Class 10 Mathematics: Statistics important questions

Statistics is a chapter of Class 10 Mathematics for the CBSE. These 16 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Practice · 1 marks

   1. If the difference of the mode and median of a data set is 24, the difference of the median and mean is: (a) 8 (b) 12 (c) 24 (d) 36

   (b) 12 Mode = 3 Median − 2 Mean, so Mode − Median = 2(Median − Mean) = 24 and Median − Mean = 12.

2. Practice · 1 marks

   2. For the distribution Class: 0–5, 5–10, 10–15, 15–20, 20–25 Frequency: 10, 15, 12, 20, 9 the sum of the lower limits of the median class and the modal class is: (a) 15 (b) 25 (c) 30 (d) 35

   (b) 25 N = 66, N/2 = 33; cumulative frequencies 10, 25, 37, … so the median class is 10–15. The modal class is 15–20. 10 + 15 = 25.

3. Very Important · 5 marks

   3. The median of the following data is 525 and the total frequency is 100. Find x and y. Class: 0–100, 100–200, 200–300, 300–400, 400–500, 500–600, 600–700, 700–800, 800–900, 900–1000 Frequency: 2, 5, x, 12, 17, 20, y, 9, 7, 4

   76 + x + y = 100, so x + y = 24. The median 525 lies in 500–600: l = 500, f = 20, cf = 36 + x, h = 100. 525 = 500 + [(50 − 36 − x)/20] × 100, so 14 − x = 5 and x = 9; y = 15.

4. Practice · 1 marks

   4. The upper limit of the modal class of this distribution is: Height (cm): below 140, below 145, below 150, below 155, below 160, below 165 Number of girls: 4, 11, 29, 40, 46, 51 (a) 165 (b) 160 (c) 155 (d) 150

   (d) 150 Class frequencies: 4, 7, 18, 11, 6, 5. The highest is 18, for 145–150.

5. Important · 3 marks

   5. The lengths of 40 leaves of a plant, measured to the nearest millimetre: Length (mm): 118–126, 127–135, 136–144, 145–153, 154–162, 163–171, 172–180 Number of leaves: 3, 5, 9, 12, 5, 4, 2 Find the mean length of the leaves.

   Class marks: 122, 131, 140, 149, 158, 167, 176. Take a = 149, d = x − 149. Σfd = −81 − 90 − 81 + 0 + 45 + 72 + 54 = −81, Σf = 40. Mean = 149 − 81/40 = 149 − 2.025 = 146.975 mm.

6. Very Important · 5 marks

   6. The median of this data is 50 and the total frequency is 90. Find p and q, and the mode. Marks: 20–30, 30–40, 40–50, 50–60, 60–70, 70–80, 80–90 Students: p, 15, 25, 20, q, 8, 10

   p + q + 78 = 90, so p + q = 12. Median 50 lies at the start of the class 50–60: 50 = 50 + [(45 − (p + 40))/20] × 10, so p = 5 and q = 7. Modal class 40–50 (f₁ = 25, f₀ = 15, f₂ = 20): Mode = 40 + [10/(50 − 35)] × 10 = 46.67.

7. Practice · 1 marks

   7. For the frequency distribution of 45 observations: Class: 0–10, 10–20, 20–30, 30–40, 40–50 Frequency: 5, 9, 15, 10, 6 the upper limit of the median class is: (A) 20 (B) 10 (C) 30 (D) 40

   (C) 30 N/2 = 22.5. Cumulative frequencies are 5, 14, 29, …, so the median class is 20–30, with upper limit 30.

8. Practice · 1 marks

   8. Which of the following gives the middle-most observation of the data? (A) Median (B) Mean (C) Range (D) Mode

   (A) Median

9. Very Important · 5 marks

   9. Find the mean and median of the data: Class: 85–90, 90–95, 95–100, 100–105, 105–110, 110–115 Frequency: 15, 22, 20, 18, 20, 25

   Take a = 102.5, h = 5, u = (x − 102.5)/5. Σf = 120, Σfu = −45 − 44 − 20 + 0 + 20 + 50 = −39. Mean = 102.5 + 5 × (−39/120) = 100.875. Cumulative frequencies: 15, 37, 57, 75, 95, 120. N/2 = 60, so the median class is 100–105 (l = 100, cf = 57, f = 18). Median = 100 + [(60 − 57)/18] × 5 ≈ 100.83.

