Probability for MATH 141
Exam Relevance for MATH 141
PDF problems appear occasionally in MATH 141. Basic setup: area under curve = 1, integrate for P(a<=X<=b).
Understanding Probability with Integration
You've seen how integrals can compute areas, volumes, and averages. Now we'll use integration for probability — answering questions like "What's the chance that a random value falls between 2 and 5?"
In calculus-based probability, we work with continuous random variables — quantities that can take any value in a range (like heights, waiting times, or temperatures), not just whole numbers.
Probability Density Functions (PDFs)
A probability density function $f(x)$ describes how probability is distributed across values. It must satisfy two conditions:
- $f(x) \geq 0$ for all $x$ (probabilities can't be negative)
- $\int_{-\infty}^{\infty} f(x) \, dx = 1$ (total probability is 100%)
Key insight: The PDF itself is NOT a probability. The area under the curve gives probability!
Finding Probability
The probability that a random variable $X$ falls between $a$ and $b$ is:
$$P(a \leq X \leq b) = \int_a^b f(x) \, dx$$
This is just the area under the PDF curve from $a$ to $b$.
Problem: A random variable $X$ has PDF $f(x) = \frac{3}{8}x^2$ for $0 \leq x \leq 2$ (and $f(x) = 0$ elsewhere). Find $P(1 \leq X \leq 2)$.
$$P(1 \leq X \leq 2) = \int_1^2 \frac{3}{8}x^2 \, dx$$
$$= \frac{3}{8} \left[\frac{x^3}{3}\right]_1^2$$
$$= \frac{3}{8} \left(\frac{8}{3} - \frac{1}{3}\right)$$
$$= \frac{3}{8} \cdot \frac{7}{3} = \frac{7}{8}$$
$$\boxed{P(1 \leq X \leq 2) = \frac{7}{8} = 0.875}$$
What this means: There's an 87.5% chance that the random variable $X$ takes a value between 1 and 2. The area under the PDF curve from $x = 1$ to $x = 2$ represents this probability.
Problem: Verify that $f(x) = 3x^2$ for $0 \leq x \leq 1$ is a valid PDF.
Check condition 1: $f(x) = 3x^2 \geq 0$ for all $x$ ✓
Check condition 2: Total probability = 1?
$$\int_0^1 3x^2 \, dx = \left[x^3\right]_0^1 = 1 - 0 = 1 \checkmark$$
$$\boxed{\text{Yes, } f(x) = 3x^2 \text{ is a valid PDF on } [0, 1]}$$
Problem: Find the value of $c$ so that $f(x) = cx$ is a valid PDF on $[0, 4]$.
For a valid PDF, we need $\int_0^4 cx \, dx = 1$.
$$c \int_0^4 x \, dx = 1$$
$$c \left[\frac{x^2}{2}\right]_0^4 = 1$$
$$c \cdot 8 = 1$$
$$\boxed{c = \frac{1}{8}}$$
Expected Value (Mean)
The expected value (or mean) of a continuous random variable is:
$$\mu = E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx$$
This is the "weighted average" of all possible values, weighted by their probabilities. Think of it as the "balance point" of the PDF.
Problem: Find the expected value of $X$ if $f(x) = 3x^2$ on $[0, 1]$.
$$E(X) = \int_0^1 x \cdot 3x^2 \, dx = \int_0^1 3x^3 \, dx$$
$$= 3 \left[\frac{x^4}{4}\right]_0^1 = \frac{3}{4}$$
$$\boxed{E(X) = \frac{3}{4} = 0.75}$$
The expected value is closer to 1 than to 0 because the PDF $3x^2$ gives more weight to larger values.
Problem: If $f(x) = \frac{1}{8}x$ on $[0, 4]$, find $E(X)$.
$$E(X) = \int_0^4 x \cdot \frac{1}{8}x \, dx = \frac{1}{8}\int_0^4 x^2 \, dx$$
$$= \frac{1}{8} \left[\frac{x^3}{3}\right]_0^4 = \frac{1}{8} \cdot \frac{64}{3} = \frac{8}{3}$$
$$\boxed{E(X) = \frac{8}{3} \approx 2.67}$$
Variance and Standard Deviation
The variance measures how spread out the distribution is:
$$\text{Var}(X) = E(X^2) - [E(X)]^2$$
where $E(X^2) = \int_{-\infty}^{\infty} x^2 \cdot f(x) \, dx$
The standard deviation is $\sigma = \sqrt{\text{Var}(X)}$.
What Does Variance Mean Intuitively?
Low variance = values cluster tightly around the mean.
Example: Heights of adult men in the US have a mean around 5'9" with standard deviation ~3 inches. Most men are between 5'6" and 6'0". The distribution is "tight."
