Lesson 13: Exponential Distribution

Calendar

Block I calendar with a red box around Tuesday 22 September, Lesson 13, Exponential Distribution.

WPR I Review

Last semester’s review, for more practice:

Substitute Instructor

Any day now I will step away to cover MA376 for a couple of weeks. While I am gone, a different instructor teaches your section.

Hour Instructor
B Mr Crow
C Dr. Thomas
D MAJ Lovas

Open Vantage

First thing today, go to the MA206 folder in Vantage.

What We Did: Lessons 1 through 12

  • Population vs sample, parameter (\(\mu\), \(\sigma\), \(p\)) vs statistic (\(\bar{x}\), \(s\), \(\hat{p}\)).
  • Random sampling buys generalization, random assignment buys causation.
  • Center: mean, median, trimmed mean. Spread: \(s^2\), \(s\), and the fourth spread \(f_s\).
  • Experiment, sample space \(\mathcal{S}\), event as a subset of \(\mathcal{S}\).
  • Union is “or”, intersection is “and”, complement is “not”.
  • Three axioms, the complement rule \(P(A') = 1 - P(A)\), and the addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
  • Equally likely outcomes: \(P(A) = N(A)/N\).
  • Product rule: \(n_1 n_2 \cdots n_k\).
  • Permutation (order matters): \(P_{k,n} = \dfrac{n!}{(n-k)!}\).
  • Combination (order does not): \(\dbinom{n}{k} = \dfrac{n!}{k!\,(n-k)!}\).
  • Conditional probability: \(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}\).
  • Multiplication rule, Law of Total Probability, and Bayes’ Theorem.
  • Independent: \(P(A \mid B) = P(A)\), tested with \(P(A \cap B) = P(A)\,P(B)\).
  • \(P(\text{at least one}) = 1 - P(\text{none})\).
  • pmf: \(p_X(x) = P(X = x)\). cdf: \(F_X(x) = P(X \le x)\), a step function.
  • Binomial: BINS, \(p(x) = \dbinom{n}{x} p^x (1-p)^{n-x}\), \(E(X) = np\), \(V(X) = np(1-p)\).
  • Poisson: a rate and a window, \(p(x; \mu) = \dfrac{e^{-\mu}\mu^x}{x!}\), \(E(X) = V(X) = \mu\).
  • \(E(X) = \sum_x x\,p_X(x)\) and \(V(X) = E(X^2) - [E(X)]^2\).
  • Probability is area under the pdf: \(P(a \le X \le b) = \int_a^b f(x)\,dx\).
  • \(P(X = c) = 0\), so \(\le\) and \(<\) give the same answer.
  • cdf: \(F(x) = P(X \le x)\), and \(P(a \le X \le b) = F(b) - F(a)\).
  • Percentile: solve \(F(c) = P(X \le c) = p\) for \(c\).
  • \(E(X) = \int x\,f(x)\,dx\) and \(V(X) = E(X^2) - [E(X)]^2\).
  • \(X \sim N(\mu, \sigma^2)\), with \(E(X) = \mu\) and \(V(X) = \sigma^2\).
  • pnorm(x, mean, sd) is the cdf \(F(x) = P(X \le x)\). qnorm runs it backwards.
  • Standardize with \(z = \dfrac{x - \mu}{\sigma}\) to land on \(Z \sim N(0, 1)\).
  • Empirical rule: 68%, 95%, 99.7% within 1, 2, 3 standard deviations.
  • \(z_\alpha\) has area \(\alpha\) to its right.

Warm-Up

Cadet sleep on a weeknight is normal with a mean of 6 hours and a standard deviation of 0.75 hours.

a) Fully specify the distribution.

\[X \sim N(\mu = 6,\ \sigma^2 = 0.75^2)\]

  • Designate the random variable. \(X\) is a cadet’s sleep on a weeknight, in hours.
  • Name the distribution. Normal.
  • Specify the parameters. \(\mu = 6\), \(\sigma = 0.75\).

b) What fraction of cadets sleep less than 5 hours?

\[z = \frac{5 - 6}{0.75} = -1.3333, \qquad P(X < 5) = F(5) = \mathbf{0.0912}\]

The sleep normal curve with the area left of 5 hours shaded.

pnorm(5, 6, 0.75) = 0.0912

c) Find \(P(5.5 < X < 7)\).

\[P(5.5 < X < 7) = F(7) - F(5.5) = \mathbf{0.6563}\]

The sleep normal curve with the area between 5.5 and 7 hours shaded.

pnorm(7, 6, 0.75) - pnorm(5.5, 6, 0.75) = 0.6563

d) Find the 75th percentile.

