Lesson 11: Continuous Random Variables

Calendar

Block I calendar with a red box around Tuesday 15 September, Lesson 11, Continuous Random Variables.

What We Did: Lessons 1 through 10

  • 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: \(P(A \cap B) = P(A \mid B)\,P(B)\).
  • Law of Total Probability: \(P(B) = \sum_{i=1}^{k} P(B \mid A_i)\,P(A_i)\).
  • Bayes’ Theorem flips the conditioning, and \(P(A \mid B) \ne P(B \mid A)\).
  • Independent: \(P(A \mid B) = P(A)\), tested with \(P(A \cap B) = P(A)\,P(B)\).
  • Independence collapses the conditional, multiplication, and addition rules.
  • Mutually exclusive is not independent. Disjoint events are as dependent as events get.
  • \(P(\text{at least one}) = 1 - P(\text{none})\), and “none” is a product under independence.
  • A random variable \(X\) assigns a number to every outcome in \(\mathcal{S}\).
  • pmf: \(p_X(x) = P(X = x)\), with \(p_X(x) \ge 0\) and \(\sum_x p_X(x) = 1\).
  • cdf: \(F_X(x) = P(X \le x)\), a step function that jumps by \(p_X(x)\) at each value.
  • For a discrete \(X\), \(\le\) and \(<\) are not interchangeable.
  • Expected value \(E(X) = \sum_x x\,p_X(x)\), variance \(V(X) = E(X^2) - [E(X)]^2\).
  • BINS: binary, independent, number of trials fixed, same \(p\).
  • pmf: \(p(x) = \dbinom{n}{x} p^x (1-p)^{n-x}\) for \(x = 0, 1, \dots, n\).
  • Fully specify: \(X \sim \text{Binom}(n,\ p)\), variable, distribution, parameters.
  • dbinom is the pmf, pbinom is the cdf.
  • \(E(X) = np\) and \(V(X) = np(1-p)\).
  • A rate, a window, and no fixed \(n\) point to the Poisson.
  • pmf: \(p(x; \mu) = \dfrac{e^{-\mu}\mu^x}{x!}\) for \(x = 0, 1, 2, \dots\), with \(\mu = \lambda t\).
  • Fully specify: \(X \sim \text{Pois}(\mu)\).
  • dpois is the pmf, ppois is the cdf.
  • \(E(X) = V(X) = \mu\).

What We’re Doing: Lesson 11

Objectives

  • Use a probability density function (pdf) to compute probabilities as areas, and recognize that any exact value has probability zero. (SLO 7)
  • Obtain the cdf from the pdf and use it to find probabilities and percentiles. (SLO 7)
  • Compute the expected value, variance, and standard deviation of a continuous random variable. (SLO 7)

Required Reading

Devore 4.1, 4.2

Break!

Family

The Takeaway for Today

NoteKey Concepts from Lesson 11
  • A continuous \(X\) fills an interval. Probability is area under the pdf: \(P(a \le X \le b) = \int_a^b f(x)\,dx\)
  • pdf: \(f(x) \ge 0\) and \(\int_{-\infty}^{\infty} f(x)\,dx = 1\). Use the second to solve for \(k\)
  • \(P(X = c) = 0\), so \(\le\) and \(<\) give the same answer
  • cdf: \(F(x) = P(X \le x) = \int_{-\infty}^{x} f(y)\,dy\), and \(P(a \le X \le b) = F(b) - F(a)\)
  • Percentile: solve \(F(c) = P(X \le c) = p\) for \(c\). The median \(\tilde{\mu}\) is the 50th percentile
  • \(E(X) = \int x\,f(x)\,dx\) and \(V(X) = E(X^2) - [E(X)]^2\). Sums become integrals

From Discrete to Continuous

Lessons 8 through 10 Discrete: pmf You can list the values. Probability sits on each value.

\(P(a \le X \le b) = \sum_{x=a}^{b} p(x)\)

Lessons 11 through 13 Continuous: pdf The values fill an interval. Probability sits over intervals.

\(P(a \le X \le b) = \int_a^b f(x)\,dx\)

Run time, fuel remaining, wait until the next call, distance a round lands from the target.

The Probability Density Function

NoteLesson 8: Probability Mass Function (pmf)

A discrete random variable \(X\) has pmf \(p(x)\) if, for any \(a \le b\), \[P(a \le X \le b) = \sum_{x=a}^{b} p(x).\] It must satisfy two conditions: \[p(x) \ge 0 \quad \text{for all } x, \qquad \sum_{\text{all } x} p(x) = 1.\]

ImportantToday: Probability Density Function (pdf)

A continuous random variable \(X\) has pdf \(f(x)\) if, for any \(a \le b\), \[P(a \le X \le b) = \int_a^b f(x)\,dx.\] It must satisfy two conditions: \[f(x) \ge 0 \quad \text{for all } x, \qquad \int_{-\infty}^{\infty} f(x)\,dx = 1.\]

Example

\[f_X(x) = \begin{cases} \dfrac{x}{16} & 2 \le x \le 6 \\ 0 & \text{otherwise} \end{cases}\]