High variance = values are spread out widely.
Example: Annual incomes have a mean around \$60,000, but the standard deviation is huge (~\$50,000+). Some people earn \$20,000, others earn \$500,000. The distribution is "spread out."
Problem: Find the variance of $X$ if $f(x) = 3x^2$ on $[0, 1]$.
Step 1: Find $E(X^2)$
$$E(X^2) = \int_0^1 x^2 \cdot 3x^2 \, dx = \int_0^1 3x^4 \, dx$$
$$= 3 \left[\frac{x^5}{5}\right]_0^1 = \frac{3}{5}$$
Step 2: Calculate variance
From Example 4, $E(X) = \frac{3}{4}$.
$$\text{Var}(X) = E(X^2) - [E(X)]^2$$
$$= \frac{3}{5} - \left(\frac{3}{4}\right)^2 = \frac{3}{5} - \frac{9}{16}$$
$$= \frac{48}{80} - \frac{45}{80} = \frac{3}{80}$$
$$\boxed{\text{Var}(X) = \frac{3}{80} = 0.0375}$$
This is a small variance — values tend to cluster near the mean of 0.75.
The Exponential Distribution
A common PDF is the exponential distribution:
$$f(x) = \lambda e^{-\lambda x} \quad \text{for } x \geq 0$$
This models waiting times (time until the next event).
- Mean: $E(X) = \frac{1}{\lambda}$
- Variance: $\text{Var}(X) = \frac{1}{\lambda^2}$
Problem: The lifetime of a battery (in years) follows an exponential distribution with $\lambda = 0.5$. Find the probability that a battery lasts more than 3 years.
$$P(X > 3) = \int_3^{\infty} 0.5 e^{-0.5x} \, dx$$
$$= \left[-e^{-0.5x}\right]_3^{\infty}$$
$$= 0 - (-e^{-1.5}) = e^{-1.5}$$
$$\boxed{P(X > 3) = e^{-1.5} \approx 0.223}$$
About 22.3% of batteries last more than 3 years.
Common Mistakes and Misunderstandings
❌ Mistake: Thinking $f(x)$ is a probability
Wrong: "The probability at $x = 2$ is $f(2) = 0.3$."
Why it's wrong: For continuous distributions, the probability of any exact value is zero! The PDF gives probability density, not probability.
Correct: $P(a \leq X \leq b) = \int_a^b f(x) \, dx$. Only the area gives probability.
❌ Mistake: Forgetting that total probability must equal 1
Wrong: Using $f(x) = x^2$ on $[0, 2]$ without checking if it integrates to 1.
Why it's wrong: $\int_0^2 x^2 \, dx = \frac{8}{3} \neq 1$, so this isn't a valid PDF!
Correct: Always verify $\int f(x) \, dx = 1$, or find the normalizing constant $c$.
❌ Mistake: Using wrong limits for expected value
Wrong: $E(X) = \int_0^{\infty} x \cdot f(x) \, dx$ when the PDF is only defined on $[0, 1]$.
Why it's wrong: You must integrate over the support of the PDF (where $f(x) \neq 0$).
Correct: If $f(x)$ is defined on $[0, 1]$, then $E(X) = \int_0^1 x \cdot f(x) \, dx$.
❌ Mistake: Confusing $E(X^2)$ with $[E(X)]^2$
Wrong: Using $\text{Var}(X) = E(X^2) - E(X)^2$ but calculating $E(X^2)$ as $(E(X))^2$.
Why it's wrong: $E(X^2) = \int x^2 f(x) \, dx$ is NOT the same as squaring $E(X)$!
Correct: Calculate $E(X^2)$ and $E(X)$ separately, then use $\text{Var}(X) = E(X^2) - [E(X)]^2$.
Probability from PDF
The probability that X falls between a and b is the area under the PDF curve from a to b.
Variables:
- $f(x)$:
- probability density function
- $a, b$:
- bounds of the interval
- $P$:
- probability (between 0 and 1)
Expected Value (Mean)
The expected value is the weighted average of all possible values, weighted by their probability densities.
Variables:
- $E(X)$:
- expected value (mean) of X
- $f(x)$:
- probability density function
- $x$:
- the random variable
Variance
Variance measures how spread out the distribution is. Calculate E(X²) and E(X) separately.
Variables:
- $Var(X)$:
- variance of X
- $E(X²)$:
- expected value of X² (integrate x² · f(x))
- $E(X)$:
- expected value of X
PDF Conditions
A valid probability density function must be non-negative and integrate to 1 (total probability = 100%).
Variables:
- $f(x)$:
- probability density function
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