Lesson 12: Continuous Random Variables

Today we start computing probabilities by integrating PDFs. As this comic shows, differentiation is mechanics — integration is art. The good news: most of our integrals today are just rectangles and triangles.

What We Did: Lessons 6–11

Sample Spaces and Events:

  • Sample space \(S\) = set of all possible outcomes
  • Event = subset of the sample space
  • Operations: Union (\(A \cup B\)), Intersection (\(A \cap B\)), Complement (\(A^c\))

Kolmogorov Axioms:

  1. \(P(A) \geq 0\)
  2. \(P(S) = 1\)
  3. For mutually exclusive events: \(P(A \cup B) = P(A) + P(B)\)

Key Rules:

  • Complement Rule: \(P(A^c) = 1 - P(A)\)
  • Addition Rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

Conditional Probability: \[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\]

Multiplication Rule: \[P(A \cap B) = P(A) \cdot P(B \mid A) = P(B) \cdot P(A \mid B)\]

Law of Total Probability: \[P(A) = P(B) \cdot P(A \mid B) + P(B^c) \cdot P(A \mid B^c)\]

Bayes’ Theorem: \[P(B \mid A) = \frac{P(B) \cdot P(A \mid B)}{P(B) \cdot P(A \mid B) + P(B^c) \cdot P(A \mid B^c)}\]

Counting Formulas:

With Replacement Without Replacement
Ordered \(n^k\) \(P(n,k) = \frac{n!}{(n-k)!}\)
Unordered \(\binom{n+k-1}{k}\) \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

Independence:

  • \(A\) and \(B\) are independent if \(P(A \cap B) = P(A) \cdot P(B)\)
  • Equivalently: \(P(A \mid B) = P(A)\)
  • Independent \(\neq\) Mutually Exclusive!

Random Variables:

  • A random variable \(X\) assigns a numerical value to each outcome in a sample space
  • Discrete RVs take finite or countably infinite values

PMF: \(p(x) = P(X = x)\) with \(p(x) \geq 0\) and \(\sum p(x) = 1\)

CDF: \(F(x) = P(X \leq x) = \sum_{y \leq x} p(y)\)

Expected Value: \(E(X) = \sum x \cdot p(x)\)

Variance: \(Var(X) = \sum (x - \mu)^2 \cdot p(x) = E(X^2) - [E(X)]^2\)

BINS Conditions:

  1. Binary outcomes (success/failure)
  2. Independent trials
  3. Number of trials is fixed (\(n\))
  4. Same probability (\(p\)) each trial

Key Formulas: If \(X \sim \text{Binomial}(n, p)\):

  • PMF: \(P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\)
  • Mean: \(E(X) = np\)
  • Variance: \(Var(X) = np(1-p)\)

R Functions: dbinom(x, size, prob) for PMF, pbinom(x, size, prob) for CDF

When to Use Poisson:

  • Counting events in a fixed interval (time, area, volume)
  • Events occur independently at a constant average rate \(\lambda\)

Key Formulas: If \(X \sim \text{Poisson}(\lambda)\):

  • PMF: \(P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \ldots\)
  • Mean: \(E(X) = \lambda\)
  • Variance: \(Var(X) = \lambda\)

R Functions: dpois(x, lambda) for PMF, ppois(x, lambda) for CDF

What We’re Doing: Lesson 12

Objectives

  • Interpret PDFs and areas as probabilities
  • Compute probabilities from PDFs and CDFs
  • Find \(E(X)\) and \(Var(X)\) for continuous random variables

Required Reading

Devore, Sections 4.1, 4.2

Break!

Reese

Cal

DMath Basketball!!

Math vs Garrison

8-3

Math vs Garrison

8-4

Army

SSG Michael Ollis — Medal of Honor (posthumous)
10th Mountain Division, 2-22 IN — KIA 28 Aug 2013, Afghanistan
Shielded a Polish officer from a suicide bomber, sacrificing his life.

