Section 4.5–4.6: Continuous Random Variables & the Normal Distribution
Continuous Random Variables & Probability Density Functions (PDFs)
Review from previous section 4.4
Last week focused on discrete random variables (probability mass functions, sums of probabilities).
Key bridge: many discrete models (e.g.
Binomial) can be approximated by continuous models (especially the normal) for large sample sizes.
Fundamental distinction
Continuous random variable X takes on every value in an interval of the real line.
Because values are uncountably infinite, P(X = x) = 0 for any single point; probabilities are only meaningful over intervals.
Probability Density Function (PDF) f(x)
Smooth curve drawn over the real-number axis; sometimes called frequency function or just probability distribution.
Area under the curve over an interval corresponds to probability:
P(a \le X \le b) = \int_a^b f(x)\,dxTotal area under entire curve is 1.
Graphical intuition: shade region from a to b → shaded area gives the desired probability.
The Normal Distribution $\mathcal N(\mu,\sigma^2)$
Central role
Describes numerous natural/industrial phenomena (heights, measurement error, IQ, bulb lifetime, etc.).
Serves as a limiting approximation for many discrete distributions (notably binomial via De Moivre–Laplace/Central Limit Theorem).
Forms bedrock of classical statistical inference (confidence intervals, hypothesis testing).
Shape & descriptive characteristics
Bell-shaped, symmetrical (a.k.a. mound-shaped).
Mean, median, and mode coincide at x = \mu (curve is perfectly balanced there).
Inflection points located one standard deviation on either side (at \mu \pm \sigma).
PDF formula f(x) = \frac{1}{\sigma\sqrt{2\pi}}\,e^{- \frac12\bigl(\frac{x-\mu}{\sigma}\bigr)^2}
Parameters:
\mu = mean (center)
\sigma = standard deviation (spread)
Transcendental constants e \approx 2.71828 and \pi \approx 3.14159 appear.
Memorization not required; recognition and parameter interpretation are essential.
Visual parameter effects (3 example curves)
\mu = 0,\; \sigma = 1 → standard curve (purple), widest of shown.
\mu = 3,\; \sigma = 1 → curve shifted right, same spread.
\mu = -4,\; \sigma = 0.5 → curve shifted left, narrower and taller because smaller \sigma.
Area–probability equivalence holds: P(c \le X \le d) = \int_c^d f(x)\,dx.
If calculus unfamiliar, mentally replace integral by area of shaded strip.
The Standard Normal Distribution $Z \sim \mathcal N(0,1)$
Special case with \mu = 0,\; \sigma = 1.
Variable usually denoted Z.
Historically, probabilities obtained from printed $Z$-tables:
E.g. P(0 \le Z \le 1.96) = 0.4750 from table (row 1.9, column 0.06).
Modern workflow: use calculators or software:
TI-84: 2nd → DIST (VARS) →
normalcdf(lower, upper).Example:
normalcdf(-1.26,1.26)returns 0.7923; table gives slightly less accurate 0.7922 (rounding).
Standardising a General Normal Variable
Any normal X \sim \mathcal N(\mu,\sigma^2) can be converted to Z via Z = \frac{X-\mu}{\sigma}
Called z-score; measures distance from mean in units of standard deviation.
Enables use of standard normal tables or functions.
Procedure for probability questions
Sketch bell curve; mark mean and shade desired area.
Convert interval bounds x1, x2 to z-scores:
z1 = \frac{x1-\mu}{\sigma},\quad z2 = \frac{x2-\mu}{\sigma}Use
normalcdf(z_1,z_2).Translate result back if interpreting in X context.
Calculator shortcut: TI-84 allows
normalcdf(lower,upper,μ,σ)directly, but understanding standardisation remains conceptually critical.
Worked Forward-Probability Examples
X \sim \mathcal N(5,10^2)
Find P(5 \le X \le 6.2).
z-scores: z1 = 0 (5 at mean), z2 = (6.2-5)/10 = 0.12.
normalcdf(0,0.12)→ 0.0478.
