Gaussian - Normal distribution properties: Worked examples

Before anything, one reminder of the three tools we will lean on, in plain words:

Recall The three moves (from the parent note)

Standardize ::: turn any $X \sim N (μ, σ^{2})$ into $Z = \frac{X - μ}{σ} \sim N (0, 1)$ so we can read one universal table. $Φ (z)$ ::: the cumulative function — the probability that a standard normal lands at or below $z$ , i.e. the area under the bell to the left of $z$ . Add / scale ::: means always add; variances add only when independent; scaling by $a$ multiplies variance by $a^{2}$ .

We use the symbol $Φ$ a lot, so pin it down with a picture.

Figure — Gaussian - Normal distribution properties

Look at the figure: the red shaded region is $Φ (z)$ . The whole area under the curve is exactly $1$ (it is a probability), so whatever is not red is the right tail $1 - Φ (z)$ . The dashed mirror line at $- z$ shows why $Φ (- z) = 1 - Φ (z)$ : the bell is a perfect mirror about $0$ , so the area left of $- z$ equals the area right of $+ z$ .

The scenario matrix

Every Gaussian exam question falls into one of these cells. Each worked example below is tagged with the cell it covers.

#	Case class	What makes it tricky	Example
A	Right tail, positive $z$	reading $1 - Φ$	Ex 1
B	Left tail / negative $z$	symmetry $Φ (- z) = 1 - Φ (z)$	Ex 2
C	Between two values	subtract two $Φ$ 's	Ex 3
D	Sum & average of Gaussians	variances add, then $/ n^{2}$	Ex 4
E	Inverse problem (given probability, find $x$ )	run standardize backwards	Ex 5
F	Degenerate $σ \to 0$	the spike / limit case	Ex 6
G	Real-world word problem	translate English → symbols	Ex 7
H	Exam twist: difference $X - Y$ & MLE from data	sign inside variance, biased vs unbiased	Ex 8

We will lean on a few landmark values of $Φ$ throughout (each is just the shaded area for that $z$ ):

$Φ (1) \approx 0.8413, Φ (2) \approx 0.9772, Φ (1.5) \approx 0.9332, Φ (1.28) \approx 0.8997.$

Cell A — Right tail, positive z

Worked example Example 1 — IQ above 130

IQ scores are $X \sim N (100, 1 5^{2})$ . Find $P (X > 130)$ .

Forecast: guess first — is it closer to $2%$ , $16%$ , or $50%$ ? (130 is two standard deviations above the mean, so it should be a small tail.)

Step 1. Standardize: $z = \frac{130 - 100}{15} = 2$ . Why this step? The table only knows $N (0, 1)$ ; we must convert "130 in IQ units" into "how many $σ$ above the mean".

Step 2. We want the right tail: $P (X > 130) = P (Z > 2) = 1 - Φ (2)$ . Why this step? $Φ$ gives area to the left; "above 130" is area to the right, so subtract from 1.

Step 3. $1 - Φ (2) = 1 - 0.9772 = 0.0228$ . Why this step? Plug the landmark value in.

Answer: $\approx 0.0228$ , i.e. 2.28% (about 1 in 44).

Verify: 130 is $2 σ$ out. The 68-95-99.7 rule says $95.4%$ lies within $\pm 2 σ$ , leaving $4.6%$ in the two tails, so one tail is $2.3%$ . Matches. ✓

Cell B — Negative z, left tail (symmetry)

Cell C — Between two values

Cell D — Sum and average

Worked example Example 4 — Averaging 5 measurements

A distance is measured 5 times, each $X_{i} \sim N (10, 0. 5^{2})$ metres, independent. Find the distribution of the average $\overset{ˉ}{X}$ and $P (\overset{ˉ}{X} > 10.4)$ .

Forecast: the average of 5 noisy readings — should its spread be bigger or smaller than a single reading's?

Step 1. Sum: $S = \sum_{i = 1}^{5} X_{i} \sim N (5 \cdot 10, 5 \cdot 0. 5^{2}) = N (50, 1.25)$ . Why this step? Means add; independent variances add. Five readings each var $0.25$ give $1.25$ .