The Takeaway for Today

NoteKey Concepts: Continuous Random Variables

Continuous vs. Discrete:

  • Discrete RVs: probabilities come from a PMF — \(P(X = x)\)
  • Continuous RVs: probabilities come from areas under a PDF — \(P(a \leq X \leq b) = \int_a^b f(x)\,dx\)

Probability Density Function (PDF): \(f(x)\) where:

  • \(f(x) \geq 0\) for all \(x\)
  • \(\int_{-\infty}^{\infty} f(x)\,dx = 1\)
  • \(P(a \leq X \leq b) = \int_a^b f(x)\,dx\)

Cumulative Distribution Function (CDF): \[F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)\,dt\]

Expected Value and Variance:

  • \(E(X) = \int_{-\infty}^{\infty} x \cdot f(x)\,dx\)
  • \(Var(X) = \int_{-\infty}^{\infty} (x - \mu)^2 \cdot f(x)\,dx = E(X^2) - [E(X)]^2\)

Critical Difference: For continuous RVs, \(P(X = c) = 0\) for any single value \(c\).

From Discrete to Continuous

The Big Shift

In Lessons 9–11, we worked with discrete random variables — they take on countable values (0, 1, 2, …), and we assign probability to each value using a PMF.

NoteDiscrete Distributions We Know
General Discrete Binomial Poisson
Notation \(X\) with PMF table \(X \sim \text{Binomial}(n, p)\) \(X \sim \text{Poisson}(\lambda)\)
Use when Any discrete setting Fixed \(n\) trials, counting successes Fixed interval, counting events
Parameters Given by PMF table \(n\) = trials, \(p\) = success probability \(\lambda\) = average rate
PMF \(p(x) = P(X = x)\) \(P(X=x) = \binom{n}{x}p^x(1-p)^{n-x}\) \(P(X=x) = \frac{e^{-\lambda}\lambda^x}{x!}\)
Support Specified by problem \(x = 0, 1, 2, \ldots, n\) \(x = 0, 1, 2, \ldots\)
Mean \(\mu = \sum x \cdot p(x)\) \(\mu = np\) \(\mu = \lambda\)
Variance \(\sigma^2 = \sum(x-\mu)^2 p(x)\) \(\sigma^2 = np(1-p)\) \(\sigma^2 = \lambda\)
R PMF Manual / table lookup dbinom(x, size, prob) dpois(x, lambda)
R CDF cumsum() on PMF pbinom(x, size, prob) ppois(x, lambda)

But many quantities in the real world are continuous:

  • Time until a piece of equipment fails
  • Weight of ammunition in a crate
  • Distance a mortar round travels
  • Body temperature of a soldier

For these, \(X\) can take any value in an interval. There are infinitely (uncountably!) many possible values, so we can’t assign probability to each one individually.

Discrete vs. Continuous Random Variables

Discrete Continuous
Values Countable (0, 1, 2, …) Any value in an interval
Examples Number of heads, number of breakdowns Time, weight, distance, temperature
How many possible values? Finite or countably infinite Uncountably infinite
Can you list all values? Yes No

Probability Density Functions (PDFs)

What is a PDF?

NoteRecall: Probability Mass Function (PMF): Discrete

A PMF \(p(x)\) for a discrete random variable \(X\) satisfies:

  1. \(p(x) \geq 0\) for all \(x\)
  2. \(\sum_{\text{all } x} p(x) = 1\) (all probabilities sum to 1)

Probabilities are computed by summing:

\[P(a \leq X \leq b) = \sum_{x=a}^{b} p(x)\]

ImportantProbability Density Function (PDF): Continuous

A probability density function \(f(x)\) for a continuous random variable \(X\) satisfies:

  1. \(f(x) \geq 0\) for all \(x\)
  2. \(\int_{-\infty}^{\infty} f(x)\,dx = 1\) (total area under the curve equals 1)

Probabilities are computed as areas under the curve:

\[P(a \leq X \leq b) = \int_a^b f(x)\,dx\]

The structure is the same — non-negative, totals to 1, use it to find probabilities over a range. The difference: sums become integrals.

Example

Consider a continuous random variable \(X\) with PDF:

\[f(x) = \begin{cases} \frac{x-1}{2} & 1 \leq x \leq 3 \\ 0 & \text{otherwise} \end{cases}\]

Notice this PDF has a slope — larger values of \(x\) have higher density. The density increases linearly from 0 at \(x = 1\) to 1 at \(x = 3\).

Is this a valid PDF?

For \(f(x)\) to be a valid PDF, we need two things:

  1. \(f(x) \geq 0\) for all \(x\)? Yes — \(\frac{x-1}{2} \geq 0\) on \([1, 3]\).