The pdf f of x equals x over 16, zero below 2, a line rising from 0.125 at x equal to 2 to 0.375 at x equal to 6, and zero above 6.

a) \(P(X < 4)\)

\[P(X < 4) = \int_2^{4} \frac{x}{16}\,dx = \frac{x^2}{32} \bigg|_2^{4} = \frac{16 - 4}{32} = 0.375\]

The pdf x over 16 with the area from 2 to 4 shaded.

b) \(P(X \le 4)\)

\[P(X \le 4) = \int_2^{4} \frac{x}{16}\,dx = 0.375\]

The pdf x over 16 with the area from 2 to 4 shaded.

c) \(P(X > 4)\)

\[P(X > 4) = \int_4^{6} \frac{x}{16}\,dx = \frac{x^2}{32} \bigg|_4^{6} = \frac{36 - 16}{32} = 0.625\]

Or from (a): \(P(X > 4) = 1 - P(X \le 4) = 1 - 0.375 = 0.625\)

The pdf x over 16 with the area from 4 to 6 shaded.

d) \(P(X \ge 4)\)

\[P(X \ge 4) = \int_4^{6} \frac{x}{16}\,dx = 0.625\]

Or from (b): \(P(X \ge 4) = 1 - P(X < 4) = 1 - 0.375 = 0.625\)

The pdf x over 16 with the area from 4 to 6 shaded.

e) \(P(4 < X < 7)\)

\(f(x) = 0\) above \(6\), so stop the integral at \(6\).

\[P(4 < X < 7) = \int_4^{6} \frac{x}{16}\,dx + \int_6^{7} 0\,dx = \frac{36 - 16}{32} = 0.625\]

The pdf x over 16 with the area from 4 to 6 shaded and nothing shaded past 6.

f) \(P(3 \le X \le 4)\)

\[P(3 \le X \le 4) = \int_3^{4} \frac{x}{16}\,dx = \frac{x^2}{32} \bigg|_3^{4} = \frac{16 - 9}{32} = 0.21875\]

The pdf x over 16 with the area from 3 to 4 shaded.

g) \(P(3 < X < 5)\)

\[P(3 < X < 5) = \int_3^{5} \frac{x}{16}\,dx = \frac{x^2}{32} \bigg|_3^{5} = \frac{25 - 9}{32} = 0.5\]

The pdf x over 16 with the area from 3 to 5 shaded.

h) \(P(X = 4)\)

\[P(X = 4) = \int_4^{4} \frac{x}{16}\,dx = 0\]

The pdf x over 16 with a single vertical line at 4 and no shaded region.

The integral over a single point has no width.

ImportantAny Exact Value Has Probability Zero

\[P(X = c) = \int_c^c f(x) \, dx = 0\] So for a continuous \(X\), \[P(a \le X \le b) = P(a < X < b) = P(a \le X < b) = P(a < X \le b).\]

Unlike Lessons 8 through 10, \(\le\) and \(<\) give the same answer.

The Cumulative Distribution Function

NoteLesson 8: cdf of a Discrete RV

The cdf of a discrete random variable \(X\) is \[F(x) = P(X \le x) = \sum_{y \,\le\, x} p(y).\]

ImportantToday: cdf of a Continuous RV

The cdf of a continuous random variable \(X\) is \[F(x) = P(X \le x) = \int_{-\infty}^{x} f(y)\,dy.\]

Fully define the cdf. Integrate the pdf from the left edge of the support.

\[F_X(x) = P(X \le x) = \int_2^x \frac{1}{16} y \, dy = \frac{y^2}{32} \bigg|_2^x = \frac{x^2 - 4}{32}\]

pdf

\[f_X(x) = \begin{cases} \dfrac{x}{16} & 2 \le x \le 6 \\ 0 & \text{otherwise} \end{cases}\]

cdf

\[F_X(x) = P(X \le x) = \begin{cases} 0 & x < 2 \\ \dfrac{x^2 - 4}{32} & 2 \le x \le 6 \\ 1 & x > 6 \end{cases}\]

All three pieces go in. \(F\) is defined for every real number.

Two panels. Left, the pdf x over 16 on 2 to 6 with the area from 2 to 4 shaded. Right, the cdf x squared minus 4 over 32, flat at 0 left of 2, curving up to 1 at x equal to 6, then flat at 1, with a dashed guide from x equal to 4 up to the curve at height 0.375.

