Statistics, Probability & Power BI - Evaluation (Actual Paper)

The joint probability is 90/500 = 0.18. Given milk, the probability of tea is P(tea | milk) = 90/200 = 0.45, using the milk baskets as the new total. Bayes' theorem reverses the conditioning: P(milk | tea) = P(tea | milk) × P(milk) / P(tea) = 0.45 × 0.40 / 0.30 = 0.60, which matches the direct count 90/150 = 0.60. The two conditional probabilities are not equal (0.45 versus 0.60) because they divide the same joint count by different totals.

In its comparison form, Bayes' theorem is P(A|B) ∝ P(B|A) × P(A): the denominator P(B) is the same for every cause, so ranking needs only prior × likelihood. Payment gateway: 0.50 × 0.20 = 0.10. Bank settlement: 0.10 × 0.60 = 0.06. Courier: 0.40 × 0.30 = 0.12, the largest. Ranking by the prior alone would favour the payment gateway and ranking by the likelihood alone would favour the bank settlement, but each ignores half of the formula, and the missing total does not block the comparison because it scales all three posteriors equally.

The question asks for a conditional probability: P(A|B) = P(A ∩ B) / P(B) = 0.12 / 0.40 = 0.30. The value 0.12 is the joint probability itself, without dividing by P(B); 0.52 adds the two probabilities; and 0.048 multiplies them. None of those is the proportion of workshop attendees who secured an offer.

By Bayes' theorem, P(A|B) = P(B|A) × P(A) / P(B) = 0.90 × 0.02 / 0.10 = 0.018 / 0.10 = 0.18. The value 0.90 is the likelihood P(B|A), a different quantity from the posterior P(A|B); 0.018 is the numerator alone, before dividing by the evidence P(B); and 0.20 divides the prior by the evidence and skips the likelihood.

A count such as the number of defective units is a discrete variable: each possible count has its own probability, listed in a PMF, and all of them sum to 1. Arrival time is a continuous variable, so the probability of any single exact value is zero and probabilities are calculated for ranges. Height is also continuous, so its PDF describes probability density, and the area under the PDF over the entire range is 1. A PDF is not used for a discrete count, and even for a continuous variable the height of the PDF at a point is a density, not a probability.

A z-score standardises a score with z = (x − μ) / σ. Meera: (78 − 66) / 8 = 1.5. Arjun: (84 − 70) / 10 = 1.4. On the standardised scale 1.5 is higher than 1.4, so Meera performed slightly better relative to her own test. Comparing raw scores (84 versus 78) or raw distances from the mean (14 versus 12) ignores that the two tests have different spreads (10 versus 8), and both scores lying within two standard deviations does not make the z-scores equal.

Payment method is nominal data, so only a measure that works on categories can be used. The mode simply reports the most frequently occurring category and is the only measure of central tendency that applies to nominal data. The mean needs numeric values that can be added and divided, the median needs the values arranged in ascending order, and the standard deviation is a measure of spread rather than a measure of central tendency.

Deviations from the mean always sum to zero, so the missing deviation is −(−3 + 5 − 1 + 2 + 4) = −7. The squared deviations are 9, 25, 1, 4, 16 and 49, which total 104. Sample variance divides by n − 1 = 5, because only five deviations are free to vary once the mean is fixed: 104 / 5 = 20.80. Dividing by n = 6 gives 17.33, which is the population formula; ignoring the lost deviation gives 55 / 5 = 11.00; and dividing by 4 gives 26.00, which removes one degree of freedom too many.

The standard error is σ/√n. With σ = 40 it is 40/√100 = 4 at n = 100 and 40/√400 = 2 at n = 400, so quadrupling the sample size halves it, and a smaller standard error means sample means cluster closer to the population mean. The population standard deviation is a property of the population and does not change with the sample size. The standard error varies with the square root of n, not with n itself, so it is halved rather than cut to a quarter.

For large samples the sample means are approximately normal whatever the shape of the population, centred on the population mean of 800. Their spread is the standard error σ/√n = 300/√225 = 300/15 = 20. About 95% of a normal distribution lies within two standard deviations, so about 95% of sample means fall within 800 ± 2 × 20, that is 760 to 840. Staying right-skewed contradicts the theorem; a spread of 300 is the population's standard deviation, not the standard error; and 780 to 820 covers only one standard error either side, about 68% rather than 95%.