Find \(c\) so that \(F(c) = P(X \le c) = 0.75\): \(\mathbf{6.51}\) hours.

The sleep normal curve with the leftmost 75 percent of the area shaded, cut off at 6.51 hours.

qnorm(0.75, 6, 0.75) = 6.5059

What We’re Doing: Lesson 13

Objectives

  • Identify the exponential distribution and relate its parameter \(\lambda\) to the mean and standard deviation (both \(1/\lambda\)). (SLO 7)
  • Compute exponential probabilities using the pdf and cdf. (SLO 7)
  • Apply the memoryless property of the exponential distribution. (SLO 7)

Required Reading

Devore 4.4

Break!

Cal

Reese

DMath Frisbee!!

Math vs USMAPS

2-0

15-0

The Takeaway for Today

NoteKey Concepts from Lesson 13
  • The exponential models a wait: the time until the next event. \(X \sim \text{Exp}(\lambda)\)
  • \(\lambda\) is a rate. The mean wait is its reciprocal: \(E(X) = \sigma = \dfrac{1}{\lambda}\), and \(V(X) = \dfrac{1}{\lambda^2}\)
  • The cdf has a closed form: \(F(x) = P(X \le x) = 1 - e^{-\lambda x}\), so \(P(X > x) = e^{-\lambda x}\)
  • pexp(x, rate) is the cdf. The second slot is the rate, not the mean
  • Percentiles solve by hand: \(c = -\ln(1-p)/\lambda\). The median sits below the mean
  • Memoryless: \(P(X \ge t + t_0 \mid X \ge t_0) = P(X \ge t)\). Time already waited does not matter
  • Poisson counts events at rate \(\lambda\). The waits between them are \(\text{Exp}(\lambda)\)

Opening: Remember This?

  • Cadets walk up to the Cadet Store counter at an average rate of \(2.5\) per hour. How many show up in the next \(2\) hours?
  • Route 9W averages \(0.4\) potholes per km. How many are on a \(10\) km stretch?
  • A cadet’s research paper averages \(0.2\) typos per page. How many are in a \(15\) page paper?
  • Emails hit your inbox at \(6\) per hour. How many arrive during a \(55\) minute class?

What was \(\lambda\) for each?

scenario \(\lambda\) \(X\) \(t\) \(\mu = \lambda t\)
Cadet Store \(2.5\)/hr cadets in 2 hours \(2\) hr \(5\) cadets
Route 9W \(0.4\)/km potholes in 10 km \(10\) km \(4\) potholes
Research paper \(0.2\)/page typos in 15 pages \(15\) pages \(3\) typos
Inbox \(6\)/hr emails in 55 minutes \(55/60\) hr \(5.5\) emails

Each count is Poisson: \(X \sim \text{Pois}(\mu = \lambda t)\).

What If We Cared About the Time Between Arrivals?

  • How long until the next cadet walks up to the Cadet Store counter?
  • How far until the next pothole on Route 9W?
  • How many pages until the next typo?
  • How long until the next email?

What is \(\lambda\) now? What is \(\mu\) now?

scenario \(\lambda\) \(X\) \(\mu = 1/\lambda\)
Cadet Store \(2.5\)/hr hours to the next cadet \(0.4\) hr (24 min)
Route 9W \(0.4\)/km km to the next pothole \(2.5\) km
Research paper \(0.2\)/page pages to the next typo \(5\) pages
Inbox \(6\)/hr hours to the next email \(1/6\) hr (10 min)

\(\lambda\) is the same rate. \(\mu\) is now the mean wait, \(1/\lambda\), and there is no window \(t\).

The Exponential Distribution

Our Example

Cadets arrive at the CIF counter at a rate of 3 per hour. How long until the next cadet?