\(P(3 \le X \le 4)\) two ways.

\[\begin{aligned} \text{pdf} \quad & \int_3^{4} \frac{1}{16} x \, dx = \frac{x^2}{32} \bigg|_3^{4} = \frac{16 - 9}{32} = 0.21875 \\ \text{cdf} \quad & F(4) - F(3) = \left.\frac{x^2 - 4}{32}\right|_{x=4} - \left.\frac{x^2 - 4}{32}\right|_{x=3} = \frac{4^2 - 4}{32} - \frac{3^2 - 4}{32} = \frac{12}{32} - \frac{5}{32} = 0.21875 \end{aligned}\]

Same questions as the pdf, now off the cdf.

\[F_X(x) = P(X \le x) = \begin{cases} 0 & x < 2 \\ \dfrac{x^2 - 4}{32} & 2 \le x \le 6 \\ 1 & x > 6 \end{cases}\]

a) \(P(X < 4)\)

\[P(X < 4) = P(X \le 4) = F(4) = \left.\frac{x^2 - 4}{32}\right|_{x=4} = \frac{4^2 - 4}{32} = \frac{12}{32} = 0.375\]

The pdf with the area from 2 to 4 shaded, next to the cdf evaluated at 4, height 0.375.

b) \(P(X \le 4)\)

\[P(X \le 4) = F(4) = \left.\frac{x^2 - 4}{32}\right|_{x=4} = \frac{4^2 - 4}{32} = \frac{12}{32} = 0.375\]

The pdf with the area from 2 to 4 shaded, next to the cdf evaluated at 4, height 0.375.

c) \(P(X > 4)\)

\[P(X > 4) = 1 - P(X \le 4) = 1 - F(4) = 1 - \left.\frac{x^2 - 4}{32}\right|_{x=4} = 1 - \frac{4^2 - 4}{32} = 1 - \frac{12}{32} = 0.625\]

The pdf with the area from 4 to 6 shaded, next to the cdf evaluated at 4, the answer being one minus that height.

d) \(P(X \ge 4)\)

\[P(X \ge 4) = 1 - P(X < 4) = 1 - F(4) = 1 - \left.\frac{x^2 - 4}{32}\right|_{x=4} = 1 - \frac{4^2 - 4}{32} = 1 - \frac{12}{32} = 0.625\]

The pdf with the area from 4 to 6 shaded, next to the cdf evaluated at 4, the answer being one minus that height.

e) \(P(4 < X < 7)\)

\(7 > 6\), so \(F(7) = 1\).

\[P(4 < X < 7) = F(7) - F(4) = 1 - \left.\frac{x^2 - 4}{32}\right|_{x=4} = 1 - \frac{4^2 - 4}{32} = 1 - \frac{12}{32} = 0.625\]

The pdf with the area from 4 to 6 shaded and nothing past 6, next to the cdf evaluated at 4 and at 7 where it is already flat at 1.

f) \(P(3 \le X \le 4)\)

\[P(3 \le X \le 4) = F(4) - F(3) = \left.\frac{x^2 - 4}{32}\right|_{x=4} - \left.\frac{x^2 - 4}{32}\right|_{x=3} = \frac{4^2 - 4}{32} - \frac{3^2 - 4}{32} = \frac{12}{32} - \frac{5}{32} = 0.21875\]

The pdf with the area from 3 to 4 shaded, next to the cdf evaluated at 3 and 4, the answer being the gap between the heights.

g) \(P(3 < X < 5)\)

\[P(3 < X < 5) = F(5) - F(3) = \left.\frac{x^2 - 4}{32}\right|_{x=5} - \left.\frac{x^2 - 4}{32}\right|_{x=3} = \frac{5^2 - 4}{32} - \frac{3^2 - 4}{32} = \frac{21}{32} - \frac{5}{32} = 0.5\]

The pdf with the area from 3 to 5 shaded, next to the cdf evaluated at 3 and 5, the answer being the gap between the heights.

h) \(P(X = 4)\)

\[P(X = 4) = F(4) - F(4) = \left.\frac{x^2 - 4}{32}\right|_{x=4} - \left.\frac{x^2 - 4}{32}\right|_{x=4} = \frac{4^2 - 4}{32} - \frac{4^2 - 4}{32} = 0\]

The pdf with a single vertical line at 4, next to the cdf evaluated at 4, the same height subtracted from itself.

Percentiles

ImportantDefinition: Percentile

The median \(\tilde{\mu}\) satisfies \[F(\tilde{\mu}) = P(X \le \tilde{\mu}) = 0.5\] The \((100p)\)th percentile is the value \(c\) that satisfies \[F(c) = P(X \le c) = p\]

Median. Find \(\tilde{\mu}\) so that \(F(\tilde{\mu}) = P(X \le \tilde{\mu}) = 0.5\).

The pdf x over 16 on 2 to 6 with the area from 2 up to an unknown point shaded and labeled area 0.5. A dashed orange line marks the unknown point, labeled mu tilde equals question mark.

\[\begin{aligned} F(\tilde{\mu}) = P(X \le \tilde{\mu}) &= 0.5 \\ \frac{\tilde{\mu}^2 - 4}{32} &= 0.5 \\ \tilde{\mu}^2 - 4 &= 16 \\ \tilde{\mu}^2 &= 20 \\ \tilde{\mu} &= \sqrt{20} = 4.4721 \end{aligned}\]

90th percentile. Find \(c\) so that \(F(c) = P(X \le c) = 0.9\).