Only the average count per hour is known (λ = 6) and the value of interest is a specific count (x = 9), which is exactly the situation for the Poisson PMF e^(−λ) × λ^x ÷ x!. The binomial PMF needs a fixed number of trials n and a constant success probability p, neither of which is available, and n = 6 trials could never produce 9 successes. A normal PDF gives a probability density rather than the probability of an exact value, and ticket counts are discrete. Swapping the roles (λ = 9, x = 6) treats the count being asked about as the known average.

Stratified sampling takes samples from each group in proportion to its representation, so 65% of 400 = 260 undergraduates and 35% of 400 = 140 postgraduates fix the mix by design. Simple random sampling gives every student an equal chance, so the programme mix would match only on average and by luck. Systematic sampling picks every k-th student by a fixed interval from a list that carries no information about programme, so the mix is left to chance, and convenience sampling is biased toward whoever is easiest to reach.

Two different groups of customers call for the two-sample independent T-test, while the same riders measured twice, before and after a workshop or on weekdays and then on weekends, are dependent samples and call for the paired T-test rather than an independent one. Association between two categorical variables, return status and city, is tested with a chi-square test, not ANOVA, which compares means across groups.

The decision rule is to reject H₀ when the P-value is below the level of significance. 0.03 is below 5%, so H₀ is rejected at that level, and 0.03 is not below 1%, so H₀ is not rejected at that level. A small P-value is evidence against H₀ but does not prove the alternate hypothesis. The P-value is the probability of results as extreme as those observed assuming H₀ is true; it is not the probability that H₀ itself is true.

The inspector's claim, that the mean fill weight is below 500 g, is what must be demonstrated, so it goes in the alternate hypothesis H₁: μ < 500 g and everything else goes in H₀: μ ≥ 500 g. Because H₁ specifies a direction, the test is one-tailed and the whole rejection region lies in the lower tail. Putting the claim in H₀ reverses the burden of evidence; a two-tailed test with H₁: μ ≠ 500 g would also react to over-filling, which is not the claim; and H₁: μ > 500 g tests the opposite direction.

t = (x̄ − μ) / (s/√n) = (17.2 − 18) / (2.4/√36) = −0.8 / 0.4 = −2.0. With 35 degrees of freedom the two-tailed 5% critical values are ±2.030, and −2.0 lies between them, so H₀ is not rejected. Comparing with ±1.96 wrongly uses the Z critical value although the population standard deviation is unknown; the sign of t does not decide the outcome, only whether it falls in the rejection region; and failing to reject H₀ never proves it true.

The P-value of 0.03 is not below α = 0.01, so the decision rule says fail to reject H₀. But H₀ is actually false because the tablet works, and failing to reject a false null hypothesis is a Type II error, whose probability is β. A decision is not automatically correct because it follows the rule, since the truth matters too; H₀ is not rejected at all, because 0.03 is greater than 0.01; and a Type I error means rejecting a true H₀, which is not what happened here.

Every pair of the 5 warehouses needs its own test: 5C2 = 5! / (2! × 3!) = 10. Running 10 separate tests multiplies the chances of a false positive, so a one-way ANOVA, which is based on the F statistic, compares all the means at once. Five tests would give only one comparison per warehouse; a chi-square test relates two categorical variables rather than comparing means; and 20 counts each pair twice, while a paired T-test needs the same subjects measured twice and cannot compare five groups.

Recall = TP/(TP + FN) = 16/20 = 80%, so the 4 false negatives are patients with the disease who were missed. Precision = TP/(TP + FP) = 16/40 = 40%, so 24 of the 40 positive predictions (60%) are false alarms. A false positive rejects a true null hypothesis (Type I error) and a false negative fails to reject a false one (Type II error). Accuracy = (16 + 156)/200 = 86% is arithmetically right, but with only 20 diseased patients out of 200 it can look strong while precision is just 40%, so it does not show that positive predictions are trustworthy.

A missed defaulter is a false negative (a Type II error), and recall = TP/(TP + FN) measures how many of the actual defaulters the model catches. Model X: TP = 96 and FN = 100 − 96 = 4, so recall = 96%; FP = 240 − 96 = 144, so precision = 96/240 = 40%. Model Y: TP = 77 and FN = 23, so recall = 77%; FP = 110 − 77 = 33, so precision = 77/110 = 70%. Weighting each missed defaulter at 10 units and each false alarm at 1 unit gives a total cost of 4 × 10 + 144 = 184 for Model X against 23 × 10 + 33 = 263 for Model Y, and Model X stays cheaper for any cost ratio above about 5.8, so "at least ten times" settles the choice: Model X's much higher recall is preferred despite its lower precision, which only means more false alarms, the cheaper error here. Accuracy is misleading on imbalanced data: Model Y scores (77 + 867)/1,000 = 94.4% against Model X's (96 + 756)/1,000 = 85.2%, yet Model Y lets 23 defaulters through instead of 4. Precision and recall can be compared across models tested on the same data.