ImportantThe Exponential pdf

\(X\) is exponential with parameter \(\lambda > 0\), written \(X \sim \text{Exp}(\lambda)\), if

\[f(x; \lambda) = \begin{cases} \lambda e^{-\lambda x} & x \ge 0 \\ 0 & \text{otherwise} \end{cases}\]

  • \(X\) is a wait: a time (or distance) until the next event.
  • \(\lambda\) is the rate of events, per unit of \(x\).
  • The support starts at \(0\) and runs forever to the right. The curve is right skewed.
NoteFully specify the CIF wait

\[X \sim \text{Exp}(\lambda = 3)\]

  • Designate the random variable. \(X\) is the hours until the next cadet arrives.
  • Name the distribution. Exponential.
  • Specify the parameter. \(\lambda = 3\) per hour.

\[f(x) = \begin{cases} 3 e^{-3x} & x \ge 0 \\ 0 & \text{otherwise} \end{cases}\]

WarningUnits

\(3\) per hour is \(1/20\) per minute. Whatever unit \(\lambda\) is in, \(x\) has to be in too.

Three exponential pdfs on shared axes, with rates 1, 3, and 6 per hour. Each starts at height lambda at 0 and decays to the right; the larger the rate, the higher the start and the faster the decay.

Mean, Variance, and Standard Deviation

ImportantMean and variance of an exponential

\[E(X) = \frac{1}{\lambda}, \qquad V(X) = \frac{1}{\lambda^2}, \qquad \sigma_X = \frac{1}{\lambda}\]

The mean and the standard deviation are the same number.

Integrate by parts. The boundary terms are \(0\) at both ends.

\[\begin{aligned} E(X) &= \int_0^{\infty} x \, \lambda e^{-\lambda x} \, dx \\ &= \Big[-x e^{-\lambda x}\Big]_0^{\infty} + \int_0^{\infty} e^{-\lambda x} \, dx \\ &= 0 + \frac{1}{\lambda} \\[1em] E(X^2) &= \int_0^{\infty} x^2 \, \lambda e^{-\lambda x} \, dx \\ &= \Big[-x^2 e^{-\lambda x}\Big]_0^{\infty} + \frac{2}{\lambda} \int_0^{\infty} x \, \lambda e^{-\lambda x} \, dx \\ &= 0 + \frac{2}{\lambda} \cdot \frac{1}{\lambda} = \frac{2}{\lambda^2} \\[1em] V(X) &= E(X^2) - [E(X)]^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2} \end{aligned}\]

CIF counter: what is the mean time between arrivals? The standard deviation?

\[E(X) = \frac{1}{\lambda} = \frac{1}{3} \text{ hour} = 20 \text{ minutes}, \qquad V(X) = \frac{1}{3^2} = \frac{1}{9}, \qquad \sigma = \frac{1}{3} \text{ hour}\]

The Cumulative Distribution Function

Unlike the normal, this integral has a closed form.

\[F(x) = P(X \le x) = \int_0^x \lambda e^{-\lambda y} \, dy = \Big[-e^{-\lambda y}\Big]_0^x = 1 - e^{-\lambda x}\]

ImportantThe Exponential cdf

\[F(x; \lambda) = P(X \le x) = \begin{cases} 0 & x < 0 \\ 1 - e^{-\lambda x} & x \ge 0 \end{cases}\]

The right tail is the part to remember: \(P(X > x) = 1 - F(x) = e^{-\lambda x}\).

CIF counter:

\[F(x) = P(X \le x) = \begin{cases} 0 & x < 0 \\ 1 - e^{-3x} & x \ge 0 \end{cases}\]

Left, the arrival wait exponential pdf with the area left of 0.25 hours shaded. Right, the exponential cdf with a dashed orange line up from 0.25 to the curve and across to 0.5276 on the y axis.

Computing Exponential Probabilities

Importantpexp is the cdf

\[\texttt{pexp(x, rate)} \;=\; F(x) \;=\; P(X \le x) \;=\; 1 - e^{-\lambda x}\]

WarningWhere cadets go wrong
  • Mean in the rate slot. A mean wait of \(1/3\) hour is a rate of \(3\): pexp(x, 3), never pexp(x, 1/3).
  • Mixed units. The rate and \(x\) must use the same time unit.
  • Leaving the rate off. pexp(0.25) quietly assumes rate = 1.