The pdf x over 16 on 2 to 6 with the area from 2 up to an unknown point near 5.7 shaded and labeled area 0.9. A dashed orange line marks the unknown point, labeled c equals question mark.

\[\begin{aligned} F(c) = P(X \le c) &= 0.9 \\ \frac{c^2 - 4}{32} &= 0.9 \\ c^2 - 4 &= 28.8 \\ c^2 &= 32.8 \\ c &= \sqrt{32.8} = 5.7271 \end{aligned}\]

Expected Value and Variance

NoteLesson 8: Discrete RV

For a discrete \(X\) with lower bound \(a\) and upper bound \(b\), \[E(X) = \mu = \sum_{x=a}^{b} x\,p(x)\] \[E(X^2) = \sum_{x=a}^{b} x^2\,p(x)\] \[V(X) = \sigma^2 = E(X^2) - [E(X)]^2, \qquad \sigma = \sqrt{V(X)}\]

ImportantToday: Continuous RV

For a continuous \(X\) with lower bound \(a\) and upper bound \(b\), \[E(X) = \mu = \int_a^b x\,f(x)\,dx\] \[E(X^2) = \int_a^b x^2\,f(x)\,dx\] \[V(X) = \sigma^2 = E(X^2) - [E(X)]^2, \qquad \sigma = \sqrt{V(X)}\]

Same \(X\), \(f(x) = x/16\) on \([2, 6]\):

\[\begin{aligned} E(X) &= \int_2^6 x \cdot \frac{1}{16} x \, dx = \frac{x^3}{48} \bigg|_2^6 = \frac{216 - 8}{48} = \frac{13}{3} = 4.3333 \\ E(X^2) &= \int_2^6 x^2 \cdot \frac{1}{16} x \, dx = \frac{x^4}{64} \bigg|_2^6 = \frac{1296 - 16}{64} = 20 \\ V(X) &= E(X^2) - [E(X)]^2 = 20 - \frac{169}{9} = \frac{11}{9} = 1.2222 \\ \sigma &= \sqrt{1.2222} = 1.1055 \end{aligned}\]

The mean (\(4.3333\)) sits below the median (\(4.4721\)). The long side of this pdf is on the left.

Class Problem

\[f_X(x) = \begin{cases} \dfrac{k}{x^2} & 1 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}\]

a) Find \(k\).

The total area is \(1\).

\[\begin{aligned} \int_1^2 \frac{k}{x^2} \, dx &= 1 \\ -\frac{k}{x} \bigg|_1^2 &= 1 \\ -\frac{k}{2} + k &= 1 \\ \frac{k}{2} &= 1 \\ k &= 2 \end{aligned}\]

\[f_X(x) = \begin{cases} \dfrac{2}{x^2} & 1 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}\]

The pdf 2 over x squared on 1 to 2 with the whole region under it shaded.

b) What is the probability \(X\) lands more than one standard deviation above the mean or more than one standard deviation below it?

\[\begin{aligned} E(X) &= \int_1^2 x \cdot \frac{2}{x^2} \, dx = \int_1^2 \frac{2}{x} \, dx = 2 \ln x \bigg|_1^2 = 2 \ln 2 - 2 \ln 1 = 1.3863 \\ E(X^2) &= \int_1^2 x^2 \cdot \frac{2}{x^2} \, dx = \int_1^2 2 \, dx = 2x \bigg|_1^2 = 4 - 2 = 2 \\ V(X) &= E(X^2) - [E(X)]^2 = 2 - 1.3863^2 = 0.0782 \\ \sigma &= \sqrt{0.0782} = 0.2796 \end{aligned}\]

\[\mu \pm \sigma = 1.3863 \pm 0.2796 = (1.1067,\ 1.6659)\]

Fully define the cdf.

\[F_X(x) = P(X \le x) = \int_1^x \frac{2}{y^2} \, dy = -\frac{2}{y} \bigg|_1^x = 2 - \frac{2}{x}\]

\[F_X(x) = P(X \le x) = \begin{cases} 0 & x < 1 \\ 2 - \dfrac{2}{x} & 1 \le x \le 2 \\ 1 & x > 2 \end{cases}\]

\[P(X < 1.1067) + P(X > 1.6659) = F(1.1067) + 1 - F(1.6659)\]

\[= \left.2 - \frac{2}{x}\right|_{x=1.1067} + 1 - \left(\left.2 - \frac{2}{x}\right|_{x=1.6659}\right)\]

\[= \left(2 - \frac{2}{1.1067}\right) + 1 - \left(2 - \frac{2}{1.6659}\right) = 0.1928 + 0.2005 = \mathbf{0.3933}\]

The pdf 2 over x squared with both tails beyond one standard deviation shaded and a triangle at the mean, next to the cdf evaluated at 1.1067 and 1.6659.

The endpoints carry zero probability.

Board Problems

Problem 1: Start From the pdf

\[f(x) = \begin{cases} \dfrac{1}{2\sqrt{x}} & 1 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}\]

  1. Fully define the cdf.

  2. Find each: \(P(X = 2)\), \(P(X < 2)\), \(P(X \le 2)\), \(P(X > 3)\), \(P(X \ge 3)\), and \(P(2 \le X \le 3)\).

  3. Find the median and the 90th percentile.

  4. Find \(E(X)\), \(V(X)\), and \(\sigma\), then find the probability \(X\) lands more than one standard deviation above or below the mean.

