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About Statistics — Class 10 Olympiad

Calculate mean, median, mode for grouped data using various methods; solve cumulative frequency problems. This topic is part of the Olympiad Class 10 mathematics syllabus (chapter: Module 11). On this page you can practice 57 questions across three difficulty levels — 19 easy, 18 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 35-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Statistics — solved examples for Class 10 Olympiad

Example 1easy

A factory measures the daily production of items. For 100 days, the average daily production was calculated as 500 items. However, it was later discovered that on 20 of these days, a faulty meter consistently under-reported the production by 10 items each day. What is the actual average daily production over these 100 days?
  1. A)498 items
  2. B)500 items
  3. C)502 items
  4. D)504 items

Step-by-step solution

  1. Let the reported total production be S_reported. The reported average is S_reported / 100 = 500. So, S_reported = 500 × 100 = 50000 items.
  2. On 20 days, the production was under-reported by 10 items each day. This means the total under-reported amount is 20 × 10 = 200 items.
  3. The actual total production, S_actual, is S_reported + 200 = 50000 + 200 = 50200 items.
  4. The actual average daily production is S_actual / 100 = 50200 / 100 = 502 items.

Answer: 502 items

Example 2medium

The mean of a grouped frequency distribution for 100 observations was calculated as 56.5. Later, it was discovered that the frequency of the class interval 50-60 was wrongly read as 20 instead of 30. The frequencies for other classes were correct: 30-40 (10), 40-50 (25), 60-70 (30), 70-80 (15). What is the corrected mean of the distribution?
  1. A)55.85
  2. B)56.36
  3. C)57.00
  4. D)58.18

Step-by-step solution

  1. The original total frequency N = 10 + 25 + 20 (incorrect) + 30 + 15 = 100.
  2. The original mean was 56.5. So, the original sum of products (Σfᵢxᵢ) = Mean × N = 56.5 × 100 = 5650.
  3. The class mark for the interval 50-60 is (50+60)/2 = 55. The incorrect contribution from this class was 20 × 55 = 1100. The correct contribution should be 30 × 55 = 1650.
  4. The change in sum of products = 1650 - 1100 = 550. The corrected sum of products = 5650 + 550 = 6200.
  5. The corrected total frequency N' = 10 + 25 + 30 (correct) + 30 + 15 = 110.
  6. The corrected mean = Corrected Σfᵢxᵢ / Corrected N' = 6200 / 110 ≈ 56.36.

Answer: 56.36

Example 3hard

A grouped frequency distribution of marks for 50 students is given. The mean marks is 30.
| Marks (Class Interval) | Number of Students (Frequency) |
| :--------------------- | :----------------------------- |
| 0-10 | 6 |
| 10-20 | 10 |
| 20-30 | p |
| 30-40 | q |
| 40-50 | 8 |
| 50-60 | 4 |
If p and q are positive integers and p < q, what is the value of q - p?
  1. A)12
  2. B)14
  3. C)16
  4. D)18

Step-by-step solution

  1. Calculate the total frequency: 6 + 10 + p + q + 8 + 4 = 28 + p + q. Since total students are 50, 28 + p + q = 50, which implies p + q = 22.
  2. Calculate the sum of (frequency × class mark): Class marks are 5, 15, 25, 35, 45, 55.
    Σ(fᵢxᵢ) = (6×5) + (10×15) + (p×25) + (q×35) + (8×45) + (4×55)
    = 30 + 150 + 25p + 35q + 360 + 220 = 760 + 25p + 35q.
  3. Use the mean formula: Mean = Σ(fᵢxᵢ) / Σfᵢ = (760 + 25p + 35q) / 50 = 30.
    760 + 25p + 35q = 1500 => 25p + 35q = 740. Dividing by 5, we get 5p + 7q = 148.
  4. Solve the system of equations: p + q = 22 and 5p + 7q = 148.
    From p + q = 22, substitute q = 22 - p into the second equation:
    5p + 7(22 - p) = 148
    5p + 154 - 7p = 148
    -2p = -6 => p = 3.
    Then q = 22 - 3 = 19.
    Check condition p < q: 3 < 19 is true.
    Finally, q - p = 19 - 3 = 16.

Answer: 16

Practice questions on Statistics

  1. Q1.easy

    Consider a grouped frequency distribution. If the median lies in the class interval 30-40, which of the following statements must be true about the cumulative frequency (cf_p) of the class preceding the median class and the frequency (f) of the median class?
    1. A)cf_p < f
    2. B)cf_p ≥ N/2
    3. C)cf_p < N/2 and cf_p + f ≥ N/2
    4. D)cf_p + f < N/2
    Show answer

    Answer: cf_p < N/2 and cf_p + f ≥ N/2

    Hint: Recall the formula for the median of grouped data and the conditions for identifying the median class. The median must fall within its class.

