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Chapter 12 · Class 11 Mathematics

Statistics — Important Questions

34 questions With answers CBSE format

SUMMARY: The chapter on Statistics in Class 11 Mathematics focuses on the collection, presentation, analysis, and interpretation of numerical data.
KEY TOPICS: Measures of central tendency, measures of dispersion, mean deviation, variance, standard deviation, analysis of frequency distributions, graphical representation of data, cumulative frequency curves, skewness, kurtosis.

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Multiple Choice Questions (MCQs) 5 questions
Q1 1 Mark

The mean of the data 2 4 6 8 10 is:

A5
B6
C7
D8
Check answerHide answer
Correct answer: Option 2 — 6
Q2 1 Mark

The median of the data 1 3 5 7 9 is:

A3
B5
C7
D9
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Correct answer: Option 2 — 5
Q3 1 Mark

The mode of the data 2 3 4 4 5 6 6 6 7 is:

A4
B5
C6
D7
Check answerHide answer
Correct answer: Option 3 — 6
Q4 1 Mark

Variance is defined as:

AMean of squared deviations from the mean
BSquare of the mean
CRange squared
DMean times standard deviation
Check answerHide answer
Correct answer: Option 1 — Mean of squared deviations from the mean
Q5 1 Mark

The standard deviation of 5 5 5 5 5 is:

A0
B1
C5
D25
Check answerHide answer
Correct answer: Option 1 — 0
Short Answer Questions 5 questions
Q6 3 Marks

Find the mean of 5 10 15 20 25.

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Mean = (5 + 10 + 15 + 20 + 25)/5 = 75/5 = 15.
Q7 3 Marks

Define mean median and mode of a data set.

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Mean: arithmetic average = sum/count. Median: middle value when data is sorted (or average of the two middle values for even count). Mode: most frequently occurring value(s).
Q8 3 Marks

Find the variance of the data 4 6 8 10 12.

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Mean = 8. Deviations: −4 −2 0 2 4 with squared deviations 16 4 0 4 16; sum = 40. Variance = 40/5 = 8.
Q9 3 Marks

Find the median of the data 7 2 9 4 5 11 3.

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Sorted: 2 3 4 5 7 9 11. Middle value (4th of 7) is 5. Median = 5.
Q10 3 Marks

If mean and standard deviation of a data are 30 and 6 respectively find the coefficient of variation.

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Coefficient of variation (CV) = (standard deviation / mean) × 100% = (6/30) × 100 = 20%.
Long Answer Questions 6 questions
Q11 6 Marks

Find the mean variance and standard deviation of the data 6 8 10 12 14 16 18.

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Mean = (6+8+10+12+14+16+18)/7 = 84/7 = 12. Deviations from mean: −6 −4 −2 0 2 4 6. Squared deviations: 36 16 4 0 4 16 36; sum = 112. Variance = 112/7 = 16. SD = √16 = 4.
Q12 6 Marks

The mean of 8 observations is 25. If two observations 11 and 19 are deleted what is the mean of the remaining 6 observations?

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Sum of original 8 = 25 × 8 = 200. After deleting 11 and 19: new sum = 200 − 11 − 19 = 170. Number of remaining observations = 6. New mean = 170/6 ≈ 28.33.
Q13 6 Marks

For the frequency distribution: x: 10 20 30 40 50; f: 4 6 8 7 5, find the mean.

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Σf = 4 + 6 + 8 + 7 + 5 = 30. Σf x = 40 + 120 + 240 + 280 + 250 = 930. Mean = Σf x / Σf = 930/30 = 31.
Q14 6 Marks

Find the variance of the first 10 natural numbers.

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Mean = (1 + 2 + ... + 10)/10 = 55/10 = 5.5. Squared deviations: (1 − 5.5)² + (2 − 5.5)² + ... + (10 − 5.5)² = 20.25 + 12.25 + 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 + 12.25 + 20.25 = 82.5. Variance = 82.5/10 = 8.25.
Q15 6 Marks

Compute the coefficient of variation of the data 6 7 10 12 13 4 8 12 with mean 9.

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Squared deviations from mean 9: (−3)² + (−2)² + 1² + 3² + 4² + (−5)² + (−1)² + 3² = 9 + 4 + 1 + 9 + 16 + 25 + 1 + 9 = 74. Variance = 74/8 = 9.25. SD = √9.25 ≈ 3.04. CV = (3.04/9) × 100 ≈ 33.78%.
Q16 6 Marks

Compare measures of central tendency: mean median and mode with the help of a table.

Assertion–Reason Questions 5 questions
Q17 1 Mark

Assertion (A): The mean of a data set is the sum of values divided by the number of values.

Reason (R): The mean represents a typical or average value of the data.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q18 1 Mark

Assertion (A): The variance of a data set is always non-negative.

Reason (R): Variance is the average of squared deviations from the mean and squares are non-negative.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q19 1 Mark

Assertion (A): The standard deviation has the same units as the original data.

Reason (R): SD is the positive square root of variance which has units that are the square of the data units.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q20 1 Mark

Assertion (A): If all values in a data set are identical the variance is 0.

Reason (R): There is no spread when all values are equal so the deviations from the mean are all zero.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q21 1 Mark

Assertion (A): The coefficient of variation is dimensionless.

Reason (R): CV is the ratio of SD to the mean expressed as a percentage so units cancel.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Statement-Based Questions 5 questions
Q22 1 Mark

Statement 1: The mean median and mode are measures of central tendency.

Statement 2: The range and variance are measures of dispersion.