CIF counter: \(X \sim \text{Exp}(\lambda = 3)\).

a) \(P(X < 0.25)\)

\[P(X < 0.25) = \int_0^{0.25} 3 e^{-3x} \, dx = F(0.25) = 1 - e^{-3(0.25)} = \mathbf{0.5276}\]

The arrival wait exponential pdf with the area left of 0.25 hours shaded.

pexp(0.25, 3) = 0.5276

b) \(P(X \le 0.25)\)

\[P(X \le 0.25) = \int_0^{0.25} 3 e^{-3x} \, dx = F(0.25) = 1 - e^{-3(0.25)} = \mathbf{0.5276}\]

The arrival wait exponential pdf with the area left of 0.25 hours shaded.

pexp(0.25, 3) = 0.5276

c) \(P(X > 0.5)\)

\[P(X > 0.5) = \int_{0.5}^{\infty} 3 e^{-3x} \, dx = 1 - F(0.5) = e^{-3(0.5)} = \mathbf{0.2231}\]

The arrival wait exponential pdf with the area right of 0.5 hours shaded.

1 - pexp(0.5, 3) = 0.2231

d) \(P(X \ge 0.5)\)

\[P(X \ge 0.5) = \int_{0.5}^{\infty} 3 e^{-3x} \, dx = 1 - F(0.5) = e^{-3(0.5)} = \mathbf{0.2231}\]

The arrival wait exponential pdf with the area right of 0.5 hours shaded.

1 - pexp(0.5, 3) = 0.2231

e) \(P(0.25 < X < 0.5)\)

\[P(0.25 < X < 0.5) = \int_{0.25}^{0.5} 3 e^{-3x} \, dx = F(0.5) - F(0.25) = e^{-0.75} - e^{-1.5} = \mathbf{0.2492}\]

The arrival wait exponential pdf with the area between 0.25 and 0.5 hours shaded.

pexp(0.5, 3) - pexp(0.25, 3) = 0.2492

f) \(P(X < 1/3)\), a wait shorter than the mean

\[P(X < 1/3) = \int_0^{1/3} 3 e^{-3x} \, dx = F(1/3) = 1 - e^{-1} = \mathbf{0.6321}\]

The arrival wait exponential pdf with the area left of the mean, one third of an hour, shaded, and an orange triangle marking the mean.

pexp(1/3, 3) = 0.6321

Not \(0.5\). The long right tail drags the mean to the right, so most waits are shorter than the mean. This is true for every exponential: \(1 - e^{-\lambda(1/\lambda)} = 1 - e^{-1}\).

g) \(P(X = 0.5)\)

\[P(X = 0.5) = \int_{0.5}^{0.5} 3 e^{-3x} \, dx = F(0.5) - F(0.5) = 0\]

pexp(0.5, 3) - pexp(0.5, 3) = 0.0000

Warningdexp is not a probability

dexp(0.5, 3) = 0.6694

That is the height of the pdf at \(0.5\), not \(P(X = 0.5)\).

h) What is the probability exactly 4 cadets arrive in the next hour?

A count in a window. \(t = 1\) hour, so \(\mu = 3(1) = 3\).

\[Y \sim \text{Pois}(\mu = 3)\]

  • Designate the random variable. \(Y\) is the number of cadets who arrive in the next hour.
  • Name the distribution. Poisson.
  • Specify the parameter. \(\mu = \lambda t = 3(1) = 3\) cadets.

\[P(Y = 4) = \frac{e^{-3} 3^4}{4!} = \mathbf{0.1680}\]

The Poisson pmf with mean 3 with the bar at 4 highlighted.

dpois(4, 3) = 0.1680

Going Backwards: Percentiles

Set the cdf equal to \(p\) and solve. No table, no R required.

\[\begin{aligned} F(c) = P(X \le c) &= p \\ 1 - e^{-\lambda c} &= p \\ e^{-\lambda c} &= 1 - p \\ c &= \frac{-\ln(1 - p)}{\lambda} \end{aligned}\]

Importantqexp is the inverse cdf

\[\texttt{qexp(p, rate)} \;=\; c = \frac{-\ln(1-p)}{\lambda}\]

Median wait at the CIF counter. \(p = 0.5\):

\[\tilde{\mu} = \frac{-\ln(0.5)}{3} = \frac{\ln 2}{3} = \mathbf{0.2310} \text{ hour} \approx 13.86 \text{ minutes}\]

qexp(0.5, 3) = 0.2310

Left, the arrival wait pdf with half the area shaded left of 0.2310 hours and an orange triangle at the mean, one third of an hour. Right, the cdf with a dashed orange line across from 0.5 and down to 0.2310.

90th percentile.

\[c = \frac{-\ln(0.1)}{3} = \mathbf{0.7675} \text{ hour} \approx 46.05 \text{ minutes}\]

qexp(0.90, 3) = 0.7675

The arrival wait exponential pdf with the leftmost 90 percent of the area shaded, cut off at 0.7675 hours.

