Results use the current categorical workbook. For income above AUD 50,000, this key treats the 50,000-75,000, 75,000-100,000, and over 100,000 bands as the intended group; the first band includes a boundary value of 50,000, so the workbook cannot establish a strictly greater-than-50,000 condition for every person.
Download the Week 2 worked Excel workbook.
Events and contingency tables
Watch me: translate a business question / Now you try: events and rules
For late delivery \(A\) and refund \(B\): late and refunded is \(A\cap B\); late or refunded is \(A\cup B\); not late is \(A^c\); refunded given late is \(P(B\mid A)\). For click \(C\) and purchase \(P\): clicked but did not purchase is \(C\cap P^c\); clicked or purchased is \(C\cup P\). The addition rule gives 0.30 + 0.12 − 0.08 = 0.34. The events overlap with probability 0.08, so they are not mutually exclusive.
Watch me: build the Marketing table
Build a PivotTable with Income in Rows, Have Tried in Columns and Count of Person ID in Values. Counts are:
| <25,000 |
108 |
109 |
217 |
| 25,000–50,000 |
148 |
198 |
346 |
| 50,000–75,000 |
69 |
112 |
181 |
| 75,000–100,000 |
22 |
40 |
62 |
| Over 100,000 |
14 |
36 |
50 |
| Total |
361 |
495 |
856 |
For percentages of grand total, divide each cell by 856; all cells together sum to 100%.
Now you try: marketing probabilities
\(P(\text{not tried})=361/856=\mathbf{0.4217}\) (marginal). \(P(\text{higher income}\cap\text{not tried})=(69+22+14)/856=\mathbf{0.1227}\) (joint). The higher-income total is 293, so the union is \((293+361-105)/856=549/856=\mathbf{0.6414}\). The union counts either event, including their overlap once.
Watch me: condition on “not tried” / Now you try: reverse the condition
\(P(\text{higher income}\mid\text{not tried})=105/361=\mathbf{0.2909}\): the denominator is all 361 who have not tried it. \(P(\text{not tried}\mid\text{higher income})=105/293=\mathbf{0.3584}\): the denominator is all 293 in the higher-income bands. These reverse questions differ because their reference groups differ. In the PivotTable, use % of Column Total for the first and % of Row Total for the second.
Watch me: tried product and gender / Now you try: assess independence
Overall trial is \(495/856=\mathbf{57.83\%}\); among women it is \(216/398=\mathbf{54.27\%}\). The sample rates differ. For living alone, the table is:
| No |
324 |
387 |
711 |
| Yes |
37 |
108 |
145 |
\(P(\text{tried}\mid\text{lives alone})=108/145=\mathbf{74.48\%}\) versus \(P(\text{tried})=\mathbf{57.83\%}\) (and \(387/711=54.43\%\) among those not living alone). Independence is not plausible as a description of this sample. Living alone is associated with trial here; these observational data do not show it causes trial.
Continuous probability
Watch me: blood pressure risk / Now you try: service time
For \(X\sim N(124,10^2)\), =NORM.DIST(117,124,10,TRUE) gives 0.2420; =1-NORM.DIST(140,124,10,TRUE) gives 0.0548; =NORM.INV(0.85,124,10) gives 134.36 for the upper-15% threshold. Sketch a Normal curve with the relevant left, right or central area shaded.
For service time \(X\sim N(8,1.5^2)\):
| \(P(X<6)\) |
=NORM.DIST(6,8,1.5,TRUE) |
0.0912 |
| \(P(X>10)\) |
=1-NORM.DIST(10,8,1.5,TRUE) |
0.0912 |
| \(P(7<X<9)\) |
=NORM.DIST(9,8,1.5,TRUE)-NORM.DIST(7,8,1.5,TRUE) |
0.4950 |
| 90th percentile |
=NORM.INV(0.9,8,1.5) |
9.9223 min |
The first two tails match by symmetry. The middle probability is an area between two cutoffs, not a single cumulative area.
Now you try: compare unlike measures
Sales at \(z=1.4\) is more unusual relative to its distribution than profit at \(z=0.6\), because it is farther from the mean in standard-deviation units. Standardising removes measurement units. It does not prove the store is well managed: costs, location, market size and other factors could explain performance.
Watch me: choose the distribution / Now you try: method selection
An individual value from a stated Normal population uses that Normal model. A standardised observation uses the standard Normal scale. For a mean with \(n=18\) and unknown population standard deviation, use a \(t\) distribution with 17 degrees of freedom if the sampling assumptions support it. A severely skewed individual outcome with no distributional assumption has not enough information for a Normal or \(t\) probability calculation.
Integrated marketing brief
Watch me: from question to conclusion / Now you try: marketing evidence brief
A valid worked choice is Live Alone. Its joint tried-and-lives-alone probability is \(108/856=\mathbf{12.62\%}\), marginal trial probability is 57.83%, and conditional trial rates are 74.48% for those living alone and 54.43% otherwise. Independence appears implausible in this sample. Three-sentence brief: “Of 856 respondents, 108 both lived alone and had tried the product. Trial was reported by 74.5% of those living alone versus 54.4% of those not living alone, an observed association. The survey does not establish a causal effect or population rates without a defensible sampling design.” A clearly labelled version of the 2×2 table above supports the brief.
Retrieval and closing checks
Have Tried is nominal categorical; Income is an ordered category in bands, even though it describes money. Each Marketing row is one respondent. The business question determines which event is the outcome and which denominator belongs in the calculation. Joint probability uses an intersection; marginal probability ignores the other variable; union uses either event; conditional probability changes the reference group. Independence would require conditional and marginal rates to agree in the population. For Excel Normal areas, use NORM.DIST for a left cumulative area, subtract from 1 for a right tail, and NORM.INV for a percentile. A \(t\) distribution becomes relevant for mean inference with unknown population SD, under suitable sampling assumptions.