Week 11 — Common continuous models

Uniform, exponential, and normal

The week question

Last week you learned the grammar of continuous random variables: a density \(f(x)\) whose area gives probability, a cdf \(F(x) = P(X \le x)\), and the rule that any single point has probability zero. That grammar is general — it works for any continuous variable. This week we stock the shelf with the three named densities that cover most of what you will actually meet, just as Week 9 stocked the discrete shelf. The week’s question is again a recognition question: given a continuous scenario, which standard model fits, and what feature of the story tells you so? Once you can name the model — Uniform, Exponential, or Normal — its density, its mean, and the way you read probabilities off it all come along for free.

We have already met the machinery these models plug into. In Week 10 you computed probabilities as areas (\(P(a \le X \le b) = \int_a^b f(x)\,dx\)) and built cdfs by integrating densities. This week we do not re-derive that machinery; we name the three densities you will reach for most, say what each one assumes, and practice matching a story to a model.

Why this matters

Naming the model is what turns a sentence like “the shuttle comes on average every fifteen minutes” into a number you can compute. If you can say “the wait is \(\text{Exponential}\) with rate \(4\) per hour,” you immediately know its density, that its mean is \(15\) minutes, and how to get \(P(\text{wait} \le 15)\) — no fresh integration setup required each time. As in Week 9, the hard part is rarely the arithmetic; it is reading the assumptions in the story carefully enough to choose the right model.

These three models also travel far beyond shuttles. The same exponential that describes a waiting time describes the lifetime of a component before it fails, the gap between earthquakes, or the time until a radioactive atom decays. The same normal that describes a commute time describes measurement error, heights in a population, or the average of many small independent effects (a fact Week 13 will make precise with the Central Limit Theorem). And the uniform — “any value in a range, all equally likely” — is the honest model for genuine no-information situations and the engine behind nearly every simulation you will run. Learning the assumptions behind each, not just its formula, is what lets you reuse them.

Learning goals

By the end of this week you should be able to:

State, in words and in symbols, the Uniform, Exponential, and Normal models, including each density and (as results cross-linked to Weeks 8 and 10) each mean.
Read a continuous scenario and identify which of the three standard models its assumptions select — and name the feature that rules the others out.
Compute a probability for our recurring commuter case: an exponential shuttle wait and a normal commute time, using the cdf and the standard-normal \(\Phi\).
Apply the 68–95–99.7 rule and the standardization \(Z = (X - \mu)/\sigma\) to read normal probabilities.
Recognize a new scenario — the time until a component fails — as exponential, and explain the memoryless property that makes it special.
Read shown R code (dunif/punif, dexp/pexp, dnorm/pnorm) so you can see how software returns these same probabilities.

Core vocabulary

Uniform\((a, b)\) — a continuous variable equally likely to land anywhere in the interval \([a, b]\); its density is flat. Written \(X \sim \text{Uniform}(a, b)\).
Exponential\((\lambda)\) — the waiting time until the next event when events arrive at a constant rate \(\lambda\) (the Poisson rate from Week 9). Written \(T \sim \text{Exponential}(\lambda)\); its mean is \(1/\lambda\).
Rate \(\lambda\) — for the exponential, the average number of events per unit time; large \(\lambda\) means short waits. Mind the parameterization: we use rate, not mean, so \(\text{Exponential}(4)\) has mean \(1/4\).
Normal\((\mu, \sigma)\) — the symmetric bell-shaped model parameterized by its mean \(\mu\) and its standard deviation \(\sigma\). Written \(X \sim \text{Normal}(\mu, \sigma)\).
Standard normal \(Z\) — the special case \(\text{Normal}(0, 1)\); its cdf is written \(\Phi(z) = P(Z \le z)\).
Standardization — the transform \(Z = (X - \mu)/\sigma\) that re-expresses any normal value as a number of standard deviations from the mean, so a single \(\Phi\) table or function handles every normal.
Memoryless property — the exponential’s signature: having already waited does not change the distribution of the remaining wait.

Concept development

Uniform: every value in a range equally likely

The simplest continuous model is the Uniform\((a, b)\). The variable can land anywhere in \([a, b]\), and no part of that interval is favored over another, so the density must be flat. Because the total area under a density is \(1\) and the interval has width \(b - a\), the height is forced to be \(1/(b - a)\):

\[ f(x) = \frac{1}{b - a}, \quad a \le x \le b, \qquad f(x) = 0 \text{ otherwise.} \]

Probabilities are then just proportions of the interval — a rectangle’s area. For \(a \le c \le d \le b\),

\[ P(c \le X \le d) = \frac{d - c}{b - a}, \]

and by symmetry the mean sits in the middle, \(E[X] = (a + b)/2\). Use the uniform when the honest description is “somewhere in this range, with no reason to prefer any part of it”: a rounding error in \([-0.5, 0.5]\), a random arrival time within a fixed window, or — crucially — the output of a computer’s random-number generator, which is the raw material every simulation in this course is built from.