NoteSide fact: the exponential is memoryless

\[P(X \ge t + t_0 \mid X \ge t_0) = P(X \ge t)\]

Time already waited does not matter. Nobody has come to the CIF counter in 0.25 hours, and the chance of waiting at least another 0.5 hours is still \(P(X \ge 0.5) = e^{-3(0.5)} = 0.2231\).

(1 - pexp(0.75, 3)) / (1 - pexp(0.25, 3)) = 0.2231
1 - pexp(0.5, 3) = 0.2231

Poisson vs Exponential

Trucks arrive at the motor pool fuel point at a rate of 2 per hour. That one rate gives two random variables.

ImportantCount or wait?
  • Count. \(Y\) is the number of trucks in an hour. \(Y \sim \text{Pois}(\mu = 2)\).
  • Wait. \(X\) is the hours until the next truck. \(X \sim \text{Exp}(\lambda = 2)\).
  • Same \(2\) per hour in both.

A three hour timeline with truck arrivals marked as dots. Above the line, the count of trucks in each one hour window. Below the line, arrows show the waits between consecutive trucks.

For each question: count or wait? Fully specify, then compute.

a) What is the probability exactly 3 trucks arrive in the next hour?

A count.

\[Y \sim \text{Pois}(\mu = 2)\]

  • Designate the random variable. \(Y\) is the number of trucks that arrive in the next hour.
  • Name the distribution. Poisson.
  • Specify the parameter. \(\mu = 2\) trucks.

\[P(Y = 3) = \frac{e^{-2} 2^3}{3!} = \mathbf{0.1804}\]

The Poisson pmf with mean 2 with the bar at 3 highlighted.

dpois(3, 2) = 0.1804

b) What is the probability the next truck arrives within half an hour?

A wait. Half an hour is \(x = 0.5\).

\[X \sim \text{Exp}(\lambda = 2)\]

  • Designate the random variable. \(X\) is the hours until the next truck arrives.
  • Name the distribution. Exponential.
  • Specify the parameter. \(\lambda = 2\) per hour.

\[P(X < 0.5) = 1 - e^{-2(0.5)} = 1 - e^{-1} = \mathbf{0.6321}\]

The exponential pdf with rate 2 per hour with the area left of 0.5 hours shaded.

pexp(0.5, 2) = 0.6321

c) What is the mean time between trucks? What is the expected number of trucks in the next hour?

Wait: \(X \sim \text{Exp}(\lambda = 2)\), so \(E(X) = 1/\lambda = \mathbf{1/2}\) hour.

Count: \(Y \sim \text{Pois}(\mu = 2)\), so \(E(Y) = \mathbf{2}\) trucks.

One is the reciprocal of the rate. The other is the rate.

d) What is the probability no trucks arrive in the next hour? Do it both ways.

“No trucks in the next hour” and “the wait is longer than 1 hour” are the same event.

\[\begin{aligned} \text{Count:} \quad & Y \sim \text{Pois}(\mu = 2), && P(Y = 0) = e^{-2} = \mathbf{0.1353} \\ \text{Wait:} \quad & X \sim \text{Exp}(\lambda = 2), && P(X > 1) = e^{-2(1)} = \mathbf{0.1353} \end{aligned}\]

Left, the Poisson pmf with mean 2 with the bar at 0 highlighted. Right, the exponential pdf with rate 2 per hour with the area beyond 1 hour shaded. Both are 0.1353.

dpois(0, 2) = 0.1353
1 - pexp(1, 2) = 0.1353

e) What is the probability at least 3 trucks arrive in the next hour?

A count.

\[Y \sim \text{Pois}(\mu = 2)\]

\[P(Y \ge 3) = 1 - P(Y \le 2) = \mathbf{0.3233}\]

The Poisson pmf with mean 2 with the bars from 3 up highlighted.

1 - ppois(2, 2) = 0.3233

f) How long until the fuel point is 90% sure a truck has arrived?

