Statistics for Data Science I · Quiz 1 · Jan 2026

**1. Read the frequencies:** Apple, Banana, Orange and Grapes have 10,8,6,6 responses, respectively. Their total is 30 students. **2. Convert counts to pie shares:** \[\frac{10}{30}=33.33\%,\quad\frac8{30}=26.67\%,\quad\frac6{30}=20\%,\quad\frac6{30}=20\%.\] These correspond to sector angles 120°,96°,72°,72°, matching Pie Chart A. **3. Verify the bar heights independently:** Bar Chart A has heights 10,8,6,6 in the same category order. Chart B instead shows 8,10,10,2, which does not match the table. **Answer: A — Pie Chart A and Bar Chart A.** A: **Correct:** Both the pie shares and bar heights reproduce the original counts. B: **Incorrect:** Pie Chart A is right, but Bar Chart B uses the wrong four frequencies. C: **Incorrect:** Bar Chart A is right, but Pie Chart B gives incorrect category proportions. D: **Incorrect:** Neither Chart B reproduces the original frequency table.

**1. Classify the category variables:** Blood groups are unordered labels, so 1 maps to Nominal (D). Exam rankings have a meaningful order without equal performance gaps, so 2 maps to Ordinal (A). **2. Classify the measurement variables:** Celsius temperature has meaningful equal-unit differences but an arbitrary zero, so 3 maps to Interval (C). Weight in kilograms has a meaningful zero and meaningful ratios, so 4 maps to Ratio (B). **3. Assemble the mapping:** 1–D, 2–A, 3–C, 4–B. **Answer: C — 1–D, 2–A, 3–C, 4–B.** A: **Incorrect:** It misclassifies rankings as interval, Celsius as ratio and weight as ordinal. B: **Incorrect:** Blood group and weight match, but rankings and Celsius are misclassified. C: **Correct:** It gives Nominal, Ordinal, Interval and Ratio in the correct order. D: **Incorrect:** It treats Celsius as ratio and weight as interval, reversing their zero-point distinction.

**1. Identify the observed and target groups:** The 150 randomly selected individuals form the sample. All residents of the district form the population about which the study seeks conclusions. **2. Identify the scale:** The diet categories run from Very Unhealthy through Very Healthy. Their codes 1–5 preserve order but do not establish equal-sized health differences, so the scale is ordinal. **3. Distinguish summary from inference:** Describing the 150 participants is descriptive. Generalizing their results to residents throughout the district is inferential. **Answer: B, C — B and C.** A: **Incorrect:** The sample is the 150 selected individuals, not every resident. B: **Correct:** The ordered health categories are ordinal even though numerical codes are used. C: **Correct:** The full district resident group is the target population. D: **Incorrect:** A conclusion about the district based on its sample is inference rather than a description restricted to the observed sample.

**1. Distinguish measurement from counting:** Luggage weight is a numerical measurement that can vary continuously. The number of books is a numerical count restricted to nonnegative integers. **2. Distinguish categories from numbers:** Poor/Fair/Good/Excellent are ordered categories, while mobile-phone brands are unordered categories. Numerical codes, if assigned to brands, would remain labels rather than numerical measurements. **Answer: A, B, D — A, B and D.** A: **Correct:** Underlying weight is continuous, even if a scale rounds the displayed measurement. B: **Correct:** The ratings are ordinal categories. C: **Incorrect:** Brand names are nominal categories, not numerical quantities. D: **Correct:** Books read are counted in discrete whole-number units in this question.

**1. Identify the variable types:** Point-biserial correlation measures the linear association between one binary categorical variable and one numerical variable. **2. Relate it to Pearson correlation:** Code the binary variable as 0 and 1, then calculate ordinary Pearson correlation with the numerical values. The resulting coefficient is the point-biserial coefficient. **3. Exclude the other pairings:** Two nominal categorical variables require an appropriate contingency-table measure. Two continuous variables can use Pearson correlation directly but do not define the point-biserial case. **Answer: A, D — A and D.** A: **Correct:** It correctly identifies a binary variable paired with a numerical variable. B: **Incorrect:** Two nominal variables are not the defining point-biserial pairing. C: **Incorrect:** The method specifically includes a binary variable, so both variables need not be continuous. D: **Correct:** It is exactly Pearson correlation with the dichotomous variable coded numerically.