Exponential: waiting time at a steady rate

In Week 9 we counted how many shuttles arrive in an hour with a Poisson at rate \(\lambda = 4\). Now ask the companion question: how long until the next one? When events arrive at a constant rate \(\lambda\) and do not coordinate, the waiting time \(T\) is Exponential\((\lambda)\), with density

\[ f(t) = \lambda e^{-\lambda t}, \quad t \ge 0, \qquad f(t) = 0 \text{ for } t < 0. \]

Integrating this density (the Week 10 move) gives a particularly clean cdf, which is what you actually use to get probabilities:

\[ F(t) = P(T \le t) = 1 - e^{-\lambda t}, \quad t \ge 0. \]

Its mean is the reciprocal of the rate,

\[ E[T] = \frac{1}{\lambda}, \]

which is exactly the intuition “rate \(4\) per hour means about one every \(1/4\) hour.” The exponential’s defining feature is the memoryless property: if you have already waited \(s\) minutes, the probability of waiting at least \(t\) more is the same as the original probability of waiting at least \(t\) — the variable has no memory of time already spent. Symbolically,

\[ P(T > s + t \mid T > s) = P(T > t). \]

This is what makes the exponential the natural model for “pure waiting at a steady rate”: a shuttle that is already five minutes late is, under this model, no more “due” than a fresh wait. (Whether that is true of real shuttles is exactly the kind of assumption you should question — but it is what the model says.)

Normal: the bell curve and standardization

The Normal\((\mu, \sigma)\) is the symmetric, bell-shaped model that shows up whenever a quantity is the sum of many small independent influences — measurement noise, biological variation, the average of many trials. Its density is

\[ f(x) = \frac{1}{\sigma \sqrt{2\pi}}\, \exp\!\left( -\frac{(x - \mu)^2}{2\sigma^2} \right), \quad -\infty < x < \infty, \]

with mean \(\mu\) at the center and standard deviation \(\sigma\) controlling the spread. There is no simple closed form for its cdf, so we lean on two practical tools instead of integrating by hand.

The first is the 68–95–99.7 rule: for any normal, about \(68\%\) of the probability lies within \(1\) standard deviation of the mean, about \(95\%\) within \(2\), and about \(99.7\%\) within \(3\). This rule of thumb lets you sanity-check a normal probability in your head before trusting a table or a computer.

The second tool is standardization. Every normal is a shifted, rescaled copy of the single standard normal \(Z \sim \text{Normal}(0, 1)\). If \(X \sim \text{Normal}(\mu, \sigma)\), then

\[ Z = \frac{X - \mu}{\sigma} \sim \text{Normal}(0, 1), \qquad P(X \le x) = \Phi\!\left( \frac{x - \mu}{\sigma} \right), \]

where \(\Phi\) is the standard-normal cdf. So you never need a separate table for each \(\mu\) and \(\sigma\): convert your value to a \(z\)-score — a count of standard deviations from the mean — and read \(\Phi(z)\) once. A score \(2\) standard deviations above the mean, for instance, has \(\Phi(2) \approx 0.977\) below it, which is just the upper end of the \(95\%\) rule.

Recognizing the model: a short checklist

When a continuous scenario lands in front of you, ask in order:

Is every value in a fixed range equally likely, with no part favored? → Uniform.
Are you measuring a waiting time (or lifetime) where events arrive at a steady rate? → Exponential (and recall its rate is the Poisson rate from Week 9).
Is the quantity a measurement that clusters symmetrically around a typical value, with extremes rare? → Normal.

As with the discrete catalogue, the checklist also tells you when no standard model fits cleanly: if the rate is not constant, the exponential is only an approximation; if a quantity is strictly positive and skewed, the normal (which allows any real value) may be a poor fit. Naming the limit of a model is part of using it honestly.

Worked examples

Worked example — Maya’s morning: the shuttle wait and the commute (recurring slice)

This is one connected morning in our recurring synthetic world; two continuous models appear back to back. (Synthetic scenario; numbers fixed for the course.)

