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NCERT Class 10 · Mathematics Based on NCERT Class 10 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter extends the study of measures of central tendency—mean, median, and mode—from ungrouped data to grouped data. It introduces methods to calculate the mean for grouped data (direct method, assumed mean method, and step-deviation method), the formula for finding the mode of grouped data, and the concept of cumulative frequency distributions for calculating the median. The one big idea a student must walk away with is that for large datasets, data is grouped into class intervals, and the mean, median, and mode can be accurately estimated using specific formulas that rely on class marks, frequencies, and cumulative frequencies.
Key Concepts (14)
Mean of grouped data
The mean (or average) of observations is the sum of the values of all the observations divided by the total number of observations. For grouped data, it is calculated using class marks as representative values of each class interval.
e.g. In a class of 30 students, the mean marks obtained is 59.3 when calculated from ungrouped data, but 62 when calculated from grouped data using class marks.
Class mark
The mid-point of a class interval, found by averaging its upper and lower limits. It serves as the representative value for all observations in that class.
e.g. For the class interval 10-25, the class mark is (10+25)/2 = 17.5.
Direct Method
A method to find the mean of grouped data by directly using the formula: mean = (sum of fixi) / (sum of fi), where xi is the class mark and fi is the frequency.
e.g. Using Table 13.3, mean = 1860.0 / 30 = 62.
Assumed Mean Method
A method to find the mean of grouped data by choosing an assumed mean 'a' (usually a central class mark), calculating deviations di = xi - a, and then using the formula: mean = a + (sum of fidi) / (sum of fi).
e.g. With a = 47.5, sum of fidi = 435, sum of fi = 30, mean = 47.5 + 435/30 = 62.
Step-deviation Method
A simplified form of the assumed mean method used when all deviations have a common factor (class size h). It uses ui = (xi - a)/h and the formula: mean = a + h * (sum of fiui) / (sum of fi).
e.g. With a = 47.5, h = 15, sum of fiui = 29, sum of fi = 30, mean = 47.5 + 15*(29/30) = 62.
Mode of grouped data
The value that occurs most often in a dataset. For grouped data, it is a value inside the modal class (the class with the maximum frequency), calculated using a specific formula.
e.g. In Example 5, the modal class is 3-5, and the mode is calculated as 3.286.
Modal class
The class interval in a grouped frequency distribution that has the highest frequency.
e.g. In Table 13.3, the class 40-55 has the highest frequency (7), so it is the modal class.
Cumulative frequency
The frequency obtained by adding the frequencies of all classes preceding a given class. It can be of 'less than type' (using upper limits) or 'more than type' (using lower limits).
e.g. In Table 13.13, the cumulative frequency for 'less than 20' is 8 (5+3).
Median of grouped data
The value of the middle-most observation in a dataset. For grouped data, it is found using the median class and the formula: median = l + [(n/2 - cf)/f] * h.
e.g. In Table 13.15, n=53, n/2=26.5, median class is 60-70, median = 60 + [(26.5-22)/7]*10 = 66.4.
Median class
The class interval whose cumulative frequency is greater than and nearest to n/2.
e.g. In Table 13.15, n/2 = 26.5, and the cumulative frequency of class 60-70 is 29, so it is the median class.
Cumulative frequency distribution of the less than type
A table showing the number of observations with values less than the upper limit of each class interval.
e.g. Table 13.13 shows the number of students with marks less than 10, less than 20, etc.
Cumulative frequency distribution of the more than type
A table showing the number of observations with values more than or equal to the lower limit of each class interval.
e.g. Table 13.14 shows the number of students with marks more than or equal to 0, more than or equal to 10, etc.
Ogive
A cumulative frequency curve, drawn by plotting cumulative frequencies against class boundaries. It can be 'less than ogive' or 'more than ogive'.
e.g. The chapter mentions ogives as cumulative frequency curves, but does not provide a specific example.
Empirical relationship between mean, median, and mode
An approximate relationship given by: 3 Median = Mode + 2 Mean.
e.g. If mean = 62 and mode = 52, then 3*Median = 52 + 2*62 = 176, so Median ≈ 58.67.
