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Introductory Statistics Unit 4: Random variables & distributions
A random variable assigns a number to each outcome of a chance process. Its distribution tells you the possible values and how likely each one is.
Covers: Discrete & continuous random variables, expected value, binomial distribution
Study guide
A random variable assigns a number to each outcome of a random process. Discrete random variables take countable values (number of heads) and are described by a probability distribution listing each value and its probability, with probabilities adding to 1. Continuous random variables take any value in an interval (time, height), and probabilities are areas under a density curve, so P(X = exact value) = 0.
The expected value (mean) of a discrete random variable is E(X) = Σ x·P(x), the long-run average. Its variance is Σ (x − μ)²·P(x), and the standard deviation is the square root. For linear transformations Y = a + bX, the mean becomes a + bμ and the standard deviation becomes |b|σ. For sums or differences of independent random variables, means add or subtract, but variances always add: Var(X ± Y) = Var(X) + Var(Y).
A binomial setting has four conditions (BINS): Binary outcomes (success or failure), Independent trials, a fixed Number of trials n, and the Same probability of success p each trial. Then P(X = k) = C(n, k)pᵏ(1 − p)ⁿ⁻ᵏ, with mean np and standard deviation √(np(1 − p)). Use technology (binompdf/binomcdf) for cumulative probabilities like P(X ≤ 3).
A geometric random variable counts trials until the first success, with P(X = k) = (1 − p)ᵏ⁻¹p and mean 1/p.
A geometric random variable has standard deviation √(1 − p)/p, and P(X > k) = (1 − p)ᵏ, because the first k trials must all be failures. If an observed outcome would be very unlikely, with a probability below about 5%, when an assumed probability is true, that is convincing evidence against the assumption. A likely outcome gives no such evidence.
If X and Y are independent and normal, X + Y and X − Y are also normal. After finding their mean and standard deviation, compute probabilities with z-scores. The standard deviation of a random variable describes how far outcomes typically are from the mean, over many repetitions.
Key ideas
- Discrete random variables have countable values; continuous ones take any value in an interval.
- Expected value is the long-run average.
- Binomial setting (BINS): Binary outcomes, Independent trials, fixed Number of trials, Same probability of success.
- Adding a constant shifts the mean only; multiplying by a constant scales both mean and SD.
- For independent random variables, variances add (never SDs).
Formulas and rules
Common mistakes
- Adding standard deviations instead of variances.
- Using the binomial formula when the trials aren't independent or the count isn't fixed.
Words to know
- Expected value of a discrete random variable
- E(X) = Σ x·P(x)
- Binomial conditions (BINS)
- Binary outcomes, Independent trials, fixed Number of trials, Same success probability.
- Binomial probability formula
- P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ
- Mean and SD of a binomial
- Mean = np, SD = √(np(1 − p))
- Variance of a difference
- Var(X − Y) = Var(X) + Var(Y) for independent X, Y.
- Linear transformation of a random variable
- For Y = a + bX: mean becomes a + b·mean, SD becomes |b|·SD.
- Random variable
- A numerical outcome of a random process.
- Discrete random variable
- Takes countable values, each with a probability.
- Continuous random variable
- Takes any value in an interval; probabilities are areas under a density curve.
- Standard deviation of a random variable
- √Σ (x − μ)²P(x)
- Mean of a sum or difference
- μ(X ± Y) = μX ± μY
- Geometric random variable
- Counts trials until the first success; mean 1/p.
- Binomial random variable
- Counts successes in n independent trials with the same p.
Sample practice questions
37 more questions like these, from level 1 to 5, are in the practice set.
X = 0, 1, 2 with probabilities 0.2, 0.5, 0.3. Find E(X).
- 1
- 0.9
- 1.5
- 1.1
Show the answer and steps
Answer: D) 1.1
- E(X) = Σ x·P(x).
- 0(0.2) + 1(0.5) + 2(0.3) = 1.1.
Binomial with n = 10, p = 0.3. Mean?
- 2.1
- 0.3
- 7
- 3
Show the answer and steps
Answer: D) 3
- Binomial mean = np.
- 10 × 0.3 = 3.
Same binomial. Standard deviation?
- 0.46
- ≈ 1.45
- 3
- 2.1
Show the answer and steps
Answer: B) ≈ 1.45
- SD = √(np(1 − p)).
- √(10 × 0.3 × 0.7) = √2.1 ≈ 1.45.