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About Statistics (Mean, Median & Mode) — Class 10 ICSE

Compute mean, median, and mode for grouped frequency distributions and draw ogives. This topic is part of the ICSE Class 10 mathematics syllabus (chapter: Chapter 8). On this page you can practice 60 questions across three difficulty levels — 20 easy, 20 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 10-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Statistics (Mean, Median & Mode) — solved examples for Class 10 ICSE

Example 1easy

The mean of the first five natural numbers is:
  1. A)22
  2. B)33
  3. C)2.52.5
  4. D)3.53.5

Step-by-step solution

  1. Sum:
    1+2+3+4+5=151+2+3+4+5 = 15
  2. Mean:
    xˉ=155=3\bar{x} = \frac{15}{5} = 3

Answer: 33

Example 2medium

Using the assumed mean method with A=25A = 25, find the mean of:

| Class | 0100-10 | 102010-20 | 203020-30 | 304030-40 | 405040-50 |
|---|---|---|---|---|---|
| fif_i | 66 | 1010 | 1212 | 88 | 44 |
  1. A)2222
  2. B)2323
  3. C)2424
  4. D)2525

Step-by-step solution

  1. Class marks:
    xi=5,15,25,35,45x_i = 5, 15, 25, 35, 45
  2. Deviations:
    di=20,10,0,10,20d_i = -20, -10, 0, 10, 20
  3. Products:
    fidi=120,100,0,80,80f_i d_i = -120, -100, 0, 80, 80
  4. Sum:
    fidi=120100+0+80+80=60\sum f_i d_i = -120-100+0+80+80 = -60
  5. Total:
    fi=40\sum f_i = 40
  6. Mean:
    xˉ=25+6040=251.5=23.5\bar{x} = 25 + \frac{-60}{40} = 25 - 1.5 = 23.5

Answer: 2323

Example 3hard

The mean of the following distribution is 57.657.6 and the total frequency is 5050. Find f1f_1 and f2f_2.

| Class | 0200-20 | 204020-40 | 406040-60 | 608060-80 | 8010080-100 |
|---|---|---|---|---|---|
| ff | 77 | f1f_1 | 1212 | f2f_2 | 88 |
  1. A)f1=8,f2=15f_1 = 8, f_2 = 15
  2. B)f1=10,f2=13f_1 = 10, f_2 = 13
  3. C)f1=6,f2=17f_1 = 6, f_2 = 17
  4. D)f1=9,f2=14f_1 = 9, f_2 = 14

Step-by-step solution

  1. From total:
    7+f1+12+f2+8=50    f1+f2=23(1)7 + f_1 + 12 + f_2 + 8 = 50 \implies f_1 + f_2 = 23 \quad \cdots(1)
  2. Class marks:
    10,30,50,70,9010, 30, 50, 70, 90
  3. Sum:
    fixi=70+30f1+600+70f2+720=1390+30f1+70f2\sum f_i x_i = 70 + 30f_1 + 600 + 70f_2 + 720 = 1390 + 30f_1 + 70f_2
  4. Mean = 57.6:
    1390+30f1+70f250=57.6\frac{1390 + 30f_1 + 70f_2}{50} = 57.6
  5. So:
    30f1+70f2=1490    3f1+7f2=149(2)30f_1 + 70f_2 = 1490 \implies 3f_1 + 7f_2 = 149 \quad \cdots(2)
  6. From (1): f1=23f2f_1 = 23 - f_2. Substituting:
    3(23f2)+7f2=149    69+4f2=149    f2=203(23-f_2) + 7f_2 = 149 \implies 69 + 4f_2 = 149 \implies f_2 = 20

Answer: f1=8,f2=15f_1 = 8, f_2 = 15

Practice questions on Statistics (Mean, Median & Mode)

  1. Q1.easy

    The mode of the data set {2,3,5,3,7,3,8}\{2, 3, 5, 3, 7, 3, 8\} is:
    1. A)22
    2. B)33
    3. C)55
    4. D)77
    Show answer

    Answer: 33

    Hint: Mode is the value that appears most frequently.

  2. Q2.easy

    The median of the data {4,7,2,9,5}\{4, 7, 2, 9, 5\} (when arranged in order) is:
    1. A)44
    2. B)55
    3. C)77
    4. D)22
    Show answer

    Answer: 55

    Hint: First arrange in ascending order, then pick the middle value.