  1. \[F(x) = P(X \le x) = \int_1^x \frac{1}{2\sqrt{y}}\,dy = \sqrt{y} \bigg|_1^x = \sqrt{x} - 1, \qquad F(x) = P(X \le x) = \begin{cases} 0 & x < 1 \\ \sqrt{x} - 1 & 1 \le x \le 4 \\ 1 & x > 4 \end{cases}\]

The pdf one over two root x on 1 to 4 next to its cdf root x minus 1.

\[\begin{aligned} P(X = 2) &= F(2) - F(2) = 0 \\ P(X < 2) &= P(X \le 2) = F(2) = \left.\left(\sqrt{x} - 1\right)\right|_{x=2} = \sqrt{2} - 1 = \mathbf{0.4142} \\ P(X \le 2) &= F(2) = \left.\left(\sqrt{x} - 1\right)\right|_{x=2} = \sqrt{2} - 1 = \mathbf{0.4142} \\ P(X > 3) &= 1 - F(3) = 1 - \left.\left(\sqrt{x} - 1\right)\right|_{x=3} = 1 - \left(\sqrt{3} - 1\right) = \mathbf{0.2679} \\ P(X \ge 3) &= 1 - P(X < 3) = 1 - F(3) = 1 - \left.\left(\sqrt{x} - 1\right)\right|_{x=3} = 1 - \left(\sqrt{3} - 1\right) = \mathbf{0.2679} \\ P(2 \le X \le 3) &= F(3) - F(2) = \left.\left(\sqrt{x} - 1\right)\right|_{x=3} - \left.\left(\sqrt{x} - 1\right)\right|_{x=2} = \left(\sqrt{3} - 1\right) - \left(\sqrt{2} - 1\right) = \mathbf{0.3178} \end{aligned}\]

The pdf one over two root x with dashed lines at 2 and 3, next to the cdf evaluated at 2 and 3.

  1. Median. Find \(\tilde{\mu}\) so that \(F(\tilde{\mu}) = P(X \le \tilde{\mu}) = 0.5\).

\[\begin{aligned} F(\tilde{\mu}) = P(X \le \tilde{\mu}) &= 0.5 \\ \sqrt{\tilde{\mu}} - 1 &= 0.5 \\ \sqrt{\tilde{\mu}} &= 1.5 \\ \tilde{\mu} &= 1.5^2 = \mathbf{2.25} \end{aligned}\]

90th percentile. Find \(c\) so that \(F(c) = P(X \le c) = 0.9\).

\[\begin{aligned} F(c) = P(X \le c) &= 0.9 \\ \sqrt{c} - 1 &= 0.9 \\ \sqrt{c} &= 1.9 \\ c &= 1.9^2 = \mathbf{3.61} \end{aligned}\]

The pdf with the area left of 3.61 shaded and the area left of 2.25 shaded darker, next to the cdf with guides from 0.5 and 0.9 across to the curve and down to 2.25 and 3.61.

\[\begin{aligned} E(X) &= \int_1^4 x \cdot \frac{1}{2\sqrt{x}}\,dx = \int_1^4 \frac{x^{1/2}}{2}\,dx = \frac{x^{3/2}}{3} \bigg|_1^4 = \frac{8 - 1}{3} = \frac{7}{3} = 2.3333 \\ E(X^2) &= \int_1^4 x^2 \cdot \frac{1}{2\sqrt{x}}\,dx = \int_1^4 \frac{x^{3/2}}{2}\,dx = \frac{x^{5/2}}{5} \bigg|_1^4 = \frac{32 - 1}{5} = 6.2 \\ V(X) &= E(X^2) - [E(X)]^2 = 6.2 - \left(\frac{7}{3}\right)^2 = 0.7556 \\ \sigma &= \sqrt{0.7556} = 0.8692 \end{aligned}\]

\(\mu \pm \sigma = 2.3333 \pm 0.8692 = (1.4641,\ 3.2026)\).

\[P(X < 1.4641) + P(X > 3.2026) = F(1.4641) + 1 - F(3.2026)\]

\[= \left.\left(\sqrt{x} - 1\right)\right|_{x=1.4641} + 1 - \left.\left(\sqrt{x} - 1\right)\right|_{x=3.2026}\]

\[= \left(\sqrt{1.4641} - 1\right) + 1 - \left(\sqrt{3.2026} - 1\right) = 0.2100 + 0.2104 = \mathbf{0.4204}\]

The pdf with both tails beyond one standard deviation shaded and a triangle at the mean, next to the cdf evaluated at both cutoffs.

Problem 2: Two Mile Run

A cadet company’s two mile run times, in minutes, follow

\[f(x) = \begin{cases} \dfrac{16 - x}{8} & 12 \le x \le 16 \\ 0 & \text{otherwise} \end{cases}\]

  1. Verify \(f\) is a legal pdf.

  2. Fully define the cdf.

  3. Find \(P(X \le 13)\), \(P(X > 14)\), and \(P(13.5 \le X \le 15)\).

  4. Find the median run time.

  5. Find \(E(X)\), \(V(X)\), and \(\sigma\), then find the probability a run time lands more than one standard deviation above or below the mean.