A wait, run backwards. The 90th percentile of \(X \sim \text{Exp}(\lambda = 2)\).

\[c = \frac{-\ln(0.1)}{2} = \mathbf{1.1513} \text{ hours}\]

The exponential pdf with rate 2 per hour with the leftmost 90 percent of the area shaded, cut off at 1.1513 hours.

qexp(0.90, 2) = 1.1513

Poisson (L10) Exponential (L13)
Random variable Count of trucks in an hour Wait until the next truck
Type Discrete, \(0, 1, 2, \dots\) Continuous, \(x \ge 0\)
\(\lambda\) Rate, 2 per hour Rate, 2 per hour
Mean \(2\) trucks \(1/2\) hour
Second argument in R The mean, \(2\) The rate, \(2\)
R-Lite ppois(y, 2) pexp(x, 2)

Distributions So Far

Four named distributions. Fully specifying one is the same three steps every time: designate the random variable in words with units, name the distribution, specify the parameters as numbers.

Binomial Poisson Normal Exponential
Type Discrete Discrete Continuous Continuous
Use it when \(x\) successes in \(n\) fixed trials \(x\) events in a window Symmetric, bell shaped measurement Wait until the next event
Fully specified \(X \sim \text{Binom}(n,\ p)\) \(X \sim \text{Pois}(\mu)\) \(X \sim N(\mu,\ \sigma^2)\) \(X \sim \text{Exp}(\lambda)\)
Parameters \(n\) trials, \(p\) success probability \(\mu = \lambda t\), expected count \(\mu\) center, \(\sigma\) spread \(\lambda\) rate
Possible values \(x = 0, 1, \dots, n\) \(x = 0, 1, 2, \dots\) \(-\infty < x < \infty\) \(x \ge 0\)
pmf or pdf \(\dbinom{n}{x} p^x (1-p)^{n-x}\) \(\dfrac{e^{-\mu}\mu^x}{x!}\) \(\dfrac{1}{\sqrt{2\pi}\,\sigma} e^{-(x-\mu)^2/(2\sigma^2)}\) \(\lambda e^{-\lambda x}\)
\(E(X)\) \(np\) \(\mu\) \(\mu\) \(1/\lambda\)
\(V(X)\) \(np(1-p)\) \(\mu\) \(\sigma^2\) \(1/\lambda^2\)
d, the pmf or pdf dbinom(x, n, p) dpois(x, mu) dnorm(x, mean, sd) dexp(x, rate)
p, the cdf \(P(X \le x)\) pbinom(x, n, p) ppois(x, mu) pnorm(x, mean, sd) pexp(x, rate)
q, the cdf backwards qbinom(area, n, p) qpois(area, mu) qnorm(area, mean, sd) qexp(area, rate)
Our example \(X \sim \text{Binom}(n = 5,\ p = 0.75)\) \(X \sim \text{Pois}(\mu = 5)\) \(X \sim N(\mu = 16,\ \sigma^2 = 2^2)\) \(X \sim \text{Exp}(\lambda = 3)\)
\(P(X = 3)\) dbinom(3, 5, 0.75) dpois(3, 5) \(0\), use an interval \(0\), use an interval
\(P(X \le 3)\) pbinom(3, 5, 0.75) ppois(3, 5) pnorm(3, 16, 2) pexp(3, 3)
\(90\)th percentile qbinom(0.90, 5, 0.75) qpois(0.90, 5) qnorm(0.90, 16, 2) qexp(0.90, 3)
WarningWhere people lose points
  • dnorm and dexp are heights, not probabilities. For a continuous variable \(P(X = x) = 0\).
  • dbinom and dpois are probabilities. \(P(X = x)\) is real for a discrete variable.
  • The normal takes \(\sigma\), not \(\sigma^2\). The Poisson takes \(\mu = \lambda t\). The exponential takes \(\lambda\), not the mean.
  • p takes an \(x\) and returns an area. q takes an area and returns an \(x\).

Board Problems

Problem 1: Range Control Calls

Calls to range control come in with a mean of 12 minutes between calls. The time until the next call is exponential.

  1. Fully specify the distribution. Give the mean and standard deviation.

  2. What is the probability the next call comes within 5 minutes?

  3. What is the probability range control goes 30 minutes without a call?

  4. Find the median wait.

  1. \(X\) is the minutes until the next call, \(X \sim \text{Exp}(\lambda = 1/12)\). \(E(X) = \sigma = 12\) minutes.

The range control call wait exponential pdf with rate 1/12 and an orange triangle at the mean, 12 minutes.

  1. \[P(X < 5) = 1 - e^{-5/12} = \mathbf{0.3408}\]

The call wait exponential pdf with the area left of 5 minutes shaded.

pexp(5, 1/12) = 0.3408

  1. \[P(X > 30) = e^{-30/12} = \mathbf{0.0821}\]

The call wait exponential pdf with the area right of 30 minutes shaded.