  2. Total area = 1?

\[\int_1^3 \frac{x-1}{2}\,dx = \frac{(x-1)^2}{4}\Bigg|_1^3 = \frac{4}{4} - \frac{0}{4} = 1 \quad \checkmark\]

You can also see this geometrically: it’s a triangle with base = 2 and height = 1, so area = \(\frac{1}{2}(2)(1) = 1\). ✓

What is \(P(X \leq 2)\)?

\[P(X \leq 2) = \int_1^2 \frac{x-1}{2}\,dx = \frac{(x-1)^2}{4}\Bigg|_1^2 = \frac{1}{4} - 0 = 0.25\]

Only 25% of the probability is below \(x = 2\) — makes sense since the PDF is low for small \(x\).

What is \(P(1.5 \leq X \leq 2.5)\)?

\[P(1.5 \leq X \leq 2.5) = \int_{1.5}^{2.5} \frac{x-1}{2}\,dx = \frac{(x-1)^2}{4}\Bigg|_{1.5}^{2.5} = \frac{2.25}{4} - \frac{0.25}{4} = \frac{2}{4} = 0.50\]

What is \(P(X = 2)\)?

\[P(X = 2) = \int_2^2 \frac{x-1}{2}\,dx = 0\]

ImportantA Critical Fact

For any continuous random variable:

\[P(X = c) = \int_c^c f(x)\,dx = 0\]

The “area” under a curve at a single point is zero.

Practical consequence:

\[P(a \leq X \leq b) = P(a < X < b) = P(a \leq X < b) = P(a < X \leq b)\]

It doesn’t matter whether we include the endpoints — the probability is the same!

This is very different from discrete RVs, where \(P(X = x)\) can be large. For continuous RVs, probability only accumulates over intervals.

The Cumulative Distribution Function (CDF)

CDF for Continuous RVs

ImportantCumulative Distribution Function

The CDF of a continuous random variable \(X\) with PDF \(f(x)\) is:

\[F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)\,dt\]

Properties:

  • \(F(x)\) is non-decreasing
  • \(\lim_{x \to -\infty} F(x) = 0\) and \(\lim_{x \to \infty} F(x) = 1\)
  • \(P(a \leq X \leq b) = F(b) - F(a)\)
  • \(f(x) = F'(x)\) — the PDF is the derivative of the CDF

CDF of Our Example

For our PDF \(f(x) = \frac{x-1}{2}\) on \([1, 3]\):

\[F(x) = \int_1^x \frac{t-1}{2}\,dt = \frac{(t-1)^2}{4}\Bigg|_1^x = \frac{(x-1)^2}{4} - \frac{(1-1)^2}{4} = \frac{(x-1)^2}{4}\]

So:

\[F(x) = \begin{cases} 0 & x < 1 \\ \frac{(x-1)^2}{4} & 1 \leq x \leq 3 \\ 1 & x > 3 \end{cases}\]

Notice the CDF is a curve (not a straight line) because the PDF has a slope. The CDF accelerates — probability accumulates faster as \(x\) increases because the PDF is taller there.

Our CDF:

\[F(x) = \begin{cases} 0 & x < 1 \\ \frac{(x-1)^2}{4} & 1 \leq x \leq 3 \\ 1 & x > 3 \end{cases}\]

What is \(P(X \leq 2)\)?

\[F(2) = \frac{(2-1)^2}{4} = \frac{1}{4} = 0.25\]

Same answer as before — the CDF just gives it to us directly without re-integrating!

What is \(P(1.5 \leq X \leq 2.5)\)?

\[P(1.5 \leq X \leq 2.5) = F(2.5) - F(1.5) = \frac{(2.5-1)^2}{4} - \frac{(1.5-1)^2}{4} = \frac{2.25}{4} - \frac{0.25}{4} = 0.50\]

Again, same answer — but using the CDF is often easier than re-doing the integral from scratch.

What is \(P(X = 2)\)?

\[P(X = 2) = F(2) - F(2) = 0\]

Still zero! The CDF confirms what we already know: for continuous RVs, the probability at a single point is always zero.

What is \(P(X > 2)\)?

\[P(X > 2) = 1 - F(2) = 1 - \frac{1}{4} = 0.75\]

75% of the probability is above \(x = 2\) — consistent with what we see in the PDF (most of the area is to the right).