  1. \(f(x) \ge 0\) on \([12, 16]\), and

\[\int_{12}^{16} \frac{16 - x}{8}\,dx = -\frac{(16 - x)^2}{16} \bigg|_{12}^{16} = 0 + \frac{16}{16} = 1\]

The pdf 16 minus x over 8 on 12 to 16, a falling line with the whole region under it shaded.

  1. \[F(x) = P(X \le x) = \int_{12}^{x} \frac{16 - y}{8}\,dy = 1 - \frac{(16 - x)^2}{16}, \qquad F(x) = P(X \le x) = \begin{cases} 0 & x < 12 \\ 1 - \dfrac{(16 - x)^2}{16} & 12 \le x \le 16 \\ 1 & x > 16 \end{cases}\]

The falling pdf on 12 to 16 next to its cdf rising to 1 at 16.

\[\begin{aligned} P(X \le 13) &= F(13) = \left.1 - \frac{(16 - x)^2}{16}\right|_{x=13} = 1 - \frac{(16 - 13)^2}{16} = 1 - \frac{9}{16} = \mathbf{0.4375} \\ P(X > 14) &= 1 - F(14) = 1 - \left(\left.1 - \frac{(16 - x)^2}{16}\right|_{x=14}\right) = 1 - \left(1 - \frac{(16 - 14)^2}{16}\right) = 1 - \frac{12}{16} = \mathbf{0.25} \\ P(13.5 \le X \le 15) &= F(15) - F(13.5) = \left.1 - \frac{(16 - x)^2}{16}\right|_{x=15} - \left.1 - \frac{(16 - x)^2}{16}\right|_{x=13.5} \\ &= \left(1 - \frac{(16 - 15)^2}{16}\right) - \left(1 - \frac{(16 - 13.5)^2}{16}\right) = \frac{15}{16} - \frac{9.75}{16} = \mathbf{0.3281} \end{aligned}\]

The falling run time pdf with dashed lines at 13, 13.5, 14, and 15, next to the cdf evaluated at those four times.

  1. Median. Find \(\tilde{\mu}\) so that \(F(\tilde{\mu}) = P(X \le \tilde{\mu}) = 0.5\).

\[\begin{aligned} F(\tilde{\mu}) = P(X \le \tilde{\mu}) &= 0.5 \\ 1 - \frac{(16 - \tilde{\mu})^2}{16} &= 0.5 \\ \frac{(16 - \tilde{\mu})^2}{16} &= 0.5 \\ (16 - \tilde{\mu})^2 &= 8 \\ 16 - \tilde{\mu} &= \sqrt{8} \\ \tilde{\mu} &= 16 - \sqrt{8} = \mathbf{13.1716} \text{ minutes} \end{aligned}\]

The run time pdf with half its area shaded up to 13.1716, next to the cdf with a guide from 0.5 across to the curve and down to 13.1716.

Take the positive root. \(\tilde{\mu}\) has to sit inside \([12, 16]\).

\[\begin{aligned} E(X) &= \int_{12}^{16} x \cdot \frac{16 - x}{8}\,dx = \frac{40}{3} = 13.3333 \\ E(X^2) &= \int_{12}^{16} x^2 \cdot \frac{16 - x}{8}\,dx = \frac{536}{3} = 178.6667 \\ V(X) &= \frac{536}{3} - \left(\frac{40}{3}\right)^2 = \frac{8}{9} = 0.8889 \\ \sigma &= 0.9428 \text{ minutes} \end{aligned}\]

\(\mu \pm \sigma = (12.3905,\ 14.2761)\).

\[P(X < 12.3905) + P(X > 14.2761) = F(12.3905) + 1 - F(14.2761)\]

\[= \left.1 - \frac{(16 - x)^2}{16}\right|_{x=12.3905} + 1 - \left(\left.1 - \frac{(16 - x)^2}{16}\right|_{x=14.2761}\right)\]

\[= \left(1 - \frac{3.6095^2}{16}\right) + \frac{1.7239^2}{16} = 0.1857 + 0.1857 = \mathbf{0.3715}\]

The run time pdf with both tails shaded and a triangle at 13.33, next to the cdf evaluated at both cutoffs.

Problem 3: Sentry Duty

A supply truck is equally likely to arrive at any point in a sentry’s \(4\) hour shift. Let \(X\) be the hours until it arrives.

\[f(x) = \begin{cases} \dfrac{1}{4} & 0 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}\]

  1. Fully define the cdf.

  2. Find \(P(X \le 1)\), \(P(1 < X < 3)\), and \(P(X = 2)\).

  3. By what time has the truck arrived with probability \(0.8\)?

  4. Find \(E(X)\) and \(\sigma\), then find the probability \(X\) lands more than one standard deviation above or below the mean.

