $10$ to infinity — an improper integral, since the upper limit is unbounded. FTC Part 2 still drives it: $\int_{10}^{\infty}\lambda e^{-\lambda t}\,dt = \big[-e^{-\lambda t}\big]_{10}^{\infty} = 0 - (-e^{-2.5}) = e^{-2.5}$. The careful handling of that $\infty$ as a limit is the business of Chapter 17; here we simply note that the antiderivative's value at the upper end goes cleanly to zero, so FTC's subtraction still works.

Question 3 — Probability in a Window: Net Change

Marcus's real operational question is about windows: what fraction of gaps fall between two and six minutes — long enough to finish the previous call, short enough to keep agents busy? A probability over an interval is a net change in the CDF (§14.7, §14.10), the difference of accumulations at the two endpoints: $$P(2 \le T \le 6) = \int_2^6 p(t)\,dt = F(6) - F(2).$$ Evaluate each: $$F(6) = 1 - e^{-1.5} = 1 - 0.2231 = 0.7769, \qquad F(2) = 1 - e^{-0.5} = 1 - 0.6065 = 0.3935.$$ Subtract: $$P(2 \le T \le 6) = 0.7769 - 0.3935 = 0.3834.$$ About 38% of gaps land in that two-to-six-minute window. Notice the structure: $F(6) - F(2)$ is precisely "$F(\text{top}) - F(\text{bottom})$," the evaluation bar of §14.5 applied to the accumulation function itself. A probability over an interval is the same subtraction Marcus would write for any definite integral — the net change in accumulated probability.

The slogan, in probability dress. "The integral of a rate of change is the net change" (§14.7) reads, in this setting, "the integral of a probability density over $[a,b]$ is the probability that the value lands in $[a,b]$." Same theorem, new vocabulary. Every CDF you ever build is an accumulation function, and every interval probability is a net change.

Question 4 — The Mean as an Integral

Finally, Marcus wants to recover the four-minute average from the model, as a consistency check that his calculus matches his data. The mean of a continuous distribution is itself an integral: $$E[T] = \int_0^\infty t\,p(t)\,dt = \int_0^\infty t\,\lambda e^{-\lambda t}\,dt.$$ This is an improper integral that needs integration by parts (Chapter 15) and a limit at infinity (Chapter 17), so Marcus flags it as out of reach with §14 tools alone — but the standard result is $E[T] = 1/\lambda = 1/0.25 = 4$ minutes. The model's mean matches the observed average of one call every four minutes exactly. The data fixed $\lambda$; FTC turned $\lambda$ into a CDF; the CDF answered every operational question; and the mean closes the loop back to the data.

Why FTC Was the Whole Story

Strip away the staffing context and this case is §14.2–§14.7 with new labels. The density is a rate. The CDF is its accumulation function. FTC Part 1 says the density is the CDF's derivative; FTC Part 2 evaluated the CDF in closed form; and every probability Marcus needed — "within four minutes," "longer than ten," "between two and six" — was either an accumulation $F(x)$ or a net change $F(b) - F(a)$. He never simulated a single call. He had a rate and FTC, and that was the entire toolkit.

Discussion Questions

Marcus found $P(T \le 4) = 0.632$, larger than the "intuitive" $0.5$. Explain in terms of the skew of the exponential density why the median gap is shorter than the mean gap. (The median solves $F(m) = 0.5$ — solve it.)
Verify FTC Part 1 directly for this model: differentiate $F(x) = 1 - e^{-0.25x}$ and confirm you recover the density $p(x) = 0.25 e^{-0.25x}$.
Question 2's long-gap probability used an improper integral. Re-express it as a limit, $\lim_{b\to\infty} \int_{10}^{b}\lambda e^{-\lambda t}\,dt$, evaluate the finite integral by FTC, then take the limit. Why does the antiderivative's upper value vanish?
Suppose the overnight rate doubled to one call every two minutes ($\lambda = 0.5$). Recompute $P(2 \le T \le 6)$. Did the window probability rise or fall, and does the direction match your intuition about a busier shift?

A Short Annotated Reading

Stewart, Calculus: Early Transcendentals, §8.5 ("Probability"). Stewart builds probability density functions, CDFs, and means as definite integrals; this case is essentially §8.5 reached early through the FTC lens, so read it next to see the mean and variance integrals done in full.
OpenStax Calculus Vol. 2, §2.2 ("Determining Volumes...") and the probability material in §6.8 ("Exponential Growth and Decay"); the latter is the free analog for the exponential model and its rate parameter $\lambda$.
Ross, A First Course in Probability, ch. 5 ("Continuous Random Variables"). The probabilist's home for the density-CDF relationship $F'(x) = p(x)$; read §5.5 on the exponential distribution to see the same $1 - e^{-\lambda x}$ derived and its memoryless property explained.