Expected Value and Variance

Same concepts as discrete — just replace sums (\(\sum\)) with integrals (\(\int\)):

Discrete Continuous
Mean \(E(X) = \sum x \cdot p(x)\) \(E(X) = \int x \cdot f(x)\,dx\)
Variance \(Var(X) = \sum (x-\mu)^2 \cdot p(x)\) \(Var(X) = \int (x-\mu)^2 \cdot f(x)\,dx\)
Shortcut \(Var(X) = E(X^2) - [E(X)]^2\) \(Var(X) = E(X^2) - [E(X)]^2\)
SD \(\sigma = \sqrt{Var(X)}\) \(\sigma = \sqrt{Var(X)}\)

Back to Our Example

Recall our PDF: \(f(x) = \frac{x-1}{2}\) on \([1, 3]\).

What is \(E(X)\)?

\[E(X) = \int_1^3 x \cdot \frac{x-1}{2}\,dx = \int_1^3 \frac{x^2 - x}{2}\,dx\]

\[= \frac{1}{2}\left[\frac{x^3}{3} - \frac{x^2}{2}\right]_1^3 = \frac{1}{2}\left[\left(\frac{27}{3} - \frac{9}{2}\right) - \left(\frac{1}{3} - \frac{1}{2}\right)\right]\]

\[= \frac{1}{2}\left[\left(9 - 4.5\right) - \left(0.333 - 0.5\right)\right] = \frac{1}{2}\left[4.5 + 0.167\right] = \frac{1}{2}\left[\frac{14}{3}\right] = \frac{7}{3} \approx 2.333\]

The mean is \(\frac{7}{3} \approx 2.33\) — shifted toward the right, which makes sense because the PDF puts more weight on larger values of \(x\).

What is \(Var(X)\)?

Using the shortcut \(Var(X) = E(X^2) - [E(X)]^2\):

Step 1: Find \(E(X^2)\):

\[E(X^2) = \int_1^3 x^2 \cdot \frac{x-1}{2}\,dx = \int_1^3 \frac{x^3 - x^2}{2}\,dx\]

\[= \frac{1}{2}\left[\frac{x^4}{4} - \frac{x^3}{3}\right]_1^3 = \frac{1}{2}\left[\left(\frac{81}{4} - \frac{27}{3}\right) - \left(\frac{1}{4} - \frac{1}{3}\right)\right]\]

\[= \frac{1}{2}\left[\left(20.25 - 9\right) - \left(0.25 - 0.333\right)\right] = \frac{1}{2}\left[11.25 + 0.083\right] = \frac{1}{2} \cdot \frac{34}{3} = \frac{17}{3} \approx 5.667\]

Step 2: Apply the shortcut:

\[Var(X) = E(X^2) - [E(X)]^2 = \frac{17}{3} - \left(\frac{7}{3}\right)^2 = \frac{17}{3} - \frac{49}{9} = \frac{51}{9} - \frac{49}{9} = \frac{2}{9} \approx 0.222\]

What is \(SD(X)\)?

\[SD(X) = \sqrt{Var(X)} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3} \approx 0.471\]

Board Problems

Problem 1

A continuous random variable \(X\) has the following PDF:

\[f(x) = \begin{cases} kx^2 & 0 \leq x \leq 3 \\ 0 & \text{otherwise} \end{cases}\]

  1. Find the value of \(k\) that makes \(f(x)\) a valid PDF.

  2. Find the CDF \(F(x)\).

  3. What is \(P(X = 2)\)?

  4. What is \(P(X < 1)\)?

  5. What is \(P(X > 2)\)?

  6. What is \(P(1 \leq X \leq 2)\)?

  7. Find \(E(X)\) and \(Var(X)\).

  8. What is the probability that \(X\) is more than 2 standard deviations above the mean?

  1. We need \(\int_0^3 kx^2\,dx = 1\):

\[k \cdot \frac{x^3}{3}\Big|_0^3 = k \cdot \frac{27}{3} = 9k = 1 \implies k = \frac{1}{9}\]

So \(f(x) = \frac{x^2}{9}\) on \([0, 3]\).

  1. For \(0 \leq x \leq 3\):

\[F(x) = \int_0^x \frac{t^2}{9}\,dt = \frac{t^3}{27}\Big|_0^x = \frac{x^3}{27}\]

So: \(F(x) = \begin{cases} 0 & x < 0 \\ x^3/27 & 0 \leq x \leq 3 \\ 1 & x > 3 \end{cases}\)

  1. \(P(X = 2) = 0\) — for any continuous random variable, the probability of any single exact value is 0.