Same distribution, P(2.9 \le X \le 7.1).
z1 = (2.9-5)/10 = -0.21, z2 = +0.21.
normalcdf(-0.21,0.21)→ 0.1663.
Light-bulb lifetime
\mu = 2000 h, \sigma = 200 h (X \sim \mathcal N(2000,200^2)).
a) P(2000 \le X \le 2400).
z1 = 0, z2 = (2400-2000)/200 = 2.
Area:
normalcdf(0,2)→ 0.4772.
b) P(X < 1470).
z = (1470-2000)/200 = -2.65.
Two approaches:
Compute left-tail directly:
normalcdf(-1e99,-2.65)→ 0.0040.Compute strip to 0 then subtract from 0.5:
normalcdf(-2.65,0) = 0.4960; 0.5 - 0.4960 = 0.0040.
Interpretation: Very unlikely (0.4 %) that a bulb fails before 1470 h.
Inverse Normal Problems (Finding Quantiles)
Given a probability (area), find z such that P(Z \le z) = p.
TI-84:
invNorm(p)(orinverse normalin menu).Input = area to left of desired point.
Output = quantile (z-value).
Examples
Determine a such that P(0 \le Z \le a) = 0.1217.
Total area to left of a: 0.5 + 0.1217 = 0.6217.
invNorm(0.6217)→ a \approx 0.3099 \approx 0.31.
Find z0 with P(Z \le z0) = 0.2858.
Left area smaller than 0.5 ⇒ quantile is negative.
invNorm(0.2858)→ z_0 \approx -0.57.
Symmetric central probability: find z0 s.t. P(-z0 \le Z \le z_0) = 0.7294.
Tail area combined: 1 - 0.7294 = 0.2706; each tail 0.1353.
Focus on left tail:
invNorm(0.1353)→ -1.1.Therefore z_0 = 1.1.
Inverse for Non-standard Normals
If P(a \le X \le b) is given and X \sim \mathcal N(\mu,\sigma^2):
Convert problem to Z via standardisation.
Use
invNormto get appropriate z-quantile(s).Convert back: x = \mu + z\sigma.
Worked Example
X \sim \mathcal N(5,10^2); find a s.t. P(5 \le X \le a) = 0.1217.
Diagram: strip to right of mean with given area.
Corresponding Z problem identical to earlier 0 \le Z \le z with area 0.1217 → z = 0.3099.
Translate: a = \mu + z\sigma = 5 + 0.3099(10) \approx 8.1.
Practical, Philosophical & Cross-Lecture Connections
Practical relevance: Engineers, scientists, and quality-control analysts rely on normal probabilities to predict defect rates (e.g., bulb life example) and design tolerance intervals.
Ethical dimension: Misinterpreting probabilities (e.g., ignoring tail risk) can lead to faulty decision-making in public policy or product safety.
Link to earlier material: The binomial distribution—treated in the discrete section—converges to the normal when number of trials is large and p not extreme (De Moivre-Laplace). Understanding standardisation prepares you for the Central Limit Theorem in later chapters.
Tool mindset: While technology replaces hand-integration, conceptual mastery (areas ↔ probabilities, standardisation, tails) is vital for correct model choice and critical evaluation of software outputs.
Summary Checklist
[ ] Recognise definition and properties of a PDF for continuous variables.
[ ] Memorise that probabilities = areas under the curve.
[ ] Identify & interpret \mu and \sigma in any normal PDF.
[ ] Convert X \sim \mathcal N(\mu,\sigma^2) ↔ Z \sim \mathcal N(0,1) via Z = (X-\mu)/\sigma.
[ ] Use
normalcdffor forward probabilities,invNormfor quantiles.[ ] Always sketch curve, mark means, endpoints, and shade area—this prevents sign mistakes & mis-ordering of bounds.