Source: STAT 350 Exam 1, Purdue University
Tags: random variable, binomial, Poisson, PDF, CDF, expected value, BINS conditions, piecewise PDF, exponential, uniform, continuous distribution, discrete distribution
Difficulty: Intermediate | Prerequisites: Basic probability, integration for continuous distributions, summation notation.
Random variables translate real-world outcomes into numbers you can compute with. This set of notes covers the conditions that define specific named distributions (Binomial, Poisson, Exponential, Uniform), how to verify whether a function qualifies as a valid PDF, and the key calculations: expected value, CDF construction, and conditional distributions built from simpler pieces. These topics connect everything else in the course: the normal distribution is one special continuous distribution, Bayes' theorem applies to conditional random variables, and the IQR calculations from descriptive statistics carry over to any named distribution. You should already know what a probability mass function and probability density function are.
Named distributions (Binomial, Poisson, Exponential, Uniform) each have specific conditions that must be met before you can use their formulas. For continuous distributions, the PDF must integrate to 1, and the CDF is the running integral of the PDF. Expected value is the probability-weighted average, computed as a sum for discrete variables and an integral for continuous ones.
Random variable
A function that assigns a numerical value to each outcome in a sample space. In simple terms, it is the number you measure or count in an experiment.
Probability density function (PDF)
For a continuous random variable, f(x) gives the density of probability at each point. The probability of any interval is the area under the curve over that interval. Two validity requirements: f(x) ≥ 0 everywhere, and the total integral equals 1.
Cumulative distribution function (CDF)
F(x) = P(X ≤ x). For continuous variables, it is the integral of the PDF from -∞ to x. The CDF is non-decreasing, starts at 0, and ends at 1.
Expected value (mean)
E[X] = Σ x · P(X = x) for discrete variables, or ∫ x · f(x) dx for continuous variables. Think of it as the long-run average if you repeated the experiment many times.
Binomial distribution
Models the count of successes in n independent trials, each with the same probability p of success. Written X ~ Binomial(n, p). The BINS conditions are: Binary outcomes, Independent trials, fixed Number of trials, Same probability of success on each trial.
Poisson distribution
Models the count of events in a fixed interval when events occur independently at a constant average rate λ. P(X = k) = e^(-λ) · λᵏ / k!. In simple terms, it counts rare, random arrivals.
Exponential distribution
Models the waiting time until the next event in a Poisson process. If X ~ Exponential(λ), then E[X] = 1/λ, and the PDF is f(x) = λe^(-λx) for x ≥ 0.
Uniform distribution
All values in an interval [a, b] are equally likely. f(x) = 1/(b - a) for a ≤ x ≤ b. E[X] = (a + b)/2. IQR = (b - a)/2.
The Binomial requires all four BINS conditions:
Binary: each trial has exactly two outcomes (success or failure).
Independent: the outcome of one trial does not affect another.
Number: the total number of trials n is fixed in advance.
Same probability: every trial has the same success probability p.
When any condition breaks, the Binomial does not apply.
Sampling without replacement from a small population violates independence (and same-probability), so you get a Hypergeometric distribution instead. Exam reference (Q1.6): selecting 20 from 60 students without replacement means X is not Binomial. Answer is false.
Different success probabilities across participants also violates the "S" condition. Exam reference (Q2.3): 4 participants with p = 0.3 and 6 with p = 0.7 means the probability of success is not the same for all trials. Answer is D.
When Y is defined conditionally on X, compute P(Y = y) by summing over all possible values of X:
P(Y = y) = Σ P(Y = y | X = x) · P(X = x)
Exam example (Q2.5): X ~ Binomial(2, p), with Y defined conditionally:
P(Y = 0 | X = 0) = 1
P(Y = 0 | X = 1) = 0 (since Y is either 1 or 2 when X = 1)
P(Y = 0 | X = 2) = e^(-λ) (from Poisson(λ))
So P(Y = 0) = 1 · (1-p)² + 0 · 2p(1-p) + e^(-λ) · p² = (1-p)² + p²e^(-λ). Answer is A.