  2. Q2.easy

    A frequency distribution is given for daily wages:
    | Daily Wages (₹) | Number of Workers |
    |-----------------|-------------------|
    | 100-120 | 10 |
    | 120-140 | 15 |
    | 140-160 | x |
    | 160-180 | 25 |
    | 180-200 | 10 |
    If the mode of this distribution is ₹165, what is the value of x?
    1. A)20
    2. B)22
    3. C)23
    4. D)24
    Show answer

    Answer: 20

    Hint: The mode is ₹165, which identifies the modal class. Use the mode formula to find the missing frequency, x.

  3. Q3.easy

    A 'less than' ogive is plotted for the marks obtained by students in a test. If a point (40, 30) lies on this ogive, what does it signify?
    1. A)30 students scored exactly 40 marks.
    2. B)40 students scored less than 30 marks.
    3. C)30 students scored less than 40 marks.
    4. D)70 students scored more than 40 marks.
    Show answer

    Answer: 30 students scored less than 40 marks.

    Hint: Recall the definition of a 'less than' cumulative frequency distribution and how it's represented on an ogive.

  4. Q4.medium

    The heights of 50 students were recorded. The 'less than' cumulative frequency table is given below. Find the median height.
    1. A)148.5 cm
    2. B)149.2 cm
    3. C)150.0 cm
    4. D)150.5 cm
    Show answer

    Answer: 150.0 cm

    Hint: Convert the 'less than' cumulative frequency distribution into a standard frequency distribution first, then apply the median formula for grouped data.

  5. Q5.medium

    The mode of the following grouped frequency distribution is 36. Find the missing frequency 'x'.
    1. A)10
    2. B)12
    3. C)14
    4. D)16
    Show answer

    Answer: 10

    Hint: Since the mode is 36, the modal class is 30-40. Use the mode formula to set up an equation and solve for 'x'.

  6. Q6.medium

    For a particular grouped frequency distribution, a 'less than' ogive and a 'more than' ogive are drawn on the same graph paper. The intersection point of the two ogives is (X, Y). Which of the following statements is true regarding X and Y?
    1. A)X represents the Mean, Y represents the Total Frequency.
    2. B)X represents the Median, Y represents N/2 (half of total frequency).
    3. C)X represents the Mode, Y represents the Maximum Frequency.
    4. D)X represents the Mean, Y represents the Frequency of the Mean class.
    Show answer

    Answer: X represents the Median, Y represents N/2 (half of total frequency).

    Hint: Consider the definition of median and how cumulative frequency curves represent the total count up to a certain point.

  7. Q7.hard

    A continuous frequency distribution has a median of 32.5 and a total frequency of 80.
    | Class Interval | Frequency |
    | :------------- | :-------- |
    | 0-10 | 5 |
    | 10-20 | x |
    | 20-30 | y |
    | 30-40 | 20 |
    | 40-50 | z |
    | 50-60 | 8 |
    | 60-70 | 5 |
    If x, y, z are positive integers and the mode of the distribution is 35, what is the value of x - y?
    1. A)4
    2. B)6
    3. C)8
    4. D)10
    Show answer

    Answer: 6

    Hint: Use the median formula to find one relationship between x, y, and z. Then use the mode formula and the given mode value to establish another relationship.

  8. Q8.hard

    A grouped frequency distribution has classes 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 with frequencies 7, 12, k, 18, 10, 5 respectively. If the mode of the distribution lies in the class 30-40, which of the following statements must be true for the integer k?
    1. A)k < 12
    2. B)12 < k < 18
    3. C)k < 18
    4. D)k > 12
    Show answer

    Answer: k < 18

    Hint: For a class to be the modal class, its frequency must be the highest, and the mode formula requires its frequency to be greater than its adjacent frequencies.

  9. Q9.hard

    For a moderately skewed distribution, the mean is 45.8 and the mode is 42.2. If, due to a re-evaluation of data, the mean shifts to 48 and the mode shifts to 45, what is the change in the median of the distribution?
    1. A)1.2
    2. B)2.0
    3. C)2.4
    4. D)3.6
    Show answer

    Answer: 2.4

    Hint: The empirical relationship between mean, median, and mode for moderately skewed distributions is Mode ≈ 3 Median - 2 Mean. Use this formula to find the initial and final medians.

These are 9 of the 57 questions available for Statistics. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.