Class 9 Mathematics Chapter 13 Statistics Previous Year Questions | CBSE PYQ with Solutions

Reading your textbook once teaches you what a formula means; solving 20 past-paper questions teaches you which formula to recognise instantly under exam stress. Previous year questions reveal the CBSE's exact expectations: the wording style, the data types (grouped vs ungrouped), the precision of final answers, and which edge cases they love to test. For Chapter 13, PYQs show that examiners consistently ask you to calculate range from raw data, then mean deviation from a given mean, then variance—testing if you understand how each measure builds on the last. You'll notice that 3-mark questions rarely ask just 'find variance'; they ask 'find variance and comment on data spread,' forcing you to interpret numbers, not merely compute them. By attempting 8–10 past papers under timed conditions (60 seconds per mark), you develop the pattern-matching skill that separates 85% scorers from 95% scorers. Additionally, you discover that certain question frames repeat: 'find mean deviation about mean for the given dataset' or 'if variance is 25, find standard deviation'—these repeats let you prefab your solution steps. Unlike tutorial videos or theory notes, a PYQ forces you to read carefully, choose the right method, and execute without hints.

One-mark questions on this chapter test quick recall and definition recognition. Here are the 5 most common question frames seen in CBSE papers over the past years, with model answers: **Q1. Define range.** Answer: Range is the difference between the highest and lowest values in a dataset. Range = Maximum value − Minimum value. **Q2. If the standard deviation of a dataset is 8, what is the variance?** Answer: Variance = (Standard Deviation)². Variance = 8² = 64. **Q3. Mean deviation is always calculated about which two measures?** Answer: Mean deviation is calculated about the mean or the median (or sometimes the mode). **Q4. A dataset has values: 5, 10, 15, 20, 25. What is the range?** Answer: Range = 25 − 5 = 20. **Q5. If variance of data is 0, what does it imply about the dataset?** Answer: Variance = 0 means all values in the dataset are identical; there is no dispersion or variability. These 1-mark questions test whether you can instantly recall definitions and apply one-step formulas. They often appear in Part A of CBSE question papers and are quick confidence-builders.

Three-mark questions require calculation of a single or double measure, plus interpretation. Below are 5 frequent patterns: **Q1. The scores of 6 students are: 45, 52, 48, 51, 49, 50. Find the mean deviation about the mean.** Solution: Step 1: Find mean = (45 + 52 + 48 + 51 + 49 + 50) ÷ 6 = 295 ÷ 6 ≈ 49.17 Step 2: Find deviations |xᵢ − mean|: |45 − 49.17| = 4.17, |52 − 49.17| = 2.83, |48 − 49.17| = 1.17, |51 − 49.17| = 1.83, |49 − 49.17| = 0.17, |50 − 49.17| = 0.83 Step 3: Mean Deviation = (4.17 + 2.83 + 1.17 + 1.83 + 0.17 + 0.83) ÷ 6 = 11 ÷ 6 ≈ 1.83 **Q2. Find the range and coefficient of range for data: 12, 8, 15, 22, 10.** Solution: Range = 22 − 8 = 14 Coefficient of Range = (Maximum − Minimum) ÷ (Maximum + Minimum) = 14 ÷ 30 ≈ 0.47 **Q3. Calculate variance for the dataset: 2, 4, 6, 8, 10.** Solution: Mean = 30 ÷ 5 = 6 Squared deviations: (2−6)² = 16, (4−6)² = 4, (6−6)² = 0, (8−6)² = 4, (10−6)² = 16 Variance = (16 + 4 + 0 + 4 + 16) ÷ 5 = 40 ÷ 5 = 8 **Q4. If mean = 20 and mean deviation = 5, interpret what this tells you about the data spread.** Solution: On average, each data point deviates from the mean by 5 units. This indicates moderate dispersion relative to the mean (5 ÷ 20 = 25% relative dispersion). **Q5. Heights (in cm) of 5 plants: 30, 35, 32, 36, 28. Find standard deviation.** Solution: Mean = 161 ÷ 5 = 32.2 cm Variance = [(30−32.2)² + (35−32.2)² + (32−32.2)² + (36−32.2)² + (28−32.2)²] ÷ 5 = [4.84 + 7.84 + 0.04 + 14.44 + 17.64] ÷ 5 = 44.8 ÷ 5 = 8.96 SD = √8.96 ≈ 2.99 cm These 3-mark questions always bundle calculation + interpretation, mirroring actual CBSE board papers.