Step 2. Average $\overset{ˉ}{X} = \frac{S}{5}$ . Scaling by $a = \frac{1}{5}$ : mean $\to \frac{50}{5} = 10$ , variance $\to (\frac{1}{5})^{2} \cdot 1.25 = \frac{1.25}{25} = 0.05$ . Why this step? $Var (a X) = a^{2} Var (X)$ — the square is why averaging shrinks spread.

So $\overset{ˉ}{X} \sim N (10, 0.05)$ , giving $σ_{\overset{ˉ}{X}} = 0.05 \approx 0.2236$ .

Step 3. $z = \frac{10.4 - 10}{0.2236} \approx 1.7889$ . Then $P (\overset{ˉ}{X} > 10.4) = 1 - Φ (1.79) \approx 1 - 0.9633 = 0.0367$ .

Answer: $\overset{ˉ}{X} \sim N (10, 0.05)$ ; $P (\overset{ˉ}{X} > 10.4) \approx 0.0367$ (3.67%).

Verify: A single reading has $σ = 0.5$ ; the average has $σ \approx 0.224$ , smaller by a factor $5 \approx 2.236$ . Averaging does tighten spread. ✓ (See Central Limit Theorem.)

Cell E — Inverse problem (find x from probability)

Worked example Example 5 — Cutoff for the top 10%

Same IQ $X \sim N (100, 1 5^{2})$ . Above what IQ score are the top 10% of people?

Forecast: the cutoff must be above 100. Roughly how many $σ$ up?

Step 1. We need $c$ with $P (X > c) = 0.10$ , i.e. $P (X \leq c) = 0.90$ , i.e. $Φ (z) = 0.90$ . Why this step? This runs the standardize move backwards: we know the area, we want the boundary.

Step 2. Read the table in reverse: $Φ (z) = 0.90 \Rightarrow z \approx 1.2816$ . Why this step? Finding $z$ from a target area is the inverse-CDF; the landmark $Φ (1.28) \approx 0.90$ pins it.

Step 3. Un-standardize: $c = μ + z σ = 100 + 1.2816 \cdot 15 \approx 119.22$ . Why this step? $z = \frac{c - μ}{σ}$ rearranged gives $c = μ + z σ$ .

Answer: IQ $\approx 119.2$ marks the top 10%.

Verify: Forward-check: $P (X > 119.22) = 1 - Φ (1.2816) \approx 1 - 0.90 = 0.10$ . ✓ It sits between $1 σ$ (115) and $2 σ$ (130), as it should for a 10% tail.

Cell F — Degenerate limit $σ \to 0$

Worked example Example 6 — What happens as the spread vanishes?

Consider $X \sim N (5, σ^{2})$ . Describe $X$ as $σ \to 0$ , and find $P (X \leq 5)$ and $P (4.9 \leq X \leq 5.1)$ in that limit.

Forecast: if measurements have zero noise, what value must $X$ take?

Step 1. As $σ \to 0$ the bell gets taller and narrower (red curve, figure), all mass piling onto $μ = 5$ . This is a degenerate distribution — a spike. Why this step? The PDF height is $\frac{1}{σ 2 π} \to \infty$ while width $\to 0$ ; area stays $1$ .

Step 2. $P (X \leq 5)$ : by symmetry, for any $σ > 0$ , exactly half the mass is at or below the mean, so $P (X \leq 5) = Φ (0) = 0.5$ . Why this step? $z = \frac{5 - 5}{σ} = 0$ regardless of $σ$ — the mean-point is always the median.

Step 3. $P (4.9 \leq X \leq 5.1) \to 1$ as $σ \to 0$ . Why this step? Once the spike is narrower than $[4.9, 5.1]$ , all mass is inside, so probability $\to 1$ .

Answer: limit is a point-mass at $5$ ; $P (X \leq 5) = 0.5$ ; $P (4.9 \leq X \leq 5.1) \to 1$ .

Verify: With $σ = 0.01$ , $z = \frac{0.1}{0.01} = 10$ , so $P (4.9 \leq X \leq 5.1) = Φ (10) - Φ (- 10) \approx 1 - 0 \approx 1$ . ✓ The bell has become a delta spike.