The shuttle wait, as an exponential. Shuttles arrive at Maya’s stop at a steady rate of \(\lambda = 4\) per hour — the same Poisson rate from Week 9 — so the time \(T\) until the next one is exponential:

\[ T \sim \text{Exponential}(\lambda = 4 \text{ per hour}), \qquad f(t) = 4 e^{-4 t}, \quad t \ge 0, \]

with time measured in hours. Its mean wait is

\[ E[T] = \frac{1}{\lambda} = \frac{1}{4}\ \text{hour} = 15\ \text{minutes}, \]

matching the intuition “one every fifteen minutes.” Now: what is the chance she waits at most \(15\) minutes — at most \(1/4\) hour? Use the cdf with \(t = 1/4\), so that \(\lambda t = 4 \times 1/4 = 1\):

\[ P(T \le 15\text{ min}) = F\!\left(\tfrac{1}{4}\right) = 1 - e^{-\lambda t} = 1 - e^{-1} \approx 0.632. \]

So a little under two-thirds of the time her wait is at or below the average — a reminder that for a right-skewed model the mean is not the median, and “at most the mean wait” is more likely than a coin flip.

The commute, as a normal. Once aboard, Maya’s commute time \(C\) (in minutes) clusters symmetrically around a typical value with rare extremes, so we model it as normal:

\[ C \sim \text{Normal}(\mu = 22,\ \sigma = 5). \]

What is the chance the commute takes at most \(30\) minutes? Standardize \(x = 30\):

\[ z = \frac{x - \mu}{\sigma} = \frac{30 - 22}{5} = \frac{8}{5} = 1.6, \qquad P(C \le 30) = \Phi(1.6) \approx 0.945. \]

So about \(94.5\%\) of mornings the commute is \(30\) minutes or less. The \(68\)–\(95\)–\(99.7\) rule confirms the ballpark: \(30\) is between \(1\sigma\) (\(27\)) and \(2\sigma\) (\(32\)) above the mean, so the cumulative probability should sit between \(0.84\) and \(0.975\) — and \(0.945\) does.

Worked example — time until a component fails, as an exponential (transfer)

Now a new context with the same model. A small electronic component fails at a constant average rate of \(\lambda = 0.5\) failures per year, with no wear-in or wear-out trend — the assumption that makes the exponential the right lifetime model. Let \(L\) be its lifetime in years:

\[ L \sim \text{Exponential}(\lambda = 0.5 \text{ per year}), \qquad f(\ell) = 0.5\, e^{-0.5 \ell}, \quad \ell \ge 0, \]

so the mean time to failure is \(E[L] = 1/\lambda = 1/0.5 = 2\) years. What is the probability the component survives past \(4\) years? Use the complement of the cdf:

\[ P(L > 4) = 1 - F(4) = e^{-\lambda \ell} = e^{-0.5 \times 4} = e^{-2} \approx 0.135. \]

About a \(13.5\%\) chance it lasts beyond \(4\) years. The memoryless property gives a striking corollary: a component that has already survived \(4\) years has the same probability of surviving another \(2\) years as a brand-new one — \(P(L > 6 \mid L > 4) = P(L > 2) = e^{-1} \approx 0.368\) — because the exponential keeps no memory of time already elapsed. This is the same machinery as Maya’s shuttle wait, applied to hardware instead of buses. (Synthetic scenario; numbers fixed for the course.)

Seeing it in software (shown, not run)

Software returns all of these probabilities directly. Below, R’s p* functions give cdf values \(P(X \le x)\) and d* functions give density heights \(f(x)\). The code is shown to read, not executed here.

# Shuttle wait T ~ Exponential(rate = 4 per hour); 15 minutes = 0.25 hour
pexp(0.25, rate = 4)            # P(T <= 15 min) = 1 - e^-1 ~ 0.632

# Commute C ~ Normal(mean = 22, sd = 5) minutes
pnorm(30, mean = 22, sd = 5)    # P(C <= 30) ~ 0.945
pnorm(1.6)                      # Phi(1.6): standardized, same answer ~ 0.945

# Component lifetime L ~ Exponential(rate = 0.5 per year)
1 - pexp(4, rate = 0.5)         # P(L > 4) = e^-2 ~ 0.135

# Density heights (NOT probabilities): f(x) for plotting the curves
dnorm(22, mean = 22, sd = 5)    # peak of the commute density, at the mean
dexp(0,  rate = 4)              # exponential density height at t = 0, equals the rate
dunif(0.1, min = 0, max = 0.25) # flat height 1/(b - a) = 4 over the interval

# Optional: simulate to confirm the two recurring-slice probabilities by frequency.
set.seed(35003)
waits <- rexp(100000, rate = 4)              # simulated shuttle waits, in hours
mean(waits <= 0.25)                          # empirical P(T <= 15 min), ~ 0.632

commutes <- rnorm(100000, mean = 22, sd = 5) # simulated commute times, in minutes
mean(commutes <= 30)                         # empirical P(C <= 30), ~ 0.945

The simulation chunk uses set.seed(35003) so it reproduces exactly when later executed; the empirical fractions should land near the exact values above. (Synthetic; seed set.)