Formulas & Equations (6)
x̄ = (Σfixi) / (Σfi)
Reads as: x bar equals sum of f i x i divided by sum of f i
Variables: x̄ = mean, Σ = summation, fi = frequency of i-th class, xi = class mark of i-th class
Use when: Use this direct method formula when xi and fi are small numbers, making calculations straightforward.
⚠ Students often forget to sum the frequencies (Σfi) in the denominator and instead use the number of classes.
x̄ = a + (Σfidi) / (Σfi)
Reads as: x bar equals a plus sum of f i d i divided by sum of f i
Variables: a = assumed mean, di = xi - a (deviation of class mark from assumed mean)
Use when: Use the assumed mean method when xi and fi are numerically large, to simplify calculations.
⚠ Students incorrectly calculate di as a - xi instead of xi - a, leading to a sign error in the final mean.
x̄ = a + h * (Σfiui) / (Σfi)
Reads as: x bar equals a plus h times sum of f i u i divided by sum of f i
Variables: a = assumed mean, h = class size, ui = (xi - a)/h (step deviation)
Use when: Use the step-deviation method when all deviations di have a common factor (class size h), to further reduce calculation work.
⚠ Students forget to multiply h with the fraction (Σfiui)/(Σfi) and simply add a to the fraction.
Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] * h
Reads as: Mode equals l plus (f1 minus f0) divided by (2 f1 minus f0 minus f2) times h
Variables: l = lower limit of modal class, h = class size, f1 = frequency of modal class, f0 = frequency of class preceding modal class, f2 = frequency of class succeeding modal class
Use when: Use this formula to find the mode of grouped data when the modal class is identified.
⚠ Students often confuse f0 and f2, swapping the frequencies of the preceding and succeeding classes.
Median = l + [(n/2 - cf) / f] * h
Reads as: Median equals l plus (n by 2 minus cf) divided by f times h
Variables: l = lower limit of median class, n = total number of observations, cf = cumulative frequency of class preceding median class, f = frequency of median class, h = class size
Use when: Use this formula to find the median of grouped data after identifying the median class.
⚠ Students use the cumulative frequency of the median class itself instead of the cumulative frequency of the class preceding the median class for cf.
3 Median = Mode + 2 Mean
Reads as: 3 times Median equals Mode plus 2 times Mean
Variables: Median, Mode, Mean are the three measures of central tendency
Use when: Use this empirical relationship to approximate one measure when the other two are known, or to verify consistency of calculated values.
⚠ Students incorrectly rearrange the formula, e.g., writing Median = (Mode + 2Mean)/3 but forgetting the division by 3.
How It Works
Calculating Mean of Grouped Data using Direct Method
  1. Step 1: Convert ungrouped data into grouped data by forming class intervals of equal width.
  2. Step 2: Find the class mark (xi) for each class interval using the formula: (Upper class limit + Lower class limit)/2.
  3. Step 3: Multiply each class mark (xi) by its corresponding frequency (fi) to get fixi.
  4. Step 4: Sum all fixi values to get Σfixi.
  5. Step 5: Sum all frequencies to get Σfi.
  6. Step 6: Apply the formula: mean = Σfixi / Σfi.
💡 The class mark is assumed to be the representative of all observations in that class, which introduces approximation; the mean obtained is an approximate mean, not the exact mean.
Calculating Mean of Grouped Data using Assumed Mean Method
  1. Step 1: Choose an assumed mean 'a' from among the class marks, preferably a central one.
  2. Step 2: Calculate the deviation di = xi - a for each class.
  3. Step 3: Multiply each deviation di by its frequency fi to get fidi.
  4. Step 4: Sum all fidi values to get Σfidi.
  5. Step 5: Sum all frequencies to get Σfi.
  6. Step 6: Apply the formula: mean = a + (Σfidi / Σfi).
💡 The value of the mean does not depend on the choice of 'a'; any class mark can be chosen as the assumed mean and the result will be the same.
Calculating Mean of Grouped Data using Step-deviation Method
  1. Step 1: Choose an assumed mean 'a' and class size 'h'.
  2. Step 2: Calculate the step deviation ui = (xi - a)/h for each class.
  3. Step 3: Multiply each ui by its frequency fi to get fiui.
  4. Step 4: Sum all fiui values to get Σfiui.
  5. Step 5: Sum all frequencies to get Σfi.
  6. Step 6: Apply the formula: mean = a + h * (Σfiui / Σfi).
💡 The step-deviation method is most convenient when all deviations di have a common factor (the class size h).