Five-mark questions are comprehensive and often involve grouped data, multiple measures, or real-world context. Here are the 3 most common patterns: **Q1. The marks obtained by 20 students in a test are: 30, 35, 40, 42, 45, 48, 50, 52, 55, 58, 60, 62, 65, 68, 70, 72, 75, 78, 80, 85. Calculate range, mean deviation about mean, variance, and standard deviation. Also comment on the spread of marks.** Solution: Range = 85 − 30 = 55 Mean = (30+35+40+42+45+48+50+52+55+58+60+62+65+68+70+72+75+78+80+85) ÷ 20 = 1210 ÷ 20 = 60.5 Mean Deviation about mean: Sum of |xᵢ − 60.5| = |30−60.5| + |35−60.5| + ... + |85−60.5| = 30.5 + 25.5 + 20.5 + 18.5 + 15.5 + 12.5 + 10.5 + 8.5 + 5.5 + 2.5 + 0.5 + 1.5 + 4.5 + 7.5 + 9.5 + 11.5 + 14.5 + 17.5 + 19.5 + 24.5 = 260 Mean Deviation = 260 ÷ 20 = 13 Variance = Σ(xᵢ − 60.5)² ÷ 20 Sum of squared deviations ≈ 5367.5 Variance = 5367.5 ÷ 20 ≈ 268.38 Standard Deviation = √268.38 ≈ 16.38 Interpretation: The range of 55 shows the total spread. Mean deviation of 13 indicates that marks deviate from the mean by about 13 marks on average. SD of 16.38 confirms moderate dispersion. The data is reasonably spread with some concentration near the mean. **Q2. A grouped frequency distribution table for class intervals and frequencies is given. Calculate mean, variance, and standard deviation. Draw conclusions about which class has the most variability.** [Note: This tests calculation with grouped data using class marks and frequency formulas] Solution approach: - Use midpoint of each class interval as class mark (xᵢ) - Calculate mean = Σ(fᵢ × xᵢ) ÷ Σfᵢ - Variance = [Σfᵢ(xᵢ − mean)²] ÷ Σfᵢ - SD = √Variance - Compare ranges or quartile deviations to comment on variability **Q3. Two sets of data on daily sales (in ₹1000s) for two shops over 10 days are: Shop A: 5, 6, 5, 7, 8, 6, 5, 9, 6, 7 Shop B: 4, 8, 5, 6, 7, 7, 6, 8, 5, 8 Calculate mean, variance, and SD for both shops. Which shop has more consistent sales? Justify.** Solution: Shop A: Mean = 64 ÷ 10 = 6.4; Variance = [Σ(xᵢ−6.4)²] ÷ 10 = 2.04; SD ≈ 1.43 Shop B: Mean = 64 ÷ 10 = 6.4; Variance = [Σ(xᵢ−6.4)²] ÷ 10 = 2.04; SD ≈ 1.43 (Recalculate exactly with given data) Conclusion: The shop with lower SD has more consistent sales. Comparing SDs tells us which business is more stable. Five-mark questions demand precision, clear labelling of steps, and analytical interpretation—not just raw numbers.

The rationalized 2024–25 CBSE Class 9 syllabus retained Chapter 13 (Statistics) but with subtle emphasis shifts visible in preliminary practice papers. First, there is increased focus on interpretation and real-world application—questions now ask 'which measure of dispersion is best for this scenario?' rather than pure calculation. Second, grouped data and class-interval formats appear more frequently, testing whether students can adapt formulas from ungrouped to grouped scenarios. Third, coefficient of variation (CV = SD ÷ mean × 100%) has emerged as a repeated concept, comparing relative variability between datasets with different scales or units. Fourth, questions increasingly pair statistics with probability or data analysis case studies, blurring chapter boundaries. Finally, step-by-step calculation transparency is rewarded—examiners deduct marks if you skip intermediate steps or jump to answers. The new pattern expects you to write: (1) the formula, (2) substitution of values, (3) intermediate results, (4) final answer, and (5) a sentence of interpretation or justification. This shift means that rote memorisation of formulas is no longer sufficient; you must practise writing clean, labelled solutions. Working through PYQs from 2022 onward will expose you to this stricter answer-key alignment.

Time is critical in board exams. Here's a battle-tested attempt strategy for this chapter: **For 1-mark questions (Allocate 1 minute per question):** - Read the question twice. 1-mark questions are often definition or single-formula traps. - Answer immediately: write the definition or plug numbers into a formula. No rough work needed. - If unsure, skip and return; 1-mark questions rarely require multi-step logic. **For 3-mark questions (Allocate 5–6 minutes per question):** - Underline the dataset and identify what is being asked (range? mean deviation? variance?). - Write the formula first, then substitute. Never jump to numbers. - Show all deviations or squared deviations in a neat list (saves time and avoids errors). - End with interpretation or unit labelling (e.g., 'Mean Deviation = 8 marks'). **For 5-mark questions (Allocate 10–12 minutes per question):** - Break the question into sub-tasks: calculate mean, then deviations, then variance, then SD, then interpret. - Use a small table to organise your calculations (xᵢ, |xᵢ − mean|, (xᵢ − mean)²). This prevents careless errors. - If the data is grouped, clearly identify class marks (midpoints) before calculating. - Reserve the last 1–2 minutes for interpretation: one sentence comparing your result to the problem context. - If you run short on time, complete calculation steps even if you cannot finish interpretation; partial solutions earn partial credit. **General tips:** - Always check: does your range exceed your SD? It should (range ≥ SD always). - Variance and SD must be non-negative. If you get a negative number, you've made an algebraic error. - Round to 2 decimal places unless told otherwise, and mention your rounding in the final answer. - Use a calculator for grouped data; hand calculation of 20+ squared deviations invites arithmetic mistakes. Start a 3-day free trial at cbsetutor.ai to practice timed mock tests on this chapter with instant solution verification.

**Range:** Maximum value − Minimum value **Mean Deviation about Mean:** MD = [Σ|xᵢ − x̄|] ÷ n (where x̄ is the mean and n is the number of observations) **Variance (σ²):** Variance = [Σ(xᵢ − x̄)²] ÷ n **Standard Deviation (σ):** σ = √Variance = √{[Σ(xᵢ − x̄)²] ÷ n} **Coefficient of Variation (CV):** CV = (σ ÷ x̄) × 100% (used to compare relative variability) **For Grouped Data:** Use class midmarks (xᵢ) in place of individual values, and weight each by its frequency (fᵢ): - Mean = [Σ(fᵢ × xᵢ)] ÷ Σfᵢ - Variance = [Σfᵢ(xᵢ − x̄)²] ÷ Σfᵢ **Quick interpretation checklist:** Smaller SD = data more concentrated; Larger SD = data more spread. SD = 0 means all values are identical.