Keerthana Classes

Maths ch13 statistics

October 04, 2026

37 marks. Read the notes, then tap Start test. Answers and your score appear only after you submit.

Quick notes

Class mark = (upper limit + lower limit)/2.

Mean of grouped data: direct method x̄ = Σfᵢxᵢ/Σfᵢ. Assumed mean method x̄ = a + Σfᵢdᵢ/Σfᵢ where dᵢ = xᵢ − a. Step-deviation method x̄ = a + (Σfᵢuᵢ/Σfᵢ) × h where uᵢ = (xᵢ − a)/h.

Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h, where l is the lower limit of the modal class (highest frequency), f₁ its frequency, f₀ and f₂ the frequencies of the classes before and after it.

Median = l + [(n/2 − cf)/f] × h, where l is the lower limit of the median class, cf the cumulative frequency of the class before it, and f its frequency.

Empirical relation: 3 Median = Mode + 2 Mean.

Part A: Multiple choice 16 questions

1. The class mark of the class 10–20 is

  1. 10
  2. 15
  3. 20
  4. 30

(10 + 20)/2 = 15.

2. The mode of grouped data is given by

  1. l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
  2. l + [(n/2 − cf)/f] × h
  3. Σfx/Σf
  4. a + Σfd/Σf

This is the mode formula.

3. The median of grouped data is given by

  1. Σfx/Σf
  2. l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
  3. l + [(n/2 − cf)/f] × h
  4. a + h × Σfu/Σf

This is the median formula.

4. The modal class is the class with the

  1. lowest frequency
  2. middle position
  3. largest class mark
  4. highest frequency

Mode means most frequent.

5. The empirical relation between mean, median and mode is

  1. 3 Median = Mode + 2 Mean
  2. 3 Mean = Mode + 2 Median
  3. Mode = Mean + Median
  4. 2 Median = Mode + Mean

This is the empirical formula.

6. If mean = 20 and median = 21, then mode is

  1. 22
  2. 23
  3. 19
  4. 24

Mode = 3 × 21 − 2 × 20 = 23.

7. The measure of central tendency that shows the most frequent value is

  1. mean
  2. median
  3. mode
  4. range

Mode is the most frequent observation.

8. The mean of the first 5 natural numbers is

  1. 2.5
  2. 15
  3. 5
  4. 3

(1 + 2 + 3 + 4 + 5)/5 = 3.

9. In the assumed mean method, dᵢ equals

  1. xᵢ − a
  2. a − xᵢ
  3. xᵢ × a
  4. xᵢ/a

Deviation from the assumed mean.

10. Cumulative frequency is used to find the

  1. mean
  2. median
  3. mode
  4. class mark

The median class is found from the cumulative frequencies.

11. If the class marks are 5, 15, 25, …, the class size is

  1. 5
  2. 15
  3. 10
  4. 20

The difference between class marks is 10.

12. The median of 2, 4, 6, 8, 10 is

  1. 4
  2. 8
  3. 5
  4. 6

The middle value is 6.

13. If Σfᵢ = 50 and Σfᵢxᵢ = 1250, the mean is

  1. 25
  2. 50
  3. 125
  4. 12.5

1250/50 = 25.

14. The upper limit of the class 20–30 is

  1. 20
  2. 30
  3. 25
  4. 10

The larger boundary.

15. The mode of 3, 5, 5, 7, 5, 8 is

  1. 3
  2. 7
  3. 5
  4. 8

5 occurs three times.

16. For n = 40, the median class is the class whose cumulative frequency is just greater than

  1. 40
  2. 10
  3. 30
  4. 20

n/2 = 20.

Part B: Fill in the blanks 8 blanks

17. The class mark of the class 30–40 is [[35]].
18. If Σfx = 600 and Σf = 30, the mean is [[20]].
19. The class with the highest frequency is called the [[modal]] class.
20. 3 Median = Mode + [[2]] Mean.
21. If median = 15 and mean = 16, then mode = [[13]].
22. The mean of 10, 20, 30, 40 and 50 is [[30]].
23. The [[median]] divides the data into two equal halves.
24. The class size of the class 0–10 is [[10]].

Part C: Questions and answers 9 questions

25. Find the mean: classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 10, 20, 10, 5.

Class marks 5, 15, 25, 35, 45. Σfx = 25 + 150 + 500 + 350 + 225 = 1250, Σf = 50.
Mean = 1250/50 = 25.

26. Find the median: classes 0–10, 10–20, 20–30, 30–40, 40–50, 50–60 with frequencies 5, 8, 20, 15, 7, 5.

n = 60, n/2 = 30. Cumulative frequencies: 5, 13, 33, 48, 55, 60. Median class 20–30, l = 20, cf = 13, f = 20, h = 10.
Median = 20 + (30 − 13)/20 × 10 = 28.5.

27. Find the mode of the same data (frequencies 5, 8, 20, 15, 7, 5).

Modal class 20–30: l = 20, f₁ = 20, f₀ = 8, f₂ = 15, h = 10.
Mode = 20 + (12/17) × 10 = 27.06 (approx).

28. Find the mean of the same data using the assumed mean method with a = 25.

dᵢ = −20, −10, 0, 10, 20, 30. Σfd = −100 − 80 + 0 + 150 + 140 + 150 = 260.
Mean = 25 + 260/60 = 29.33 (approx).

29. The mean of the distribution below is 18. Find the missing frequency f.
Classes 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25 with frequencies 7, 6, 9, 13, f, 5, 4.

Class marks 12, 14, 16, 18, 20, 22, 24. Σf = 44 + f, Σfx = 752 + 20f.
(752 + 20f)/(44 + f) = 18 → 752 + 20f = 792 + 18f → f = 20.

30. The median of the distribution is 28.5 and the total frequency is 60. Find x and y.
0–10: 5, 10–20: x, 20–30: 20, 30–40: 15, 40–50: y, 50–60: 5.

45 + x + y = 60, so x + y = 15. Median class 20–30 with cf = 5 + x.
28.5 = 20 + [(30 − 5 − x)/20] × 10 → 8.5 = (25 − x)/2 → x = 8, so x = 8, y = 7.

31. Find the mode: 0–20: 6, 20–40: 8, 40–60: 10, 60–80: 12, 80–100: 6, 100–120: 5, 120–140: 3.

Modal class 60–80: l = 60, f₁ = 12, f₀ = 10, f₂ = 6, h = 20.
Mode = 60 + [2/(24 − 10 − 6)] × 20 = 60 + 5 = 65.

32. If the mode of some data is 24 and the mean is 30, find the median.

3 Median = Mode + 2 Mean = 24 + 60 = 84.
Median = 28.

33. When is the median a better measure of central tendency than the mean?

When the data has a few extreme values (very high or very low). The mean gets pulled by them, but the median does not. For example, the median is better for typical incomes.

Part D: Word puzzles 4 puzzles

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