Finding Mode of Grouped Data
  1. Step 1: Identify the modal class — the class interval with the highest frequency.
  2. Step 2: Note the lower limit (l) of the modal class, class size (h), frequency of modal class (f1), frequency of class preceding modal class (f0), and frequency of class succeeding modal class (f2).
  3. Step 3: Substitute these values into the formula: Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] * h.
💡 The class intervals must be continuous before applying the formula; if they are not, adjust them by adding/subtracting half the gap between consecutive classes.
Finding Median of Grouped Data
  1. Step 1: Construct a cumulative frequency table (less than type) for the data.
  2. Step 2: Calculate n/2, where n is the total number of observations.
  3. Step 3: Identify the median class — the class whose cumulative frequency is greater than and nearest to n/2.
  4. Step 4: Note the lower limit (l) of the median class, cumulative frequency (cf) of the class preceding the median class, frequency (f) of the median class, and class size (h).
  5. Step 5: Substitute these values into the formula: Median = l + [(n/2 - cf) / f] * h.
💡 The cumulative frequency used in the formula is of the class preceding the median class, not of the median class itself.
Constructing Cumulative Frequency Distribution of Less Than Type
  1. Step 1: List the upper limits of each class interval.
  2. Step 2: For the first class, the cumulative frequency is its own frequency.
  3. Step 3: For each subsequent class, add its frequency to the cumulative frequency of the previous class.
  4. Step 4: The final cumulative frequency equals the total number of observations (n).
💡 The cumulative frequency for a class represents the number of observations with values less than the upper limit of that class.
Constructing Cumulative Frequency Distribution of More Than Type
  1. Step 1: List the lower limits of each class interval.
  2. Step 2: Start with the total number of observations (n) for the first class (more than or equal to the lowest limit).
  3. Step 3: Subtract the frequency of each class from the cumulative frequency of the previous class to get the cumulative frequency for the next class.
💡 The cumulative frequency for a class represents the number of observations with values more than or equal to the lower limit of that class.
Exam Traps — Don't Make These Mistakes
⚠ Using wrong cumulative frequency in median formula
✗ Students write cf as the cumulative frequency of the median class itself.
→ The formula requires the cumulative frequency of the class preceding the median class, not the median class itself. Using the median class's cumulative frequency gives a smaller value for (n/2 - cf), leading to an incorrect median.
✓ Identify the median class first, then find the cumulative frequency of the class just before it. For example, if median class is 60-70 with cf=29, use cf of class 50-60 which is 22.
⚠ Confusing f0 and f2 in mode formula
✗ Students swap the frequencies of the class preceding and succeeding the modal class when substituting into the mode formula.
→ The formula is Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] * h. f0 is the frequency of the class before the modal class, and f2 is the frequency of the class after. Swapping them changes the denominator and yields a wrong mode.
✓ Always label: f1 = frequency of modal class, f0 = frequency of class immediately before modal class, f2 = frequency of class immediately after modal class.
⚠ Forgetting to multiply by h in step-deviation method
✗ Students write mean = a + (Σfiui)/(Σfi) without multiplying by h.
→ The step-deviation formula is mean = a + h * (Σfiui)/(Σfi). The term (Σfiui)/(Σfi) is the mean of the step deviations, which must be multiplied by the class size h to get the actual deviation from the assumed mean.
✓ Always include h: mean = a + h * (Σfiui)/(Σfi). For example, if a=47.5, h=15, Σfiui=29, Σfi=30, then mean = 47.5 + 15*(29/30) = 62.
⚠ Using non-continuous class intervals for mode and median
✗ Students directly apply the mode or median formula to data with gaps between class intervals, e.g., 0-10, 10-20, etc., without converting to continuous classes.
→ The formulas for mode and median assume continuous class intervals. If there is a gap (e.g., 0-10 and 10-20 have a common boundary but are not continuous if the upper limit of one is not the lower limit of the next), the calculation of l and h becomes ambiguous.
