Triola Elementary Statistics · 11th Edition
Chapter 5: Discrete Probability Distributions
Every key formula from Triola Elementary Statistics Chapter 5, in one searchable page. Click a card to study it — worked examples included.
5.2 · RANDOM VARIABLES
Random variable x
A numerical value from a random procedure. Discrete x: countable values (number of girls in 5 births). Each value has a probability P(x) — listed in a table or drawn as a probability histogram.
Requirements for a probability distribution
∑P(x) = 1 · 0 ≤ P(x) ≤ 1 for every x
Sums like 0.999 or 1.001 are OK — rounding error, not failure.
KEY NOTES
- In a probability histogram, each bar has width 1 and area = P(x) — so area ↔ probability.
5.2 · MEAN, VARIANCE & SD
Mean (also the expected value E)
μ = ∑[x · P(x)]
Weight each value by its probability — this is a weighted average.
Variance
σ² = ∑[(x − μ)² · P(x)] · σ² = ∑[x² · P(x)] − μ² (shortcut)
Standard deviation
σ = √∑[x² · P(x)] − μ²
WORKED EXAMPLE
x: 0, 1, 2 P(x): 0.25, 0.50, 0.25
μ = 0 + 0.50 + 0.50 = 1.0.
σ² = (0−1)²(.25) + 0 + (2−1)²(.25) = 0.50 ⇒ σ = 0.71.
μ = 0 + 0.50 + 0.50 = 1.0.
σ² = (0−1)²(.25) + 0 + (2−1)²(.25) = 0.50 ⇒ σ = 0.71.
5.2 · UNUSUAL RESULTS
Range rule of thumb
Usual: within μ ± 2σ · Unusual: outside μ ± 2σ
Pea pods: μ = 3.8, σ = 1.0 ⇒ usual = 1.8 to 5.8; 1 green pod in 5 ⇒ unusual.
Rare event rule (probabilities)
Unusually high: P(x or more) ≤ 0.05
Unusually low: P(x or fewer) ≤ 0.05
Unusually low: P(x or fewer) ≤ 0.05
KEY NOTES
- The 0.05 cutoff is a convention, not a law — 0.01 also appears.
- Watch the tail: “or more” for high, “or fewer” for low — not just P(x) alone.
5.3 · BINOMIAL REQUIREMENTS
All 4 must hold
1. Fixed number of trials n. 2. Trials are independent — one outcome doesn’t change the others. 3. Two categories per trial (success/failure). 4. Constant p — the probability of success is the same every trial.
Notation
n = trials, x = # successes, p = P(success), q = 1 − p = P(failure).
KEY NOTES
- 5% guideline: if n ≤ 5% of the population, treat selections as independent even without replacement.
- Sampling without replacement from a small population breaks the independence requirement.
5.3 · BINOMIAL FORMULA
P(x) = n!(n − x)! x! · px · qn−x
Ways to choose the x successes × probability of exactly x successes and n − x failures.
WORKED EXAMPLE
5 coin tosses, p = 0.5. P(exactly 3 heads):
5C3 = 10; p³q² = 0.03125
⇒ P(3) = 10 · 0.03125 = 0.3125.
5C3 = 10; p³q² = 0.03125
⇒ P(3) = 10 · 0.03125 = 0.3125.
For large n, the nCx arithmetic is cumbersome — use Table A-1 or technology.
5.3 · BINOMIAL MEAN & SD
No tables needed!
μ = np · σ² = npq · σ = √(npq)
WORKED EXAMPLE
10 coin tosses, p = 0.5:
μ = 10 · 0.5 = 5; σ² = 10 · 0.5 · 0.5 = 2.5; σ = 1.58.
Usual range: 5 ± 2(1.58) = 1.84 to 8.16 — 0 or 1 heads in 10 is unusual.
μ = 10 · 0.5 = 5; σ² = 10 · 0.5 · 0.5 = 2.5; σ = 1.58.
Usual range: 5 ± 2(1.58) = 1.84 to 8.16 — 0 or 1 heads in 10 is unusual.
KEY NOTES
- These shortcuts apply only to binomial distributions — μ = np is not a general formula.
WATCH OUT!
Mistakes that cost points
- Check both requirements: ∑P(x) = 1 and every P(x) in [0, 1]
- E = μ — expected value IS the mean of the distribution
- Binomial: verify all 4 requirements before touching the formula
- q = 1 − p — don’t forget the failure probability
- Rare-event test uses tails: P(x or more), P(x or fewer)
- μ = np and σ = √(npq) are binomial-only shortcuts
No formulas match your search. Try a different keyword.