**1. Restrict the denominator to people with heart disease:** That row contains 120 people with diabetes and 80 without diabetes, giving 200 heart-disease cases. **2. Identify the overlapping cases:** Of those 200 people, 120 also have diabetes. **3. Calculate the conditional percentage:** \[\Pr(\text{Diabetes}\mid\text{Heart disease})\times100=\frac{120}{120+80}\times100=60\%.\] Dividing by all 1000 participants would answer a different, joint-percentage question. **Answer: 60.**

**1. Restrict the denominator to people without diabetes:** This column contains 80 people with heart disease and 620 without it, totaling 700. **2. Calculate the conditional percentage:** \[\Pr(\text{Heart disease}\mid\text{No diabetes})\times100=\frac{80}{80+620}\times100=11.428571\ldots\%.\] **3. Round only at the end:** The requested two-decimal answer is 11.43. The denominator is the no-diabetes group, not the heart-disease row or all 1000 people. **Answer: 11.43.**

**1. Recover the total number of purchased products:** Kabir bought 8 products with relative frequency 0.25, so \(N=8/0.25=32\). **2. Solve for the shared unknown:** The five counts sum to \[(b+5)+(b-2)+(b+1)+8+b=4b+12=32.\] Hence \(b=5\). **3. Add the requested customers’ counts:** Sneha bought \(b-2=3\) and Nikita bought \(b=5\), so together they bought \(3+5=8\) products. **Answer: 8.**

**1. Find total purchases:** Kabir’s 8 purchases represent 0.25 of the total, so \(N=8/0.25=32\). **2. Find Rohan’s count:** Summing the five counts gives \(4b+12=32\), so \(b=5\). Rohan’s count is \(b+5=10\). **3. Calculate Rohan’s relative frequency:** \[p=\frac{10}{32}=0.3125\approx0.31.\] The denominator is the number of products, not the number of customers. **Answer: 0.31.**

**1. Recover the total and unknown:** Kabir’s row gives \(N=8/0.25=32\). The sum of all counts is \(4b+12=32\), so \(b=5\). **2. Identify the requested rows:** q is Sneha’s relative frequency and r is Isha’s. Their counts are b−2=3 and b+1=6. **3. Add before rounding:** \[q+r=\frac{3+6}{32}=\frac9{32}=0.28125\approx0.28.\] Avoid rounding each frequency first, which can distort their sum. **Answer: 0.28.**

**1. Sort the nine observations:** 15,23,25,29,31,35,42,54,60. **2. State the inclusive linear-percentile convention:** The one-based percentile position is \(h=1+(n-1)p\). Thus \(h_{0.25}=1+8(0.25)=3\) and \(h_{0.75}=1+8(0.75)=7\), giving \(Q_1=x_{(3)}=25\) and \(Q_3=x_{(7)}=42\). **3. Subtract the quartiles:** \[IQR=Q_3-Q_1=42-25=17.\] **Answer: 17.** **Convention note:** This answer uses the inclusive linear-percentile convention, consistent with the supplied key. The median-of-halves method excluding the central observation would give Q1=24, Q3=48 and IQR=24. A finite-sample quartile question should specify its convention; these methods are not interchangeable.

**1. Read the four paired values:** Baking times X are 12,15,20,25 minutes and texture scores Y are 6,7,9,10. Their means are 18 and 8. **2. Calculate the centered sums:** \[S_{xx}=(-6)^2+(-3)^2+2^2+7^2=98,\quad S_{yy}=(-2)^2+(-1)^2+1^2+2^2=10.\] The cross-product sum is \(S_{xy}=12+3+2+14=31\). **3. Calculate and interpret the coefficient:** \[r=\frac{31}{\sqrt{98(10)}}\approx0.9903.\] It is strongly positive: longer baking times are associated with higher texture scores in these four observations, supporting B and D. **Answer: B, D — B and D — r≈0.99, strong positive correlation.** **Convention note:** Association in four observed items does not by itself establish that increasing baking time causes a better texture for every item or beyond this observed range. A: **Incorrect:** The calculated coefficient is approximately 0.99, not 0.30. B: **Correct:** A coefficient near 0.99 indicates a strong positive linear association. C: **Incorrect:** The centered cross-product sum and coefficient are positive, not negative. D: **Correct:** The positive sign indicates that texture scores tend to increase with baking time in the observed pairs.

**1. Calculate the recorded total:** The ten recorded values 120,135,140,150,160,170,180,190,15,20 sum to 1280. **2. Correct the stated over-recording:** Days 5,6,7 are each 15 too high, so subtract 3×15=45. Corrected total so far: 1280−45=1235. **3. Restore the omitted final digits:** Day 9 changes from 15 to 150, adding 135; day 10 changes from 20 to 200, adding 180. Thus the fully corrected total is 1235+135+180=1550. **4. Calculate the corrected mean:** \[\bar x_{corrected}=\frac{1550}{10}=155\text{ products/day}.\] **Answer: 155.**

**1. Check the order and count:** The values 110,135,140,145,152,166,170,190,200,210 are already in ascending order, and n=10 is even. **2. Identify the two middle positions:** The central observations are the fifth and sixth values, 152 and 166. **3. Average the central pair:** \[\operatorname{Median}=\frac{152+166}{2}=159.\] The median need not itself be one of the observations. **Answer: 159.**