1 - pexp(30, 1/12) = 0.0821

  1. \[\tilde{\mu} = \frac{-\ln(0.5)}{1/12} = 12 \ln 2 = \mathbf{8.32} \text{ minutes}\]

The call wait exponential pdf with half the area shaded left of 8.32 minutes.

qexp(0.5, 1/12) = 8.3178

Problem 2: Night Vision Tubes

The image intensifier tube in a set of night vision goggles fails at a rate of \(\lambda = 0.0004\) per hour of use.

  1. Fully specify the distribution. Give the mean and standard deviation.

  2. What is the probability a new tube lasts more than 3000 hours?

  3. A tube already has 1000 hours on it. What is the probability it lasts at least another 3000?

  4. The unit wants a replacement schedule so that only 5% of tubes fail first. How many hours is that?

  1. \(X\) is the hours until the tube fails, \(X \sim \text{Exp}(\lambda = 0.0004)\). \(E(X) = \sigma = 1/0.0004 = 2500\) hours.

  2. \[P(X > 3000) = e^{-0.0004(3000)} = e^{-1.2} = \mathbf{0.3012}\]

The tube life exponential pdf with the area right of 3000 hours shaded.

1 - pexp(3000, 0.0004) = 0.3012

  1. Memoryless. The 1000 hours do not matter.

\[P(X \ge 4000 \mid X \ge 1000) = \frac{e^{-0.0004(4000)}}{e^{-0.0004(1000)}} = e^{-1.2} = \mathbf{0.3012}\]

The tube life exponential pdf with the region beyond 1000 hours lightly shaded and the region beyond 4000 hours shaded darker.

(1 - pexp(4000, 0.0004)) / (1 - pexp(1000, 0.0004)) = 0.3012

  1. The 5th percentile.

\[c = \frac{-\ln(0.95)}{0.0004} = \mathbf{128.23} \text{ hours}\]

The tube life exponential pdf with the leftmost 5 percent of the area shaded, cut off at 128 hours.

qexp(0.05, 0.0004) = 128.2330

That schedule would pull tubes almost new. Because of (c), replacing a working tube early buys nothing: an old tube is as good as a new one.

Problem 3: MEDEVAC Requests

MEDEVAC requests reach a battalion aid station at a rate of 2 per hour.

  1. What is the probability of no requests in the next 45 minutes? Do it both ways.

  2. What is the mean wait between requests, in minutes?

  3. What is the probability of exactly 3 requests in the next 2 hours?

  4. What is the probability the next request comes within 10 minutes?

  1. \(t = 0.75\) hours.

\[\begin{aligned} \text{Poisson:} \quad & X \sim \text{Pois}(\mu = 2(0.75) = 1.5), && P(X = 0) = e^{-1.5} = \mathbf{0.2231} \\ \text{Exponential:} \quad & T \sim \text{Exp}(\lambda = 2), && P(T > 0.75) = e^{-2(0.75)} = \mathbf{0.2231} \end{aligned}\]

Left, the Poisson pmf with mean 1.5 with the bar at 0 highlighted at height 0.2231. Right, the exponential pdf with rate 2 per hour with the area beyond 0.75 hours shaded, also 0.2231.

dpois(0, 1.5) = 0.2231
1 - pexp(0.75, 2) = 0.2231

  1. \(1/\lambda = 1/2\) hour \(= \mathbf{30}\) minutes.

  2. A count in a window, so Poisson with \(\mu = 2(2) = 4\).

\[P(X = 3) = \frac{e^{-4} 4^3}{3!} = \mathbf{0.1954}\]

The Poisson pmf with mean 4 with the bar at 3 highlighted.

dpois(3, 4) = 0.1954

  1. A wait, so exponential. 10 minutes is \(1/6\) hour.

\[P(T < 1/6) = 1 - e^{-2/6} = \mathbf{0.2835}\]

The exponential pdf with rate 2 per hour with the area left of one sixth of an hour shaded.

pexp(1/6, 2) = 0.2835

Problem 4: Which Distribution?

Fully specify each random variable, then write the R-Lite for the probability asked.

  1. Each of the 12 cadets in a squad passes the ACFT with probability 0.9, independently. The number who pass. Find \(P(\text{all } 12 \text{ pass})\).

