Probability Distributions
Interactive review for Module 2. Select a distribution to explore its properties, visualize the PDF/PMF, and solve practice problems.
Uniform
Equally likely outcomes over an interval.
Normal
The Bell Curve. Central to statistics.
Exponential
Waiting time between events.
Poisson
Count of events in a fixed interval.
CLT
Why everything becomes Normal.
Uniform Distribution \( X \sim U(a, b) \)
Visualization
Practice Problem: Line Segment Ratio
A point \( X \) is chosen uniformly on \([0, L]\). What is the probability that the ratio of the shorter to the longer segment is less than \( 1/4 \)?
Shorter:
Longer:
Ratio:
Theoretical Solution:
Condition: \( \frac{\min(X, L-X)}{\max(X, L-X)} < \frac{1}{4} \)
Implies \( X < L/5 \) or \( X > 4L/5 \).
Probability = \( \frac{L/5 + L/5}{L} = \frac{2}{5} = 0.4 \)
Normal Distribution \( X \sim N(\mu, \sigma^2) \)
Interactive Bell Curve
Calculator & Practice
Calculate \( P(X > x) \) or \( P(a < X < b) \).
Practice Context (Q6):
Package weights: \( \mu=500, \sigma=20 \).
- Try calculating \( P(X > 530) \)
- Try calculating \( P(480 < X < 520) \)
Exponential Distribution \( X \sim \text{Exp}(\lambda) \)
PDF Visualization \( \lambda e^{-\lambda x} \)
Memoryless Property
The probability of waiting \( t \) more minutes doesn't depend on how long you've already waited.
Example (Bus Arrival):
Rate \( \lambda = 0.1 \) (avg wait 10 mins).
If you've waited 5 mins, what's the prob of waiting 10 more?
P(X > 15 | X > 5) = P(X > 10)
\( e^{-0.1(10)} = e^{-1} \approx 0.368 \)
Calculate Probability:
Poisson Distribution \( X \sim \text{Pois}(\lambda) \)
PMF Visualization
Practice: Wrong Calls
Average 3 wrong calls per day (\( \lambda = 3 \)).
Probability of exactly \( k \) calls:
Conditional Probability Problem:
Today, 1 wrong call received. Probability of at least 2 more?
We want \( P(X \ge 3 | X \ge 1) \).
\( = \frac{P(X \ge 3)}{P(X \ge 1)} = \frac{1 - P(0) - P(1) - P(2)}{1 - P(0)} \)
\( P(0) = e^{-3} \approx 0.0498 \)
\( P(1) = 3e^{-3} \approx 0.1494 \)
\( P(2) = 4.5e^{-3} \approx 0.2240 \)
Num: \( 1 - 8.5e^{-3} \approx 0.5768 \)
Denom: \( 1 - e^{-3} \approx 0.9502 \)
Result: \( \approx 0.607 \)
Central Limit Theorem
Simulation Setup
Sum of \( n \) independent dice rolls.
Number of dice rolled per trial.
Theory: As \( n \) increases, the distribution of the sum approaches Normal.
Mean \( \approx n \times 3.5 \)
Var \( \approx n \times 2.92 \)