  5. The truck has not arrived after \(1\) hour. What is the probability it takes at least \(3\)?

  1. \[F(x) = P(X \le x) = \begin{cases} 0 & x < 0 \\ \dfrac{x}{4} & 0 \le x \le 4 \\ 1 & x > 4 \end{cases}\]

The flat pdf at one quarter on 0 to 4 next to its straight line cdf.

\[\begin{aligned} P(X \le 1) &= F(1) = \left.\frac{x}{4}\right|_{x=1} = \frac{1}{4} = \mathbf{0.25} \\ P(1 < X < 3) &= F(3) - F(1) = \left.\frac{x}{4}\right|_{x=3} - \left.\frac{x}{4}\right|_{x=1} = \frac{3}{4} - \frac{1}{4} = \mathbf{0.5} \\ P(X = 2) &= F(2) - F(2) = \mathbf{0} \end{aligned}\]

The flat pdf with dashed lines at 1, 2, and 3 hours, next to the straight line cdf evaluated at those three times.

  1. Find \(c\) so that \(F(c) = P(X \le c) = 0.8\).

\[\begin{aligned} F(c) = P(X \le c) &= 0.8 \\ \frac{c}{4} &= 0.8 \\ c &= \mathbf{3.2} \text{ hours} \end{aligned}\]

The flat pdf with the area from 0 to 3.2 shaded, next to the straight line cdf with a guide from 0.8 across to the curve and down to 3.2 hours.

\[\begin{aligned} E(X) &= \int_0^4 \frac{x}{4}\,dx = \frac{x^2}{8} \bigg|_0^4 = 2 \\ E(X^2) &= \int_0^4 \frac{x^2}{4}\,dx = \frac{x^3}{12} \bigg|_0^4 = \frac{16}{3} \\ V(X) &= \frac{16}{3} - 4 = \frac{4}{3}, \qquad \sigma = 1.1547 \end{aligned}\]

\(\mu \pm \sigma = (0.8453,\ 3.1547)\).

\[P(X < 0.8453) + P(X > 3.1547) = F(0.8453) + 1 - F(3.1547)\]

\[= \left.\frac{x}{4}\right|_{x=0.8453} + 1 - \left.\frac{x}{4}\right|_{x=3.1547} = \frac{0.8453}{4} + 1 - \frac{3.1547}{4} = 0.2113 + 0.2113 = \mathbf{0.4226}\]

The flat pdf with both tails shaded and a triangle at 2 hours, next to the cdf evaluated at both cutoffs.

  1. Lesson 6. \(\{X \ge 3\}\) sits inside \(\{X \ge 1\}\).

\[P(X \ge 3 \mid X \ge 1) = \frac{P(X \ge 3)}{P(X \ge 1)} = \frac{1 - F(3)}{1 - F(1)} = \frac{1 - \left.\frac{x}{4}\right|_{x=3}}{1 - \left.\frac{x}{4}\right|_{x=1}} = \frac{1 - \frac{3}{4}}{1 - \frac{1}{4}} = \frac{1/4}{3/4} = \mathbf{0.3333}\]

The flat pdf with the region from 1 to 4 shaded once and the region from 3 to 4 shaded twice, so it reads darker.

Problem 4: Reading the cdf

\(X\) is the fraction of a fuel tank a vehicle burns on a patrol. You are handed only the cdf:

\[F(x) = P(X \le x) = \begin{cases} 0 & x < 0 \\ 3x^2 - 2x^3 & 0 \le x \le 1 \\ 1 & x > 1 \end{cases}\]

  1. Recover the pdf.

  2. Find \(P(X > 0.7)\) and \(P(0.25 \le X \le 0.75)\).

  3. Find the median.

  4. Find \(E(X)\) and \(\sigma\).

  1. Differentiate.

\[f(x) = F'(x) = \begin{cases} 6x - 6x^2 & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}\]

The mound shaped pdf 6x minus 6x squared on 0 to 1 next to its S shaped cdf.

\[\begin{aligned} P(X > 0.7) &= 1 - F(0.7) = 1 - \left.\left(3x^2 - 2x^3\right)\right|_{x=0.7} = 1 - \left(3(0.7)^2 - 2(0.7)^3\right) = 1 - (1.47 - 0.686) = \mathbf{0.216} \\ P(0.25 \le X \le 0.75) &= F(0.75) - F(0.25) = \left.\left(3x^2 - 2x^3\right)\right|_{x=0.75} - \left.\left(3x^2 - 2x^3\right)\right|_{x=0.25} \\ &= \left(3(0.75)^2 - 2(0.75)^3\right) - \left(3(0.25)^2 - 2(0.25)^3\right) = 0.84375 - 0.15625 = \mathbf{0.6875} \end{aligned}\]

The mound shaped pdf with dashed lines at 0.25, 0.7, and 0.75, next to the S shaped cdf evaluated at those three values.