A common mistake

The most frequent error this week is mixing up the exponential’s rate and mean. We parameterize the exponential by its rate \(\lambda\), so \(\text{Exponential}(4)\) has mean \(1/4\), not \(4\). In R this is the rate = argument; passing the mean where a rate belongs (or vice versa) silently gives the reciprocal of the answer you want. Always ask: is the number I have an events-per-unit-time (a rate) or a time-per- event (a mean)? For the shuttle, \(4\) per hour is the rate; \(15\) minutes is the mean.

A second slip is treating a density value \(f(x)\) as a probability. The bell curve’s height \(dnorm(22, 22, 5) \approx 0.080\) is not the chance the commute equals exactly \(22\) minutes — that probability is zero, exactly as Week 10 insisted. Density is probability per unit length; only the area between two values is a probability. And for the uniform, the flat height can even exceed \(1\) (it is \(4\) on \([0, 0.25]\)) precisely because it is a density, not a probability.

Low-stakes self-checks (ungraded)

These are for your own practice — ungraded, no submission, just a way to test whether the recognition skill is sticking. Try them before peeking at the reasoning.

A bus is equally likely to arrive at any moment in the next \(10\) minutes. Which model fits, and what is the probability it arrives in the first \(3\) minutes? (Uniform\((0, 10)\); \(P = 3/10 = 0.3\), a proportion of the interval.)
For Maya’s commute \(C \sim \text{Normal}(22, 5)\), roughly what fraction of mornings is the commute between \(17\) and \(27\) minutes, without a calculator? (Those are \(\mu \pm 1\sigma\), so about \(68\%\) by the rule.)
A help desk receives calls at a steady rate of \(\lambda = 6\) per hour. What model describes the wait until the next call, and what is the mean wait? (Exponential\((6)\); mean \(1/6\) hour \(= 10\) minutes.)
Why is \(P(C = 22)\) equal to \(0\) for the normal commute, even though \(22\) is its mean and the most likely region? (It is continuous: any single exact value has zero area, so zero probability; only intervals carry probability.)
A component’s lifetime is \(\text{Exponential}(0.5)\) per year. Without recomputing, is the probability it survives its next year larger if it is brand new or if it has already lasted \(3\) years? (The same — memoryless; past survival does not change the remaining-wait distribution.)

You can check the probability statements by simulation using the shown rexp/rnorm/runif chunks above: draw many values and compare the empirical fraction to the exact answer.

Reading and source pointer

This week tracks Grinstead & Snell, Chapter 5 — Important Distributions, specifically the continuous distributions (uniform, exponential, and normal), for the catalogue of named models and the parameterizations we use. The standardization \(Z = (X - \mu)/\sigma\) and the \(68\)–\(95\)–\(99.7\) framing for reading normal probabilities are supported by the review of continuous distributions and the normal in MIT OCW 18.05. Our parameterizations follow the course ledger: \(\text{Uniform}(a, b)\); \(\text{Exponential}(\lambda)\) as a rate with mean \(1/\lambda\); \(\text{Normal}(\mu, \sigma)\) as mean and standard deviation. The means quoted here are stated results carried from the expectation work in Week 8 and the density/area machinery in Week 10, not re-derived. These notes are the course’s own synthesis, grounded in but not copied from the sources. All scenario data are synthetic, with seeds set in any simulation.

Formula-verification status

verified: false. The densities, cdfs, means, and numeric probabilities on this page are drafted and provisional: the course math gate is BLOCKED pending human sign-off. Treat every formula and number here — the exponential cdf \(1 - e^{-\lambda t}\), the mean \(1/\lambda\), the standardization, and the worked values \(1 - e^{-1} \approx 0.632\) and \(\Phi(1.6) \approx 0.945\) — as provisional pending human verification, not as confirmed results.

Public vs. graded

These notes, the examples, and the practice here are public and ungraded — study material only. No graded prompts, answer keys, rubrics, point values, or due dates appear on this site. Graded checkpoints, quizzes, homework, labs, the midterm, the project, and the final live in Blackboard (the LMS), which is authoritative for due dates, submissions, and grades. If this page and Blackboard ever disagree, follow Blackboard.

Looking ahead

So far every random variable — discrete or continuous — has been a single quantity. Next week we let two vary together: joint distributions and dependence. We return to the rain-and-lateness pair from Weeks 3–4, now as random variables \(X\) (rain) and \(Y\) (late), and measure how they move together with covariance and correlation. After that, Week 13 brings the normal back in force: the Central Limit Theorem explains why the bell curve we used for Maya’s commute is so common — it is what the average of many independent pieces looks like, no matter their individual shape.