✓ Convert to continuous classes by subtracting 0.5 from lower limits and adding 0.5 to upper limits. For example, 0-10 becomes -0.5-10.5, but in practice, if classes are like 0-10, 10-20, they are already continuous. Only adjust if there is a gap, e.g., 0-10 and 12-20.
⚠ Misinterpreting 'less than' cumulative frequency as frequency of a class
✗ Students treat the cumulative frequency of a 'less than' distribution as the frequency of that class interval.
→ In a 'less than' cumulative frequency table, the value given is the total number of observations below the upper limit, not the number in that specific class. For example, 'less than 145: 11' means 11 girls have height less than 145 cm, not that 11 girls are in the 140-145 class.
✓ To find the frequency of a class interval from a 'less than' table, subtract the cumulative frequency of the previous class from the cumulative frequency of the current class. For 140-145, frequency = 11 - 4 = 7.
Compare & Contrast
Approach to calculating mean of grouped data
Direct Method
Assumed Mean Method
The direct method uses the actual class marks (xi) and computes Σfixi directly, while the assumed mean method simplifies calculations by choosing an assumed mean 'a' and working with deviations (di = xi - a), then adding 'a' back to the mean of deviations.
Level of simplification
Assumed Mean Method
Step-deviation Method
The assumed mean method uses deviations di = xi - a, while the step-deviation method further simplifies by dividing these deviations by the class size h to get ui = (xi - a)/h, making calculations even easier when all di have a common factor.
Sensitivity to extreme values
Mean
Median
The mean takes into account all observations and is affected by extreme values, whereas the median is the middle-most value and is not affected by extreme values, making it a better measure when data has outliers.
Purpose of use
Mode
Mean
The mode identifies the most frequent value or most popular item (e.g., most watched TV program), while the mean gives the average value and is used when all observations are considered equally important (e.g., average marks of a class).
Exam Focus
Marks pattern: 1-2 MCQs (1 mark each), 1-2 Short Answer questions (2-3 marks each), 1 Long Answer question (5 marks) — total approximately 8-10 marks
Most asked: Calculating mean of grouped data using direct method, assumed mean method, or step-deviation method · Finding mode of grouped data using the formula · Finding median of grouped data using the formula and cumulative frequency distribution · Constructing cumulative frequency tables (less than and more than type) · Empirical relationship between mean, median, and mode
Don't skip: Long answer question (5 marks) requiring calculation of mean, median, and mode from a grouped frequency distribution table, often with missing frequencies that need to be found using given median or mean.; Short answer question (3 marks) on finding the mode or median from a given grouped data table.
One-Line Revision (read aloud before the exam)
  1. The mean of grouped data can be found by the direct method: x̄ = Σfixi / Σfi.
  2. The assumed mean method uses deviations from an assumed mean: x̄ = a + Σfidi / Σfi.
  3. The step-deviation method further simplifies calculations: x̄ = a + h * Σfiui / Σfi, where ui = (xi - a)/h.
  4. The class mark is the mid-point of a class interval: (Upper limit + Lower limit)/2.
  5. The mode of grouped data is given by: Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] * h.
  6. The modal class is the class interval with the highest frequency.
  7. The median of grouped data is given by: Median = l + [(n/2 - cf) / f] * h.
  8. The median class is the class whose cumulative frequency is greater than and nearest to n/2.
  9. Cumulative frequency of a class is the sum of frequencies of all classes up to that class.
  10. A cumulative frequency distribution can be of 'less than type' (using upper limits) or 'more than type' (using lower limits).
  11. The empirical relationship between the three measures is: 3 Median = Mode + 2 Mean.
  12. The mean is affected by extreme values, while the median is not.
  13. The mode is the most frequently occurring value in a dataset.
  14. For grouped data, the mean obtained is approximate because the class mark is assumed to represent all observations in that class.
  15. When class intervals are not continuous, convert them to continuous classes before applying mode or median formulas.