  1. Find \(\tilde{\mu}\) so that \(F(\tilde{\mu}) = P(X \le \tilde{\mu}) = 0.5\).

\[\begin{aligned} F(\tilde{\mu}) = P(X \le \tilde{\mu}) &= 0.5 \\ 3\tilde{\mu}^2 - 2\tilde{\mu}^3 &= 0.5 \end{aligned}\]

A cubic, so check the middle of the support. The pdf \(6x - 6x^2\) is symmetric about \(0.5\).

\[\left.\left(3x^2 - 2x^3\right)\right|_{x=0.5} = 3(0.5)^2 - 2(0.5)^3 = 0.75 - 0.25 = 0.5, \qquad \tilde{\mu} = \mathbf{0.5}\]

The symmetric mound shaped pdf with its left half shaded, next to the S shaped cdf with a guide from 0.5 across to the curve and down to 0.5.

\[\begin{aligned} E(X) &= \int_0^1 x(6x - 6x^2)\,dx = \left[2x^3 - \tfrac{3}{2}x^4\right]_0^1 = 0.5 \\ E(X^2) &= \int_0^1 x^2(6x - 6x^2)\,dx = \left[\tfrac{3}{2}x^4 - \tfrac{6}{5}x^5\right]_0^1 = 0.3 \\ V(X) &= 0.3 - 0.25 = 0.05, \qquad \sigma = 0.2236 \end{aligned}\]

The mound shaped pdf fully shaded with a triangle marking the mean at 0.5.

Problem 5: Stretch

\(X\) is the hours a generator runs before it faults. Every generator runs at least \(1\) hour. Its cdf is

\[F(x) = P(X \le x) = \begin{cases} 0 & x < 1 \\ 1 - \dfrac{1}{x^2} & x \ge 1 \end{cases}\]

  1. Find the pdf.

  2. Find \(P(X > 2)\) and \(P(2 < X \le 4)\).

  3. Find the median and the 90th percentile.

  4. Find \(E(X)\).

  5. Find \(V(X)\).

  1. \(f(x) = F'(x) = \dfrac{2}{x^3}\) for \(x \ge 1\), and \(0\) otherwise.

The pdf 2 over x cubed falling from 2 at x equal to 1, next to its cdf climbing toward 1.

\[\begin{aligned} P(X > 2) &= 1 - F(2) = 1 - \left(\left.1 - \frac{1}{x^2}\right|_{x=2}\right) = 1 - \left(1 - \frac{1}{2^2}\right) = \frac{1}{4} = \mathbf{0.25} \\ P(2 < X \le 4) &= F(4) - F(2) = \left.1 - \frac{1}{x^2}\right|_{x=4} - \left.1 - \frac{1}{x^2}\right|_{x=2} = \left(1 - \frac{1}{4^2}\right) - \left(1 - \frac{1}{2^2}\right) = \frac{15}{16} - \frac{3}{4} = \mathbf{0.1875} \end{aligned}\]

The pdf 2 over x cubed with dashed lines at 2 and 4, next to the generator cdf evaluated at 2 and 4 hours.

  1. Median. Find \(\tilde{\mu}\) so that \(F(\tilde{\mu}) = P(X \le \tilde{\mu}) = 0.5\).

\[\begin{aligned} F(\tilde{\mu}) = P(X \le \tilde{\mu}) &= 0.5 \\ 1 - \frac{1}{\tilde{\mu}^2} &= 0.5 \\ \frac{1}{\tilde{\mu}^2} &= 0.5 \\ \tilde{\mu}^2 &= 2 \\ \tilde{\mu} &= \sqrt{2} = \mathbf{1.4142} \end{aligned}\]

90th percentile. Find \(c\) so that \(F(c) = P(X \le c) = 0.9\).

\[\begin{aligned} F(c) = P(X \le c) &= 0.9 \\ 1 - \frac{1}{c^2} &= 0.9 \\ \frac{1}{c^2} &= 0.1 \\ c^2 &= 10 \\ c &= \sqrt{10} = \mathbf{3.1623} \end{aligned}\]

The pdf 2 over x cubed with the area left of 3.1623 shaded and the area left of 1.4142 shaded darker, next to the cdf with guides from 0.5 and 0.9 down to those two values.

  1. \[E(X) = \int_1^{\infty} x \cdot \frac{2}{x^3}\,dx = \int_1^{\infty} \frac{2}{x^2}\,dx = -\frac{2}{x} \bigg|_1^{\infty} = \mathbf{2}\]

The pdf 2 over x cubed shaded with a triangle marking the mean at 2 hours.

  1. \[E(X^2) = \int_1^{\infty} x^2 \cdot \frac{2}{x^3}\,dx = \int_1^{\infty} \frac{2}{x}\,dx = 2 \ln x \bigg|_1^{\infty} = \infty\]

\(E(X^2)\) diverges, so \(V(X)\) is infinite. The mean exists, the variance does not.

The curve 2 over x shaded from 1 out to 40, the area still growing at the right edge.

Before You Leave

Today

  • Probability for a continuous \(X\) is area under the pdf
  • Total area \(1\) pins down \(k\)
  • \(P(X = c) = 0\), so endpoints do not matter
  • The cdf turns every probability into \(F(b) - F(a)\), and percentiles into solving \(F(c) = p\)
  • \(E(X)\) and \(V(X)\) are the Lesson 8 formulas with integrals

Any questions?

Next Lesson

Lesson 12: Normal Distribution

  • Identify the normal and standard normal (\(z\)) distributions
  • Standardize values with \(z = (x-\mu)/\sigma\) and compute probabilities using the \(z\) table
  • Determine percentiles and \(z\) critical values for the normal distribution

Reading: Devore 4.3

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