Cell G — Real-world word problem

Worked example Example 7 — Coffee machine fill line

A machine pours coffee with volume $V \sim N (250, 8^{2})$ ml. Cups hold 260 ml. What fraction of pours overflow the cup?

Forecast: the target 250 is 10 ml below the 260 brim — one-and-a-bit $σ$ . Small or large tail?

Step 1. Translate English → symbols: overflow means $V > 260$ . Why this step? Word problems are 90% translation; "overflow" is a right tail.

Step 2. Standardize: $z = \frac{260 - 250}{8} = 1.25$ . Why this step? Convert the 260 ml brim into $σ$ -units.

Step 3. $P (V > 260) = 1 - Φ (1.25) \approx 1 - 0.8944 = 0.1056$ . Why this step? Right tail = $1 - Φ$ .

Answer: about 10.6% of pours overflow.

Verify: Units: ml canceled inside $z$ (ml/ml), leaving a pure number — correct, since $z$ must be dimensionless. And 1.25σ tail ≈ 10.6% is between the $1 σ$ tail (15.9%) and $2 σ$ tail (2.3%). ✓

Cell H — Exam twist: difference of Gaussians + MLE from data

Worked example Example 8 — Two parts and a fitted model

(a) Rod A length $\sim N (20, 0. 3^{2})$ mm, rod B $\sim N (19.5, 0. 4^{2})$ mm, independent. Find $P (A > B)$ . (b) You then measure 4 gaps and get data ${2, 4, 4, 6}$ . Compute the MLE $\overset{μ}{^}$ and biased MLE $\overset{σ}{^}^{2}$ .

Forecast (a): A is on average longer than B by 0.5 mm. So $P (A > B)$ should be a bit above $50%$ — but by how much?

Step 1 (a). Define $D = A - B$ . Then $D \sim N (20 - 19.5, 0. 3^{2} + 0. 4^{2}) = N (0.5, 0.25)$ . Why this step? $A > B ⟺ D > 0$ . Means subtract, but variances still add — subtracting a variable does not subtract its noise. ( $σ_{D} = 0.25 = 0.5$ .)

Step 2 (a). $P (D > 0) = P (Z > \frac{0 - 0.5}{0.5}) = P (Z > - 1) = Φ (1) = 0.8413$ . Why this step? Right tail past $- 1$ equals the left area up to $+ 1$ by symmetry.

Answer (a): $P (A > B) \approx 0.8413$ (84.1%).

Step 3 (b). $\overset{μ}{^} = \frac{2 + 4 + 4 + 6}{4} = \frac{16}{4} = 4$ . Why this step? MLE of the mean is just the sample mean (parent Step 3).

Step 4 (b). Deviations: $(2 - 4)^{2} = 4, (4 - 4)^{2} = 0, (4 - 4)^{2} = 0, (6 - 4)^{2} = 4$ . Sum $= 8$ . Biased $\overset{σ}{^}^{2} = \frac{8}{4} = 2$ . Why this step? MLE variance divides by $n$ , not $n - 1$ (biased estimator, parent Step 4).

Answer (b): $\overset{μ}{^} = 4$ , $\overset{σ}{^}_{MLE}^{2} = 2$ .

Verify (b): unbiased version divides by $n - 1 = 3$ : $8/3 \approx 2.667 > 2$ , confirming the biased one is smaller — exactly Bessel's correction. ✓ See Maximum Likelihood Estimation.

Recall Rapid self-test

IQ $\sim N (100, 1 5^{2})$ : $P (X > 130)$ ? ::: $1 - Φ (2) \approx 0.0228$ Why does averaging 5 readings shrink $σ$ by $5$ ? ::: variance divides by $n^{2}$ but there are $n$ of them, net $/ n$ on variance $\Rightarrow / n$ on $σ$ Var of $A - B$ (independent, $0.3, 0.4$ )? ::: $0.09 + 0.16 = 0.25$ (add, never subtract) MLE variance divides by? ::: $n$ (biased); unbiased uses $n - 1$