MCQs — 10 Questions with Answers

Q1. A teacher groups the marks of 50 students into class intervals and calculates the mean using class marks (midpoints). A student claims this mean will always equal the mean calculated using individual ungrouped marks. Is this claim correct, and why?

  • A. No, grouping introduces an assumption that all values in a class cluster at the midpoint, which may not reflect actual distribution ✓
  • B. Yes, because the sum of all frequencies remains the same
  • C. Yes, because class marks are averages of class limits
  • D. No, because class intervals cannot be used for mean calculation

Answer: A — Grouping data loses information about actual individual values; the midpoint assumption introduces approximation error, making the grouped mean different from the ungrouped mean, as shown in the chapter's Example 1 vs Table 13.3 (59.3 vs 62).

Q2. In the assumed mean method, a student chooses the class mark a = 92.5 (near the upper end) instead of a central value like 47.5. How does this choice affect the validity of the formula x̄ = a + d̄?

  • A. The formula becomes invalid because 'a' must always be the midpoint of all class marks
  • B. The formula remains valid, but calculations become more tedious due to larger deviations (di) ✓
  • C. The formula is valid only if frequencies are symmetrically distributed around a = 92.5
  • D. The formula cannot be applied when a is not centrally located

Answer: B — The assumed mean method works for any choice of 'a' because the mathematical relationship x̄ = a + d̄ holds universally; however, choosing 'a' near the center minimizes absolute deviations and reduces computational effort, not because the formula requires it.

Q3. Assertion (A): When calculating the mean of grouped data using the direct method, multiplying each frequency by its corresponding class mark is mathematically equivalent to assuming all data in that class is concentrated at the class mark. Reason (R): The class mark represents the average of the upper and lower class limits. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A
  • B. Both A and R are true but R is not the correct explanation of A ✓
  • C. A is true but R is false
  • D. A is false but R is true

Answer: B — A is true (the method assumes concentration at class mark), and R is true (class mark is defined as the average of limits), but R does not explain WHY we make this assumption—we do so for computational convenience and to represent the class centrally, not because of how class mark is calculated.

Q4. A student observes that in Table 13.4, deviations (di) are positive for classes above 47.5 and negative for classes below it. She concludes that the assumed mean must always be chosen as the middle class mark. Is this conclusion justified?

  • A. Yes, because deviations must balance around the assumed mean
  • B. No, the sign pattern of deviations depends on which value is chosen as 'a', not on whether 'a' is the middle class mark ✓
  • C. Yes, because the formula requires a = (upper class mark + lower class mark)/2
  • D. No, because deviations can be positive or negative regardless of the choice of 'a'

Answer: B — The sign pattern of di naturally mirrors which class marks are above or below the chosen 'a'; choosing the middle value is a convenience to reduce calculation work, not a requirement for the method to function correctly.

Q5. Assertion (A): If two datasets with the same frequencies but different class intervals are grouped and their means calculated, the grouped means will always be different. Reason (R): Different class intervals produce different class marks, which affect the calculation of the mean. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true ✓

Answer: D — A is false because different groupings of the same raw data may yield the same grouped mean if class marks happen to align with the distribution; R is true—different class marks do affect the mean—but the conclusion in A overgeneralizes.

Q6. When using the assumed mean method with di = xi – a, a student obtains Σ(fidi) = 0. What does this result tell us about the relationship between 'a' and the true mean x̄?

  • A. It means a = x̄, because zero deviation from the mean occurs only when the assumed mean equals the true mean ✓
  • B. It means the data is symmetric, so the median and mode must also equal a
  • C. It is inconclusive; zero mean deviation does not guarantee a = x̄
  • D. It means all frequencies are equal

Answer: A — From the formula x̄ = a + d̄, if Σ(fidi) = 0, then d̄ = 0, which implies x̄ = a + 0 = a, making the assumed mean equal to the true mean.

Q7. Assertion (A): The assumed mean method and the direct method always produce identical means for the same grouped dataset. Reason (R): Both methods use the same class marks and the mathematical formula for mean. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both A and R are true: the assumed mean method is an algebraic rearrangement of the direct method (x̄ = a + Σ(fidi)/Σfi is derived from x̄ = Σ(fixi)/Σfi), so they must yield identical results, and this is explained by the fact that both use the same underlying formula applied to the same class marks.