10. Very Important · 5 marks

    10. The monthly expenditure on milk in 200 families is: Expenditure (₹): 1000–1500, 1500–2000, 2000–2500, 2500–3000, 3000–3500, 3500–4000, 4000–4500, 4500–5000 Families: 24, 40, 33, x, 30, 22, 16, 7 Find x and the mean expenditure.

    172 + x = 200, so x = 28. Class marks 1250, 1750, …, 4750; Σfx = 30000 + 70000 + 74250 + 77000 + 97500 + 82500 + 68000 + 33250 = 5,32,500. Mean = 532500/200 = ₹2662.50.

11. Practice · 1 marks

    11. The median of 9 distinct observations is 20.5. If each observation is increased by 2, the median of the new set: (A) increases by 2 (B) decreases by 2 (C) doubles (D) stays the same

    (A) It increases by 2: every value, including the middle one, goes up by 2.

12. Very Important · 5 marks

    12. The mode of this distribution is 55. Find x, and then the mean. Class: 0–15, 15–30, 30–45, 45–60, 60–75, 75–90 Frequency: 10, 7, x, 15, 10, 12

    Modal class 45–60: l = 45, h = 15, f₁ = 15, f₀ = x, f₂ = 10. 55 = 45 + [(15 − x)/(30 − x − 10)] × 15 gives 10(20 − x) = 15(15 − x), so x = 5. Class marks 7.5, 22.5, 37.5, 52.5, 67.5, 82.5; Σf = 59, Σfx = 2872.5. Mean = 2872.5/59 ≈ 48.69.

13. Very Important · 5 marks

    13. Heights of 51 girls (cm): less than 140: 4; less than 145: 11; less than 150: 29; less than 155: 40; less than 160: 46; less than 165: 51. Find the median height. If the mode is 148.05, find the mean using the empirical formula.

    Classes: 135–140: 4, 140–145: 7, 145–150: 18, 150–155: 11, 155–160: 6, 160–165: 5. N/2 = 25.5, so the median class is 145–150 (l = 145, cf = 11, f = 18, h = 5). Median = 145 + [(25.5 − 11)/18] × 5 = 145 + 4.03 ≈ 149.03 cm. Mode = 3 Median − 2 Mean, so Mean = (3 × 149.03 − 148.05)/2 ≈ 149.52 cm.

14. Practice · 1 marks

    14. The mean age of a combined group of men and women is 35 years. The mean ages of men and women are 38 and 30 years. The percentage of women in the group is: (A) 15 (B) 25.5 (C) 35 (D) 37.5

    (D) 37.5 With x men and y women: (38x + 30y)/(x + y) = 35, so 3x = 5y and x : y = 5 : 3. Women = 3/8 = 37.5%.

15. Practice · 1 marks

    15. The mean and median of some data are 35.5 and 32. The mode is: (A) 23 (B) 26 (C) 25 (D) 30

    (C) 25 Mode = 3 × Median − 2 × Mean = 96 − 71 = 25.

16. Very Important · 4 marks

    16. Weekly test marks (out of 50) of Class X students: Marks: 0–10, 10–20, 20–30, 30–40, 40–50 Students: 3, 6, 12, 15, 14 (i) Which class has the most common performance (modal class)? (ii) Find the median class. (iii) (A) Find the mean. OR (B) Find the mode.

    (i) 30–40. (ii) Cumulative frequencies: 3, 9, 21, 36, 50. N/2 = 25, so the median class is 30–40. (iii) (A) Class marks 5, 15, 25, 35, 45; Σfx = 15 + 90 + 300 + 525 + 630 = 1560. Mean = 1560/50 = 31.2. OR (B) Mode = 30 + [(15 − 12)/(30 − 12 − 14)] × 10 = 30 + 7.5 = 37.5.