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Correct answer: Option 1 — Both statements are true.
Q23 1 Mark

Statement 1: The mean is sensitive to outliers.

Statement 2: The median is more robust to extreme values than the mean.

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Correct answer: Option 1 — Both statements are true.
Q24 1 Mark

Statement 1: Variance is the square of the standard deviation.

Statement 2: Both variance and SD measure the spread of the data.

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Correct answer: Option 1 — Both statements are true.
Q25 1 Mark

Statement 1: The mean of a frequency distribution equals Σ(f x)/Σf.

Statement 2: The same formula works for ungrouped data when each value has frequency 1.

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Correct answer: Option 1 — Both statements are true.
Q26 1 Mark

Statement 1: Coefficient of variation is useful for comparing dispersion of data sets with different units.

Statement 2: A higher CV indicates greater relative variability.

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Correct answer: Option 1 — Both statements are true.
Case Study / Passage Questions 3 questions
Q27 3 Marks
A teacher records the marks of 10 students in a Maths test as 35 42 47 50 52 55 58 60 65 70. The teacher wants to find the mean median range and standard deviation.
  1. The mean of the marks is:
    A52
    B53
    C53.4
    D55
  2. The median of the marks is:
    A52
    B53
    C53.4
    D55
  3. Compute the range and standard deviation.
Show answersHide answers
1. Option 3 — 53.4
2. Option 3 — 53.4
3. Mean = (35 + 42 + 47 + 50 + 52 + 55 + 58 + 60 + 65 + 70)/10 = 534/10 = 53.4. For median: with 10 (even) values sorted, median = (5th + 6th)/2 = (52 + 55)/2 = 53.5. Range = 70 − 35 = 35. Standard deviation requires squared deviations from the mean.
Q28 3 Marks
Two teams' run scores in 5 matches are: Team A: 30 50 40 60 70; Team B: 50 50 50 50 50. The captain wants to compare consistency using SD and CV.
  1. The team with smaller standard deviation is:
    ATeam A
    BTeam B
    CBoth equal
    DCannot decide
  2. The team that is more consistent is:
    ATeam A
    BTeam B
    CBoth equal
    DCannot decide
  3. Compute the coefficients of variation for both teams.
Show answersHide answers
1. Option 2 — Team B
2. Option 2 — Team B
3. Team A: mean = 50; deviations −20 0 −10 10 20; squared 400 0 100 100 400; variance = 1000/5 = 200; SD = √200 ≈ 14.14. Team B: all values equal mean; SD = 0. Team B is more consistent (SD = 0).
Q29 3 Marks
A survey of monthly expenditure (in ₹ 000) of 5 households gives the values 4 5 6 7 8. The analyst wants the variance and standard deviation.
  1. The variance equals:
    A2
    B4
    C6
    D8
  2. The standard deviation equals:
    A√2
    B2
    C√5
    D5
  3. Compute the coefficient of variation.
Show answersHide answers
1. Option 1 — 2
2. Option 1 — √2
3. Mean = (4 + 5 + 6 + 7 + 8)/5 = 6. Squared deviations: 4 1 0 1 4; sum = 10. Variance = 10/5 = 2. SD = √2 ≈ 1.414.
Table-Based Questions 4 questions
Q30 3 Marks

Study summary statistics for a dataset:

StatisticValue
Sample size20
Mean50
Variance25
Standard deviation5
Coefficient of variation10%
  1. The standard deviation equals:
    A5
    B10
    C25
    D50
  2. The coefficient of variation equals:
    A5%
    B10%
    C20%
    D50%
  3. Compute the variance if SD doubled to 10.
Show answersHide answers
1. Option 1 — 5
2. Option 2 — 10%
3. SD = √Variance = √25 = 5. CV = (SD/Mean) × 100% = (5/50) × 100 = 10%. CV is dimensionless and useful for comparing dispersions across datasets with different units or scales.
Q31 3 Marks

Study a frequency distribution:

xf
103
205
307
404
501
  1. The mean of the distribution equals:
    A20
    B25
    C28
    D30
  2. The frequency at x = 20 equals:
    A20
    B3
    C5
    D7
  3. Compute the mode of the distribution.
Show answersHide answers
1. Option 3 — 28
2. Option 3 — 5
3. Mean = Σf x / Σf = (10·3 + 20·5 + 30·7 + 40·4 + 50·1)/(3 + 5 + 7 + 4 + 1) = (30 + 100 + 210 + 160 + 50)/20 = 550/20 = 27.5. Approximate option: 28.
Q32 6 Marks

For the dataset 4, 6, 8, 10, 12, compute (i) the mean, (ii) the variance, (iii) the standard deviation, (iv) the coefficient of variation.

StepValue
Sample size n5
Mean?
Variance?
SD?
CV%?
Q33 6 Marks

For the frequency distribution given, compute the mean, median (approx) and modal class.

xFrequency f
104
206
308
407
505
Picture-Based Questions 1 question
Q34 3 Marks

Study the histogram of class marks and answer:

Statistics figure
  1. The modal class (highest bar) is:
    A20-30
    B30-40
    C40-50
    D50-60
  2. The total frequency (number of students) is:
    A50
    B60
    C75
    D100
  3. Describe the shape of the distribution and the location of the mode.
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1. Option 3 — 40-50
2. Option 3 — 75
3. Total = 3 + 8 + 14 + 20 + 16 + 9 + 5 = 75 students. The modal class is 40-50 with frequency 20. The distribution is roughly symmetric and bell-shaped, with mean and median both close to the modal class — typical of well-designed exams.
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