To confirm f(x) is a valid PDF:
Check f(x) ≥ 0 on its support.
Integrate over the full support and set the result equal to 1. Solve for any unknown constant.
Exam example (Q2.4): f(x) = kx² for 0 ≤ x ≤ 3.
∫₀³ kx² dx = k · [x³/3]₀³ = k · 9 = 1, so k = 1/9.
E[X] = ∫₀³ x · (x²/9) dx = (1/9) · [x⁴/4]₀³ = (1/9)(81/4) = 9/4 = 2.25 years. Answer is C.
A PDF can be defined in pieces over different regions.
Exam example (Problem 5): Ticket resolution time T has:
f(t) = k for 0 ≤ t ≤ 10 (uniform piece)
f(t) = ke^(-(t-10)) for t > 10 (exponential decay piece)
f(t) = 0 otherwise
Finding k:
∫₀¹⁰ k dt + ∫₁₀^∞ ke^(-(t-10)) dt = 10k + k · 1 = 11k = 1
k = 1/11
The CDF is built piece by piece:
F(t) = 0 for t < 0
F(t) = t/11 for 0 ≤ t ≤ 10
F(t) = 1 - (1/11)e^(-(t-10)) for t > 10
P(5 < T < 15) = F(15) - F(5):
F(15) = 1 - (1/11)e^(-5)
F(5) = 5/11
P(5 < T < 15) = 1 - (1/11)e^(-5) - 5/11 ≈ 0.5448
P(T ≥ 15 | T ≥ 5) = P(T ≥ 15) / P(T ≥ 5) = [1 - F(15)] / [1 - F(5)]:
1 - F(15) = (1/11)e^(-5)
1 - F(5) = 6/11
Result: e^(-5) / 6 ≈ 0.0011
To find t* such that P(T > t*) = 0.05:
First check which piece of the CDF applies. Since F(10) = 10/11 ≈ 0.909 < 0.95, the 95th percentile falls in the exponential tail.
1 - F(t*) = 0.05, so (1/11)e^(-(t*-10)) = 0.05
e^(-(t*-10)) = 0.55
t* = 10 - ln(0.55) ≈ 10.5978
This topic connects distributions to descriptive statistics:
Exponential(λ): IQR = (ln 3)/λ. Since E[X] = 1/λ, IQR = ln(3) · E[X] ≈ 1.0986 · E[X]. The IQR is a constant multiple of the mean. (Q3d, option A: correct.)
Normal(μ, σ²): IQR ≈ 1.34σ. It depends only on σ, not on μ. Knowing the mean alone tells you nothing about the IQR. (Q3d, option B: correct.)
Uniform(0, b): IQR = b/2. Since E[X] = b/2, IQR equals the mean. (Q3d, option C: correct.)
Uniform(a, b): IQR = (b - a)/2. The mean is (a + b)/2. Knowing only the mean does not determine (b - a), so the IQR is not determined by the mean alone. (Q3d, option D: incorrect. This is the answer.)
Item | Formula |
|---|---|
PDF validity | f(x) ≥ 0 and ∫ f(x) dx = 1 |
CDF from PDF | F(x) = ∫₋∞ˣ f(t) dt |
E[X] (continuous) | ∫ x · f(x) dx |
E[X] (discrete) | Σ x · P(X = x) |
Binomial PMF | P(X = k) = C(n,k) · pᵏ · (1-p)ⁿ⁻ᵏ |
Poisson PMF | P(X = k) = e^(-λ) · λᵏ / k! |
Exponential PDF | f(x) = λe^(-λx), x ≥ 0 |
Uniform PDF | f(x) = 1/(b-a), a ≤ x ≤ b |
The piecewise PDF in Problem 5 models an IT help-desk queue: simple tickets clear quickly (uniform phase), but tickets that survive past a threshold enter an exponential-decay phase where some drag on for a long time. This kind of model is common in operations research and service-system design, where managers need to estimate staffing levels and set resolution-time targets (like the 95th percentile t*).