Q8. A dataset is grouped into class intervals of width 10. A student argues that using class marks (midpoints) introduces less error than using the lower class limit as the representative value. Evaluate this claim.

  • A. The claim is incorrect; both methods introduce the same magnitude of error on average
  • B. The claim is correct; class marks are centrally located, so deviations from actual values are generally smaller and more balanced ✓
  • C. The claim is correct only if the data within each class is uniformly distributed
  • D. The claim is incorrect; the lower class limit is always more representative

Answer: B — Using the class mark minimizes error because it balances deviations above and below the representative point, whereas the lower limit would systematically underestimate for most classes.

Q9. Assertion (A): In the assumed mean method, if Σ(fidi) is negative and large in magnitude, the true mean x̄ will be significantly less than the assumed mean 'a'. Reason (R): The formula x̄ = a + d̄ shows that x̄ is obtained by adding the mean deviation to the assumed mean. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — A is true: if Σ(fidi) < 0, then d̄ < 0, so x̄ = a + d̄ < a; R is true and correctly explains A because the formula directly shows how a negative mean deviation pulls the true mean below the assumed mean.

Q10. A teacher presents two grouped frequency tables for the same raw data with different class widths. A student claims that the table with smaller class width will always produce a more accurate mean. Is this claim universally valid?

  • A. Yes, because smaller class width means class marks are closer to actual individual values
  • B. No, accuracy also depends on how data is distributed within each class; a smaller width only guarantees reduced range within a class, not reduced error ✓
  • C. Yes, because the assumed mean method becomes more efficient with smaller widths
  • D. No, because smaller class widths increase computational difficulty

Answer: B — While smaller class width generally improves approximation by bringing class marks closer to data, the actual error reduction depends on the distribution of data within classes; a small width alone does not ensure accuracy if data is clustered away from midpoints.

Flashcards

What is the formula for mean of grouped data using direct method?

Mean = Σ(fi × xi) ÷ Σfi, where fi is frequency and xi is class mark.

How do you find the class mark of a class interval?

Class mark = (Upper class limit + Lower class limit) ÷ 2.

What is the assumed mean method and when is it used?

Choose a central class mark as 'a', find deviations di = xi − a, then mean = a + Σ(fi × di) ÷ Σfi; used when xi values are large.

What is the step deviation method formula?

ui = (xi − a) ÷ h, then mean = a + h × Σ(fi × ui) ÷ Σfi, where h is class size.

Why is the mean of grouped data approximate, not exact?

Because grouped data assumes all frequencies in a class are concentrated at the class mark, not at actual individual values.

If a student scores exactly at the upper class limit, which class does the score belong to?

The score belongs to the next higher class interval, not the current one.

Does the choice of assumed mean 'a' affect the final mean value?

No, the final mean is the same regardless of which xi is chosen as 'a'; Activity 1 proves this.

Why do we divide di by class size h in the step deviation method?

To convert all deviations to smaller numbers that are multiples of the class size, reducing calculation effort.

What is Σ notation and what does it represent?

Σ (Greek letter sigma) represents summation; Σfi × xi means add up all products of fi and xi.

How does grouped data differ from ungrouped data in finding mean?

Ungrouped data uses actual individual values; grouped data uses class marks to represent all values in each class interval.

Important Board Questions

Define class mark. How is it calculated for a class interval 50−70? [2 marks]

State that class mark is the midpoint of a class interval; use the formula (upper limit + lower limit) / 2 and substitute 50 and 70 to get 60.

A frequency distribution has class intervals 0−10, 10−20, 20−30 with frequencies 5, 8, 7 respectively. Using the direct method, find the mean. [3 marks]

Find class marks (5, 15, 25), calculate fi × xi for each class, sum to get Σ(fi × xi) = 390 and Σfi = 20, then apply mean = 390/20 = 19.5.