  2. The staff duty desk gets calls at 1.5 per hour. The number of calls in an 8 hour shift. Find \(P(\text{at most } 10 \text{ calls})\).

  3. Same desk. The minutes until the next call. Find \(P(\text{a call within } 10 \text{ minutes})\).

  4. Packed rucksack weights have a mean of 45 lb and a standard deviation of 3 lb, bell shaped. The weight of one ruck. Find \(P(\text{under } 50 \text{ lb})\).

  5. A generator runs a mean of 400 hours between failures. The hours until the next failure. Find \(P(\text{it runs past its mean})\).

  1. \(X \sim \text{Binom}(n = 12,\ p = 0.9)\), the number of the 12 who pass. dbinom(12, 12, 0.9) = 0.2824

  2. \(X \sim \text{Pois}(\mu = 1.5(8) = 12)\), the number of calls in the shift. ppois(10, 12) = 0.3472

  3. \(X \sim \text{Exp}(\lambda = 1/40)\), minutes until the next call. \(1.5\) per hour is \(1/40\) per minute. pexp(10, 1/40) = 0.2212

  4. \(X \sim N(\mu = 45,\ \sigma^2 = 3^2)\), the weight in pounds. pnorm(50, 45, 3) = 0.9522

  5. \(X \sim \text{Exp}(\lambda = 1/400)\), hours until failure. 1 - pexp(400, 1/400) = 0.3679

Five small panels, one per part: a binomial pmf with the bar at 12 highlighted, a Poisson pmf with bars 0 through 10 highlighted, an exponential pdf shaded left of 10 minutes, a normal pdf shaded left of 50 pounds, and an exponential pdf shaded right of 400 hours.

Problem 5: Stretch, Solve for the Rate

Nobody tells you \(\lambda\). The S4 reports that 20% of radio batteries die within their first 6 hours. Assume battery life is exponential.

  1. Solve for \(\lambda\). What is the mean battery life?

  2. Find the probability a battery lasts more than a full 24 hour day.

  3. A battery has run 12 hours. What is the probability it dies in the next 6?

  1. Set the cdf at 6 equal to 0.2.

\[\begin{aligned} F(6) = P(X \le 6) &= 0.2 \\ 1 - e^{-6\lambda} &= 0.2 \\ e^{-6\lambda} &= 0.8 \\ \lambda &= \frac{-\ln(0.8)}{6} = \mathbf{0.0372} \text{ per hour} \end{aligned}\]

\[E(X) = \frac{1}{\lambda} = \mathbf{26.89} \text{ hours}\]

  1. \[P(X > 24) = e^{-24\lambda} = \left(e^{-6\lambda}\right)^4 = 0.8^4 = \mathbf{0.4096}\]

The battery life exponential pdf with the leftmost 20 percent shaded below 6 hours in orange and the area beyond 24 hours shaded in blue.

1 - pexp(24, -log(0.8)/6) = 0.4096

  1. Memoryless. The first 12 hours do not matter, so this is the same as a new battery dying in its first 6: \(\mathbf{0.2}\).

\[P(X \le 18 \mid X \ge 12) = \frac{F(18) - F(12)}{1 - F(12)} = \frac{0.1280}{0.64} = \mathbf{0.2}\]

(pexp(18, -log(0.8)/6) - pexp(12, -log(0.8)/6)) / (1 - pexp(12, -log(0.8)/6)) = 0.2000

Before You Leave

Today

  • \(X \sim \text{Exp}(\lambda)\) models a wait, and \(E(X) = \sigma = 1/\lambda\)
  • \(P(X > x) = e^{-\lambda x}\), and pexp takes the rate, not the mean
  • The median \(\ln 2 / \lambda\) sits below the mean
  • Memoryless: time already waited does not change what comes next
  • Poisson counts the events, exponential measures the waits between them

Any questions?

Next Lesson

Lesson 14: Exploratory Data Analysis (EDA)

  • Summarize project data with appropriate graphical and numerical descriptive methods (histograms, boxplots, scatterplots)
  • Compute and interpret the sample correlation coefficient \(r\)
  • Distinguish correlation from causation
  • Communicate exploratory findings and pose questions for further analysis

Reading: Supplement S3 (Canvas)

Upcoming Graded Events

  • WebAssign 4.4: due at the start of Lesson 14
  • WPR I: Lesson 16 (covers Lessons 1-13)
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