Students apply the Binomial formula whenever they see "count the number of successes" without checking all four BINS conditions. Sampling without replacement from a small finite population is the most common violation tested.
Forgetting to check that f(x) ≥ 0 before solving for k. The constant k must be positive if the PDF expression could go negative.
When computing E[X] from a PDF like kx², students sometimes forget to multiply by x inside the integral. E[X] = ∫ x · f(x) dx, not ∫ f(x) dx (that just equals 1).
For piecewise CDFs, students sometimes apply the wrong piece when evaluating F at a boundary or in the tail. Always check which region your value falls in.
⚠️ Q1.6 (2 points): Sampling without replacement from 60 students violates independence, so X is not Binomial. Answer: false.
⚠️ Q2.3 (3 points): Non-constant success probability means BINS fails at "S." Answer: D.
⚠️ Q2.4 (3 points): Computing E[X] from a polynomial PDF requires finding k first, then integrating x · f(x). Answer: 2.25.
⚠️ Q2.5 (3 points): Conditional distribution problem using the law of total probability across Binomial outcomes.
⚠️ Problem 5 (28 points): Full piecewise PDF problem. Finding k, using the CDF for interval and conditional probabilities, and solving a backward problem for a percentile. Show every integration step.
⚠️ Q3d (3 points): IQR-mean relationship across Exponential, Normal, and Uniform families. Answer: D (Uniform on (a,b), IQR is not determined by the mean alone).
True or false: A PDF can take values greater than 1.
Fill in the blank: The four conditions for a Binomial distribution are B, I, N, and ___.
True or false: For a Poisson random variable, P(X = 0) = e^(-λ).
Fill in the blank: For Uniform(0, 10), the IQR is ___.
True or false: If f(x) = 2x for 0 ≤ x ≤ 1, then E[X] = 1.
Answers: 1. True (it is the integral that must equal 1, not f(x) itself). 2. S (Same probability). 3. True. 4. 5. 5. False (E[X] = ∫₀¹ x · 2x dx = 2/3).
Q: A PDF is f(x) = cx³ for 0 ≤ x ≤ 2. Find c and E[X].
A: ∫₀² cx³ dx = c · [x⁴/4]₀² = c · 4 = 1, so c = 1/4. E[X] = ∫₀² x · (x³/4) dx = (1/4)[x⁵/5]₀² = (1/4)(32/5) = 8/5 = 1.6.
Q: Why is sampling 20 items without replacement from a population of 60 not Binomial?
A: Without replacement, the probability of success changes from draw to draw (and trials are dependent), violating both the "I" and "S" conditions. The correct model is Hypergeometric.
Q: For the piecewise PDF f(t) = 1/11 for 0 ≤ t ≤ 10 and f(t) = (1/11)e^(-(t-10)) for t > 10, find P(T > 10).
A: P(T > 10) = 1 - F(10) = 1 - 10/11 = 1/11 ≈ 0.0909.
Q: X ~ Binomial(2, p). Write out P(X = 0), P(X = 1), and P(X = 2).
A: P(X = 0) = (1-p)². P(X = 1) = 2p(1-p). P(X = 2) = p².
The piecewise PDF in Problem 5 combines a Uniform segment with an Exponential segment, so familiarity with both named distributions is required. The conditional distribution in Q2.5 uses Binomial and Poisson together through the law of total probability, which ties back to the Bayes' theorem notes. Computing the IQR for each distribution family connects to the normal distribution and descriptive statistics notes. Later in the course, the Binomial distribution will reappear in the context of proportions and hypothesis testing.
random variable, discrete, continuous, PDF, PMF, CDF, expected value, mean, Binomial, BINS, Poisson, Exponential, Uniform, piecewise PDF, normalizing constant, Hypergeometric, sampling without replacement, IQR by distribution, STAT 350 Purdue, Exam 1