LGT5101 Cheatsheet · Statistics for Management

0 · Exam Map syllabus · Lec 0

Assignment 1 15% · Assignment 2 25% · participation 10% · final 50%.

“Finally, there is an open-book final exam (50%).” (LGT5101 Statistics for Management syllabus, Semester 1 2026/27) · “Final Exam 50% Lectures 1-13 TBD (Open-book)” (LGT5101 Lecture 0 course overview, Semester 1 2026/27)

The captured materials do not state duration, sections, question count or a more detailed materials rule. Show formula → substitution → result → interpretation: Assignment 1 explicitly asks for formulas and calculation steps wherever possible.

1 · Start With Scope C1

Five-line opening: experimental unit (one row) · target population · observed sample · variables · intended inference. Then record source, filters and missing values. A complete branch roster is a population for that branch but a sample for a company-wide claim.

Opening audit

One row = one level. Do not turn a delivery row into a customer or depot claim. Description organises observed units; inference extends beyond them and needs a sampling argument. Record any grouping because it loses detail.

Check → If “code 4 is twice code 2” is meaningless, the digits are labels. Generalisation beyond observed rows needs representativeness and stable conditions.

2 · Frequencies & Displays C2

frequency ledgerfᵢ = count in class i
relative frequency rᵢ = fᵢ / n
checks: Σfᵢ = usable n · Σrᵢ = 1 (rounding aside)

Histogram boundary rule: intervals are adjacent, non-overlapping and collectively cover the values; state which endpoint is included. The lecture guidance normally suggests 5–15 classes, depending on the number of values. Compare reasonable widths: broad bins can hide a cluster; narrow bins can manufacture noise.

Build the display in order

1. Name the usable denominator after filters and missing values.
2. List mutually exclusive, collectively exhaustive classes.
3. Count fᵢ; divide by n when relative frequency is required.
4. Reconcile totals before drawing.
5. Label title, axes, units and count/relative-frequency scale.

2b · Read, Then Question C2 workflow

Describe quantitative pattern in the fixed order centre → spread → shape → unusual features. Add title, labelled axes, frequency scale, denominator, observation window and missing-value rule. Preserve natural order for ordinal categories.

3 · Centre, Spread, Shape C3

samplex̄ = Σxᵢ / n
s² = Σ(xᵢ − x̄)² / (n − 1) · s = √s²

populationμ = Σxᵢ / N
σ² = Σ(xᵢ − μ)² / N · σ = √σ²

positionrange = max − min
z = (x − mean) / standard deviation

Arithmetic checks: order before median; Σ(xᵢ−x̄)=0 apart from rounding; squared deviations ≥0; variance uses squared units, SD original units. Keep precision until the final line. Scope, not dataset size, chooses n−1 versus N.

Choose the lead statistic

Use the median when a skewed distribution or distant values make the mean unrepresentative of an ordinary case. Retain the mean when total labour, cost or customer impact matters because every magnitude contributes. Mode answers “most common”, including for categories. Pair every centre with spread: equal means can conceal very different consistency.

3b · Distribution Rules C3

Right skew: long high tail, typically mean > median. Left skew: long low tail, typically mean < median. Multiple clusters can make any single centre incomplete.

4 · Event Probability C4

event ledgerP(Aᶜ) = 1 − P(A)
P(A∪B) = P(A)+P(B)−P(A∩B)
P(A|B) = P(A∩B)/P(B)
P(A∩B) = P(B)P(A|B)

independent onlyP(A|B)=P(A) ⇔ P(A∩B)=P(A)P(B)

Basic outcome probabilities lie in [0,1] and total 1. Label the source: equally likely theoretical outcomes, empirical relative frequency or informed judgement. A conditional difference is association, not proof of causation.

Region-first audit

1. Name experiment, sample space and events.
2. Translate the wording; draw a region, table or tree.
3. Apply the rule; check bounds and subsets.
4. Interpret using the correct reference group.

5 · Discrete Variables C5

pmf lists P(X=x): each mass ∈[0,1], ΣP(X=x)=1. cdf F(k)=P(X≤k) accumulates left to right, never decreases and ends at 1; each jump recovers the mass.

discrete boundariesP(X=k)=F(k)−F(k−1)
P(X>k)=1−F(k)
P(X≥k)=1−F(k−1)

momentsE(X)=ΣxP(X=x)
Var(X)=Σ(x−μ)²P(X=x)
=E(X²)−[E(X)]² · SD=√Var(X)

Expectation is a long-run weighted centre; it need not be an attainable single outcome. Use a four-column ledger x, P, xP, x²P; check mass before computing moments.

For integer X, strict and inclusive thresholds cannot be exchanged: a single point can carry positive mass. Recover p(k) from the jump F(k)−F(k−1). If listed masses total 0.90, the model is incomplete even though each entry looks plausible.