  3. Q3.easy

    For a grouped frequency distribution, the mean is given by:
    1. A)xˉ=fixifi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}
    2. B)xˉ=fixi\bar{x} = \frac{\sum f_i}{\sum x_i}
    3. C)xˉ=fixi\bar{x} = \sum f_i x_i
    4. D)xˉ=xin\bar{x} = \frac{\sum x_i}{n}
    Show answer

    Answer: xˉ=fixifi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

    Hint: Mean = (sum of all fx) / (sum of all frequencies).

  4. Q4.medium

    Find the mode of the following distribution:

    | Class | 102010-20 | 203020-30 | 304030-40 | 405040-50 | 506050-60 |
    |---|---|---|---|---|---|
    | ff | 55 | 88 | 1212 | 66 | 44 |

    Using the mode formula: l+f1f02f1f0f2×hl + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h
    1. A)33.3333.33
    2. B)34.1734.17
    3. C)3535
    4. D)36.2536.25
    Show answer

    Answer: 34.1734.17

    Hint: Modal class has highest frequency. Identify f0f_0, f1f_1, f2f_2 and use the formula.

  5. Q5.medium

    Find the median from the following cumulative frequency distribution:

    | Less than | 1010 | 2020 | 3030 | 4040 | 5050 |
    |---|---|---|---|---|---|
    | cf | 33 | 1212 | 2727 | 3737 | 4040 |
    1. A)2525
    2. B)26.6726.67
    3. C)27.7827.78
    4. D)28.528.5
    Show answer

    Answer: 26.6726.67

    Hint: First find n2\frac{n}{2}, identify the median class, then apply the median formula.

  6. Q6.medium

    Using the step deviation method with A=35A = 35 and h=10h = 10, find the mean of:

    | Class | 102010-20 | 203020-30 | 304030-40 | 405040-50 | 506050-60 |
    |---|---|---|---|---|---|
    | fif_i | 44 | 88 | 1010 | 1212 | 66 |
    1. A)35.535.5
    2. B)3636
    3. C)3737
    4. D)37.537.5
    Show answer

    Answer: 3737

    Hint: Find ui=xiAhu_i = \frac{x_i - A}{h}, then xˉ=A+h×fiuifi\bar{x} = A + h \times \frac{\sum f_i u_i}{\sum f_i}.

  7. Q7.hard

    Three groups of students have means 4040, 5050, and 6060 with 2020, 3030, and 5050 students respectively. The combined mean of all 100100 students is:
    1. A)5050
    2. B)5151
    3. C)5252
    4. D)5353
    Show answer

    Answer: 5252

    Hint: Combined mean = (n1*mean1 + n2*mean2 + n3*mean3) / (n1 + n2 + n3).

  8. Q8.hard

    In a distribution, the mean is 4545, variance is 100100. If each observation is first multiplied by 22 and then 55 is added, what is the new mean and variance?
    1. A)Mean =95= 95, Variance =400= 400
    2. B)Mean =95= 95, Variance =200= 200
    3. C)Mean =90= 90, Variance =400= 400
    4. D)Mean =95= 95, Variance =100= 100
    Show answer

    Answer: Mean =95= 95, Variance =400= 400

    Hint: If y=ax+by = ax + b, then yˉ=axˉ+b\bar{y} = a\bar{x} + b and Var(y)=a2Var(x)\text{Var}(y) = a^2 \text{Var}(x).

  9. Q9.hard

    The median of the following data is 3232. Find xx and yy if total frequency is 100100.

    | Class | 0100-10 | 102010-20 | 203020-30 | 304030-40 | 405040-50 | 506050-60 |
    |---|---|---|---|---|---|---|
    | ff | 1010 | xx | 2525 | 3030 | yy | 1010 |
    1. A)x=9,y=16x=9, y=16
    2. B)x=12,y=13x=12, y=13
    3. C)x=15,y=10x=15, y=10
    4. D)x=8,y=17x=8, y=17
    Show answer

    Answer: x=9,y=16x=9, y=16

    Hint: Use two conditions: total frequency = 100, and median = 32 gives an equation from the median formula.

These are 9 of the 60 questions available for Statistics (Mean, Median & Mode). Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.