Dimensions slice and organise the data and become the chart's categories; measures are numeric values that get aggregated with operations such as sum, average or count. Sales is aggregated, so it is a measure. Postal Code is used to categorise rows, so it is a dimension even though its values look numeric, and Order Year is a dimension when it splits the bars into groups; context decides the role. Summing postal codes would be meaningless, so treating Postal Code as a measure is wrong.

A calendar table related to the sales table by date supplies year, quarter and month for slicing. Slicers for region, product and salesperson filter the report, and its visuals respond to the selected values. Profit margin percentage is profit divided by total sales, formatted as a percentage. Total orders is a count of distinct orders, not a sum of the Quantity column, because one order can contain many units.

The managers accept data that is up to an hour old and the source also serves billing, so the default choice is an extract: data is downloaded and pre-computed at scheduled intervals, dashboards render quickly from stored snapshots, and the warehouse is protected from 300 people triggering live queries. A live connection queries the source on every view, which is slower and loads the infrastructure, so it is kept for real-time needs; claiming that live renders faster reverses the trade-off; and an extract does not query the source on each view.

After the slicer keeps only customers C1 and C3, the remaining Customer_ID values are C1, C1, C3, C1: four rows. COUNTROWS(Orders) counts rows, so Count of Orders is 4. DISTINCTCOUNT(Orders[Customer_ID]) counts unique values, C1 and C3, so Distinct Customers is 2. The pair 6 and 3 ignores the slicer, 4 and 4 treats DISTINCTCOUNT like COUNTROWS, and 2 and 4 swaps the two measures.

Quantity × Unit Price = 6 × 250 = 1,500. Subtracting the discount of 150 and the shipping cost of 100 gives 1,500 − 150 − 100 = 1,250. The value 1,750 adds the two deductions instead of subtracting them, 1,350 subtracts only the discount, and 1,400 subtracts only the shipping cost.

Power Query cleans and reshapes data before it is loaded, so turning 'CUST-001' into 1 (for example with find-and-replace or by splitting the column) belongs there and lets the tables join, because matching depends on the values rather than the column names. DAX works after the data is loaded, and SUMX performs a row-by-row calculation before summing, which suits a profit measure that responds to report filters. The reverse assignment misuses both tools, Power Query can find-and-replace and split column values, and DAX SUMX does iterate over rows.

3D graphs are hard to read on a 2D surface, and labelling bars directly is preferred to a legend that makes readers juggle between legend and graph. Gray for context with a single accent colour builds a visual hierarchy and directs attention to the message. About 8% of people are colour-blind, so meaning should never rely on colour alone and red-green pairings should be avoided. Starting the axis near the smallest bar distorts the perception of magnitude, since axes should start at zero.

Executives need the conclusion and the key number, with a title that states the 'so what'. One line chart that uses an accent colour for the months that matter and gray for context, under a headline title, delivers a single message that can be grasped in about ten seconds. The 3D chart with gridlines and a legend adds clutter and dilutes the message; the rainbow palette gives every series equal weight so nothing stands out; and a full table under a topic title forces the reader to hunt for the message.

A pair plot draws every numeric variable against every other numeric variable, and its diagonal shows each variable's own distribution; hue= colours the points by a categorical variable, which adds a dimension. A facet grid is the opposite: the analyst must choose which categorical variable to split on with col= or row=, and it compares one relationship across categories rather than discovering all relationships. With 100 or more categories a facet grid produces far too many panels, which is exactly its trade-off.

The pair plot lays out a 5 × 5 grid, but corner=True removes the redundant mirrored panels above the diagonal, leaving the 5 diagonal panels (each variable's own distribution) and the 10 below it: 15 panels. The relplot with col=region and row=plan draws one panel per combination: 3 × 2 = 6. The total is 15 + 6 = 21. Forgetting corner=True gives 25 + 6 = 31; dropping the diagonal gives 10 + 6 = 16; and adding the rows and columns (3 + 2 = 5) instead of multiplying them gives 15 + 5 = 20.