5b · Binomial Model C5

Earn the shortcut: fixed n · two outcomes after grouping · constant p · independent trials. Sampling without replacement from a small population can violate independence.

X ∼ Binomial(n,p)P(X=k)=C(n,k)pᵏ(1−p)ⁿ⁻ᵏ
E(X)=np · Var(X)=np(1−p)
SD=√[np(1−p)]

6 · Continuous & Normal C6

For a continuous variable, probability is area. Total area =1; one exact point has zero area, so open/closed endpoints agree. Density height alone is not probability.

X ∼ Normal(μ,σ)z=(x−μ)/σ
left: P(X<x)=Φ(z)
right: P(X>x)=1−Φ(z)
between: Φ(zᵤ)−Φ(zₗ)

Forward: boundary → area. Draw, shade, standardise, select left/right/between, interpret in original units. Inverse: area → boundary. Convert wording to a left-tail p, then x₀=NORM.INV(p,μ,σ). “Exceeded on 5%” means p=0.95.

Function translation

The normal model is symmetric and determined by μ and σ; mean, median and mode coincide. Keep unrounded z-values in the calculation and attach the original unit to a returned boundary. An empirical-rule check is a plausibility screen, not a replacement for the cdf.

6b · Normal Approximation C6

For a binomial count, the captured course rule requires np>10 and n(1−p)>10. Then use μ=np and σ²=np(1−p). A large n alone is insufficient when p is near 0 or 1. The smooth curve approximates integer bars; apply the continuity correction to the event boundary.

Continuity direction: an inclusive integer boundary moves half a unit outward when mapped to continuous area; sketch the bars and curve so the included integers determine the corrected boundary. State that the answer is an approximation and, when feasible, compare it with the exact binomial result.

7 · Worked: Display Choice C2 example

Q. A service desk records 80 tickets: self-service 28, chat 24, phone 18, in-person 10, plus quantitative resolution time. Choose displays and show the channel shares.

Answer. Use separated bars for channel categories; a pie is possible because the four classes partition one whole, but bars support closer comparison. Use equal, non-overlapping, touching histogram bins for resolution time. State boundaries and read centre, spread, shape and unusual features.

Marks are in the checks

The frequency checksum is 28+24+18+10=80. The relative-frequency checksum is .350+.300+.225+.125=1.000. State that the bar length encodes a category’s share of the usable 80 tickets. For the time histogram, exact observations are deliberately sacrificed to reveal concentration, spread, skew, gaps and possible clusters.

Trap → Do not use a histogram for coded categories or separated bars for adjacent quantitative intervals.

7b · Set-up Mini-Drill C1 example

A delivery file records service tier, minutes late, complaint status and depot code. Unit: one completed delivery. Variables: tier + depot = nominal; complaint = binary categorical; minutes late = quantitative ratio. It describes observed deliveries; inference to future deliveries needs representative operating conditions.

Boundary: describe sampled deliveries; inference to future work requires comparable conditions.

8 · Worked: Centre & Spread C3 example

Q. Packing times (min): 18, 19, 20, 20, 21, 22, 34. Calculate centre, sample spread and the standard score of 34.

1. Centres: Σx=154, so x̄=154/7=22 min. Ordered middle (4th)=20; median=20; mode=20. Range=34−18=16 min.
2. Spread: Σ(x−22)²=178. Sample s²=178/(7−1)=29.667 min²; s=√29.667≈5.447 min.
3. Standardise: z=(34−22)/5.447≈2.203. The job is 2.203 sample SD above the mean, unusual but not beyond the course’s three-SD extreme rule.
4. Interpret: the high time creates a right tail, so the mean exceeds the median. Median better represents an ordinary job; mean still matters for total resources.

Decision split: “ordinary packing job” points to the median because it resists the high value; “total staffing minutes” keeps the mean because all minutes contribute. The statistics answer different management questions rather than competing to be universally correct.

Rule drills

An invoice taking 57 minutes under mean 42 and sample SD 6 has z=(57−42)/6=2.5: it is 2.5 sample SD above the mean. A z-score locates; it is not automatically a tail probability. For an approximately bell-shaped distribution, compare with the 68–95–99.7 rule. If shape is unknown, use the weaker Chebyshev guarantee rather than the empirical rule.

For a population, replace x̄, n and s with μ, N and σ and divide the squared-deviation total by N. Scope—not the number of rows—selects the denominator.

9 · Worked: Union + Given C4 example

Q. P(Urgent)=0.28, P(Late)=0.18 and P(Urgent∩Late)=0.08. Find the union, lateness given urgency and test independence.

1. Union: P(U∪L)=0.28+0.18−0.08=0.38.
2. Condition: “given urgent” makes U the denominator: P(L|U)=0.08/0.28=0.2857.
3. Independence: compare conditional with marginal: 0.2857≠0.18, so U and L are not independent.
4. Meaning: urgency and lateness are associated in this model; this does not establish that urgency causes delay.