  2. \(P(X < 1) = F(1) = \frac{1^3}{27} = \frac{1}{27} \approx 0.037\)

  3. \(P(X > 2) = 1 - F(2) = 1 - \frac{8}{27} = \frac{19}{27} \approx 0.704\)

  4. \(P(1 \leq X \leq 2) = F(2) - F(1) = \frac{8}{27} - \frac{1}{27} = \frac{7}{27} \approx 0.259\)

  5. \(E(X) = \int_0^3 x \cdot \frac{x^2}{9}\,dx = \int_0^3 \frac{x^3}{9}\,dx = \frac{x^4}{36}\Big|_0^3 = \frac{81}{36} = \frac{9}{4} = 2.25\)

\(E(X^2) = \int_0^3 x^2 \cdot \frac{x^2}{9}\,dx = \int_0^3 \frac{x^4}{9}\,dx = \frac{x^5}{45}\Big|_0^3 = \frac{243}{45} = \frac{27}{5} = 5.4\)

\(Var(X) = E(X^2) - [E(X)]^2 = 5.4 - (2.25)^2 = 5.4 - 5.0625 = 0.3375\)

\(SD(X) = \sqrt{0.3375} \approx 0.581\)

  1. 2 standard deviations above the mean: \(\mu + 2\sigma = 2.25 + 2(0.581) = 3.412\)

Since \(3.412 > 3\) and the PDF is only defined on \([0, 3]\), we have \(P(X > 3.412) = 0\).

The probability of being more than 2 standard deviations above the mean is 0.

Problem 2: Sentry Duty Wait Time

A sentry expects a supply truck sometime during their 4-hour shift. The truck is equally likely to arrive at any point during the shift. Let \(X\) = time (in hours) until the truck arrives, with PDF:

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

  1. Verify this is a valid PDF.

  2. What is \(P(X \leq 1)\) — the truck arrives in the first hour?

  3. What is \(P(1 \leq X \leq 3)\)?

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

  5. The sentry gets relieved early at hour 3. What is the probability the truck has already arrived?

  6. What is the probability the truck arrives in the last 30 minutes of the shift?

  1. \(\int_0^4 \frac{1}{4}\,dx = \frac{x}{4}\Big|_0^4 = 1\) ✓

  2. \(P(X \leq 1) = \int_0^1 \frac{1}{4}\,dx = \frac{1}{4} = 0.25\)

  3. \(P(1 \leq X \leq 3) = \int_1^3 \frac{1}{4}\,dx = \frac{2}{4} = 0.50\)

  4. \(E(X) = \int_0^4 x \cdot \frac{1}{4}\,dx = \frac{x^2}{8}\Big|_0^4 = \frac{16}{8} = 2\) hours

\(E(X^2) = \int_0^4 x^2 \cdot \frac{1}{4}\,dx = \frac{x^3}{12}\Big|_0^4 = \frac{64}{12} = \frac{16}{3}\)

\(Var(X) = \frac{16}{3} - 4 = \frac{4}{3}\)

\(SD(X) = \sqrt{4/3} \approx 1.155\) hours

  1. \(P(X \leq 3) = \int_0^3 \frac{1}{4}\,dx = \frac{3}{4} = 0.75\)

  2. \(P(3.5 \leq X \leq 4) = \int_{3.5}^4 \frac{1}{4}\,dx = \frac{0.5}{4} = 0.125\)

Problem 3: Fuel Consumption

A tactical vehicle’s fuel consumption rate (gallons per mile) on a certain terrain follows the PDF:

\[f(x) = \begin{cases} 6x(1-x) & 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}\]

  1. Verify this is a valid PDF.

  2. What is the probability the fuel consumption rate is less than 0.5 gallons per mile?

  3. What is the probability the rate is between 0.25 and 0.75?

  4. Find \(E(X)\) — the expected fuel consumption rate.

  5. Find \(Var(X)\) and \(SD(X)\).

  6. Is it more likely that the rate is above or below 0.5? Explain.

  1. \(\int_0^1 6x(1-x)\,dx = \int_0^1 (6x - 6x^2)\,dx = \left[3x^2 - 2x^3\right]_0^1 = 3 - 2 = 1\) ✓

  2. \(P(X < 0.5) = \int_0^{0.5} 6x(1-x)\,dx = \left[3x^2 - 2x^3\right]_0^{0.5} = 0.75 - 0.25 = 0.50\)

  3. \(P(0.25 \leq X \leq 0.75) = \int_{0.25}^{0.75} 6x(1-x)\,dx = \left[3x^2 - 2x^3\right]_{0.25}^{0.75}\)

