Triola Elementary Statistics · 11th Edition
Chapter 6: Normal Probability Distributions
Every key formula from Triola Elementary Statistics Chapter 6, in one searchable page. Click a card to study it — worked examples included.
6.2 · THE STANDARD NORMAL
μ = 0, σ = 1; total area under the curve = 1
Table A-2 gives the cumulative area from the LEFT.
Notation
P(Z < a) = area left of a · P(a < Z < b) = area between
P(Z > a) = 1 − P(Z < a) · P(Z = a) = 0
P(Z > a) = 1 − P(Z < a) · P(Z = a) = 0
Continuous: endpoints never matter. Symmetric: P(Z > −a) = P(Z < a).
68–95–99.7 rule (§3-3)
68% within 1σ · 95% within 2σ · 99.7% within 3σ
WORKED EXAMPLE
Find P(Z < 1.27): Table A-2, row 1.2 + col 0.07 ⇒
P(Z < 1.27) = 0.8980.
P(Z < 1.27) = 0.8980.
6.3 · Z-SCORES: STANDARDIZING
z = x − μσ (Formula 6-2; round z to 2 decimals)
x = μ + zσ (inverse form)
KEY NOTES
- A z score = how many σs a value sits above (+) or below (−) the mean.
Procedure: x value ⇒ probability
Sketch the curve and shade → convert each boundary x to z → Table A-2 for the area.
WORKED EXAMPLE
Men’s heights: μ = 69.0 in, σ = 2.8 in. For x = 80:
z = (80 − 69.0)/2.8 = 3.93 (extremely tall).
z = (80 − 69.0)/2.8 = 3.93 (extremely tall).
Unusual x: |z| > 2 is a flag (< 5% of values).
6.3 · FINDING z FROM AREA
Inverse procedure
Sketch, shade the known area → find that area in the body of Table A-2 → read z from the margins.
zα notation
zα = z score with area α to its right; z0.025 = 1.96
Common critical values
Middle 95%: ±1.96 · 95th percentile: z = 1.645
For 95% confidence look up 0.9750 in the body, not 0.95.
WORKED EXAMPLE
Find the 95th percentile: area 0.95 in the body of Table A-2 ⇒ closest is 0.9500 ⇒
z = 1.645.
z = 1.645.
μx̄ = μ, σx̄ = σ√n (standard error of the mean)
The distribution of x̄ is normal if n > 30 or the population is normal.
z = x̄ − μσ/√n
Individual value: use σ. Sample mean: use σ/√n.
WORKED EXAMPLE
Men’s weights: μ = 172 lb, σ = 29 lb. P(x̄ > 175), n = 20:
σx̄ = 29/√20 = 6.48; z = (175 − 172)/6.48 = 0.46;
P = 1 − 0.6772 = 0.3228 (0.3218 by technology).
σx̄ = 29/√20 = 6.48; z = (175 − 172)/6.48 = 0.46;
P = 1 − 0.6772 = 0.3228 (0.3218 by technology).
KEY NOTES
- Sampling without replacement and n > 5% of N? Multiply σx̄ by √((N−n)/(N−1)).
6.6 · NORMAL APPROX.
Requirements
np ≥ 5 and nq ≥ 5
Normal parameters
μ = np, σ = √(npq)
Continuity correction
Whole x → interval [x − 0.5, x + 0.5]
“At least 1150” starts at 1149.5.
WORKED EXAMPLE
Survey: n = 40,000, p = 0.03; P(x ≥ 1150):
μ = 1200, σ = 34.12; z = (1149.5 − 1200)/34.12 = −1.48;
P = 1 − 0.0694 = 0.9306.
μ = 1200, σ = 34.12; z = (1149.5 − 1200)/34.12 = −1.48;
P = 1 − 0.0694 = 0.9306.
6.7 · ASSESSING NORMALITY
Three checks
- Histogram: reject normality if it departs dramatically from a bell shape
- Outliers: reject normality if there is more than one outlier
- Normal quantile plot: normal if points lie reasonably close to a straight line with no other systematic pattern
A normal quantile plot graphs (x, z): each x from the data vs the z score expected from the standard normal. Use the criteria loosely for small samples, strictly for large ones.
Mistakes that cost points
- z scores are on the axes; areas are in the table body — don’t swap them
- Sample mean? Divide σ by √n — the #1 error in §6.5
- CLT needs n > 30 or a normal population
- Continuity correction: “at least x” starts at x − 0.5
- Binomial approx needs np ≥ 5 and nq ≥ 5