The completed table exposes every denominator: inside the urgent row, late occupies .08/.28; inside the late column, urgent occupies .08/.18. These are different conditional questions. It also makes “neither”=.62 and “exactly one”=.20+.10=.30 visible without a new formula.

Tree alternative: begin with U (.28) and Uᶜ (.72); second-stage late probabilities are conditional on those branches. Multiply along each full path; add only paths that cannot both occur. If the late branch probability were identical after U and Uᶜ, that would visually express independence.

9b · Worked: Count in Stages C4 drill

Select 3 from 8 analysts, then name one selected analyst chair. Group first: C(8,3)=56. Role second: ×3. Total=168 outcomes. Treating the whole selection as one permutation would count orders that do not create distinct committees.

10 · Worked: Binomial Damage C5 example

Q. Six parcels have independent damage probability 0.08. Find P(exactly 2 damaged) and summarise the count X.

1. Define: success=damaged parcel; X counts damaged parcels.
2. Model: fixed 6, damaged/not, constant .08, independent ⇒ X∼Binomial(6,.08).
3. Exact mass: P(X=2)=C(6,2)(.08)²(.92)⁴=0.0687737.
4. Centre: E(X)=np=6(.08)=0.48 parcel.
5. Spread: Var(X)=6(.08)(.92)=0.4416; SD=√.4416≈0.6645 parcel.

The non-integer expectation is a long-run average across repeated six-parcel audits, not a fractional parcel prediction for one audit.

Cumulative boundary drill

If X counts failed inspections in 10 independent items with failure probability .04, “more than one” is X>1, so write P(X>1)=1−P(X≤1)=1−F(1) before software. “At least one” would instead be 1−F(0). The one-step shift is the discrete-boundary mark.

Sampling without replacement from a small batch would make each damage probability depend on earlier draws. If p varies by parcel route or packing line, a single constant-p binomial model also fails.

Distribution reconstruction: a full pmf table adds columns xP(X=x) and x²P(X=x). Sum them for E(X) and E(X²), then use Var(X)=E(X²)−[E(X)]². Convert pmf to cdf by running sums; recover any mass from the successive jump. These longer routes should agree with np and np(1−p) when the model is binomial.

Trap → TRUE accumulates through k; FALSE returns the exact mass at k.

11 · Worked: Normal Interval C6 example

Q. Delivery time X is normal with μ=52 min, σ=8 min. Find P(44<X<65).

1. Sketch: shade between 44 and 65; both boundaries are known ⇒ forward calculation.
2. Standardise: zₗ=(44−52)/8=−1; zᵤ=(65−52)/8=1.625.
3. Areas: Φ(1.625)=0.9479; Φ(−1)=0.1587.
4. Subtract: 0.9479−0.1587=0.7892.
5. Interpret: the model assigns about 78.92% probability to delivery times between 44 and 65 minutes.

Software route

NORM.DIST(65,52,8,TRUE)
− NORM.DIST(44,52,8,TRUE)
= 0.9479 − 0.1587 = 0.7892

Keep 1.625 unrounded while obtaining the cdf. The answer is a model probability, so the interpretation begins “the model assigns”, names the event 44–65 and returns to minutes. A left-tail result of .9479 alone would answer X<65, not the requested interval.

11b · Inverse Drill C6 drill

Daily output is normal with μ=1,200 units and σ=90. A level exceeded on only 5% of days leaves 95% to its left: solve P(X<x₀)=0.95 with NORM.INV(0.95,1200,90). Direction check: 0.95>0.5, so x₀ must exceed 1,200.

Model gate: check a histogram for approximate symmetry, one peak and process stability before treating the cdf digits as meaningful. For continuous X, P(X=boundary)=0, so < and ≤ have the same area.

12 · Method Chooser C1–C6

13 · High-Frequency Traps fast scan

1. Scope first: n−1 versus N; observed description versus inference.
2. Bar ≠ histogram: categories separated; measurement intervals touch.
3. Mean alone is incomplete: report spread and shape; do not delete a high z automatically.
4. Given flips the denominator: P(A|B) uses P(B), not P(A).
5. Exclusive ≠ independent: non-empty exclusive events cannot co-occur.
6. Discrete boundaries matter: ≥k uses 1−F(k−1); >k uses 1−F(k).
7. Binomial needs all four conditions: n, two outcomes, constant p, independence.
8. Cdf is left-tail: right tail is 1−cdf; between is a difference.
9. Inverse takes left area: upper 5% ⇒ input .95.
10. Interpret: result + event + unit + management meaning; association is not causation.