\(= (3(0.5625) - 2(0.421875)) - (3(0.0625) - 2(0.015625)) = (1.6875 - 0.84375) - (0.1875 - 0.03125) = 0.84375 - 0.15625 = 0.6875\)

  1. \(E(X) = \int_0^1 x \cdot 6x(1-x)\,dx = \int_0^1 (6x^2 - 6x^3)\,dx = \left[2x^3 - \frac{3x^4}{2}\right]_0^1 = 2 - 1.5 = 0.5\)

  2. \(E(X^2) = \int_0^1 x^2 \cdot 6x(1-x)\,dx = \int_0^1 (6x^3 - 6x^4)\,dx = \left[\frac{3x^4}{2} - \frac{6x^5}{5}\right]_0^1 = 1.5 - 1.2 = 0.3\)

\(Var(X) = 0.3 - 0.25 = 0.05\)

\(SD(X) = \sqrt{0.05} \approx 0.224\)

  1. Exactly equally likely! The PDF \(f(x) = 6x(1-x)\) is symmetric about \(x = 0.5\), and we showed \(P(X < 0.5) = 0.50\). So \(P(X > 0.5) = 0.50\) as well.

Problem 4: Marksmanship Score

A cadet’s marksmanship score on a standardized assessment follows a continuous distribution with PDF:

\[f(x) = \begin{cases} \frac{1}{40} & 60 \leq x \leq 100 \\ 0 & \text{otherwise} \end{cases}\]

Let \(X\) = the cadet’s score.

  1. Verify this is a valid PDF.

  2. What is \(P(X \geq 90)\) — the probability of scoring 90 or above?

  3. What is the probability the score is between 70 and 85?

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

  5. A score of 80 or above is “Expert.” What percentage of cadets qualify as Expert?

  6. Given that a cadet scored at least 75, what is the probability they scored at least 90? (Hint: use conditional probability with the CDF)

  1. \(\int_{60}^{100} \frac{1}{40}\,dx = \frac{x}{40}\Big|_{60}^{100} = \frac{100}{40} - \frac{60}{40} = \frac{40}{40} = 1\) ✓

  2. \(P(X \geq 90) = \int_{90}^{100} \frac{1}{40}\,dx = \frac{10}{40} = 0.25\)

  3. \(P(70 \leq X \leq 85) = \int_{70}^{85} \frac{1}{40}\,dx = \frac{15}{40} = 0.375\)

  4. \(E(X) = \int_{60}^{100} x \cdot \frac{1}{40}\,dx = \frac{x^2}{80}\Big|_{60}^{100} = \frac{10000}{80} - \frac{3600}{80} = \frac{6400}{80} = 80\)

\(E(X^2) = \int_{60}^{100} x^2 \cdot \frac{1}{40}\,dx = \frac{x^3}{120}\Big|_{60}^{100} = \frac{1000000}{120} - \frac{216000}{120} = \frac{784000}{120} \approx 6533.33\)

\(Var(X) = 6533.33 - 80^2 = 6533.33 - 6400 = 133.33\)

\(SD(X) = \sqrt{133.33} \approx 11.55\)

  1. \(P(X \geq 80) = \int_{80}^{100} \frac{1}{40}\,dx = \frac{20}{40} = 0.50\) — 50% qualify as Expert.

  2. \(P(X \geq 90 \mid X \geq 75) = \frac{P(X \geq 90)}{P(X \geq 75)} = \frac{10/40}{25/40} = \frac{10}{25} = 0.40\)

Before You Leave

Today

  • Continuous RVs have probabilities given by areas under the PDF, not by the PMF
  • PDF: \(f(x) \geq 0\) and \(\int f(x)\,dx = 1\) — but \(f(x)\) is density, not probability
  • CDF: \(F(x) = P(X \leq x)\) — use it to compute \(P(a \leq X \leq b) = F(b) - F(a)\)
  • \(P(X = c) = 0\) for continuous RVs — endpoints don’t matter
  • \(E(X)\) and \(Var(X)\) use integrals instead of sums, but the formulas are analogous

Any questions?

Next Lesson

Lesson 13: Normal Distribution

  • Standardize and use normal probabilities
  • Find normal quantiles for given tail areas
  • Assess plausibility using normal models

Upcoming Graded Events

  • WebAssign 4.1–4.2 - Due before Lesson 13
  • Lesson 15 - Review (Lessons 1–14)
  • WPR I - Lesson 16