A dataset of 30 observations has class intervals 0−5, 5−10, 10−15, 15−20, 20−25 with frequencies 4, 6, 8, 7, 5. Using the assumed mean method with a = 12.5, find the mean and explain why this method reduces calculation effort compared to the direct method. [5 marks]

Calculate class marks (2.5, 7.5, 12.5, 17.5, 22.5), find deviations di from a = 12.5, multiply by frequencies to get Σ(fi × di), apply mean = a + Σ(fi × di)/Σfi = 12.5 + 25/30 ≈ 13.33; explain that deviations are smaller numbers reducing arithmetic complexity compared to multiplying large xi values directly.

Assertion-Reason Questions for Statistics

Q1. Assertion (A):

The mean of grouped data using the assumed mean method remains the same regardless of which class mark is chosen as the assumed mean 'a'.

Reason (R):

The actual mean is a fixed numerical value that depends only on the data values and their frequencies, not on the computational method or choice of assumed mean.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the assertion is true. The mean is an intrinsic property of the dataset and cannot change based on computational convenience variables. The assumed mean is merely a tool to simplify calculations, not to alter the result.

Q2. Assertion (A):

For a class interval 35-45, the class mark is always 40 regardless of the class size.

Reason (R):

The class mark is calculated as the average of upper and lower class limits: (35+45)/2 = 40.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both statements are true, but the reason is not the correct explanation for the assertion. The class mark is always 40 for this interval because of how it's mathematically defined, not because of class size. The reason focuses on calculation rather than the fundamental principle that class mark depends only on the interval boundaries, not on their width.

Q3. Assertion (A):

The step deviation method (using ui) is more convenient than the assumed mean method when class sizes are unequal.

Reason (R):

Step deviations reduce numerical calculations by dividing deviations by the class size h.

Answer: D — A is false but R is true — The assertion is false. The step deviation method works best when all class sizes are equal (uniform h). When class sizes are unequal, using a common h in the formula ui = (xi - a)/h becomes problematic and the method loses its computational advantage. The reason is partially true but doesn't support the assertion accurately. Neither statement correctly addresses the limitations of the method with unequal class sizes.

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Case-Based Questions (CBQ) for Statistics

Case Study:

A school conducted a physical fitness test for 100 students of Class X. The test measured the time (in seconds) each student could hold a push-up position. The data was recorded in grouped form as follows: 10-20 seconds (8 students), 20-30 seconds (12 students), 30-40 seconds (20 students), 40-50 seconds (35 students), 50-60 seconds (18 students), 60-70 seconds (7 students). The fitness trainer needs to find the average holding time to assess overall class performance.

Q1. What is the class mark for the class interval 40-50?

  • A. 40
  • B. 45 ✓
  • C. 50
  • D. 10

Answer: B

Q2. Using the direct method, if Σfixi = 3900, what is the mean holding time?

  • A. 35 seconds
  • B. 39 seconds ✓
  • C. 40 seconds
  • D. 45 seconds

Answer: B

Q3. If the assumed mean is taken as a = 45, what is the deviation di for the class 20-30?

  • A. -25
  • B. -20
  • C. -15 ✓
  • D. 5

Answer: C

Q4. The class interval with the highest frequency is:

  • A. 40-50 ✓
  • B. 30-40
  • C. 50-60
  • D. 60-70

Answer: A

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Previous Year Questions (PYQ) for Statistics

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 31/4/1 · 1 mark

While calculating mean of a grouped frequency distribution using step deviation method x̄ = a + (h/f)Σfu it was found that x̄ = 62, a = 47.5, h = 5. The value of ū is:

  • 3
  • 14.5
  • 2.9
  • 3.1

CBSE 2026 · Set 31/4/1 · 1 mark

The Air Quality Index (AQI) is a scale from 0 to 500 that indicates air quality, with higher numbers signifying more pollution and greater health concerns. Mansi collected the daily data of AQI of her city for a month and presented it as given below: AQI Range: 1 – 100, 101 – 200, 201 – 300, 301 – 400, 401 – 500. Number of Days: 3, 9, 12, 4, 2

Convert the data to continuous frequency distribution.

CBSE 2026 · Set 31/4/2 · 1 mark

While calculating mean of a grouped frequency distribution using step deviation method x̄ = a + (h/Σf) × Σ(u × f), it was found that x̄ = 62, a = 47.5, h = 5. The value of ū is:

  • 3
  • 14.5
  • 2.9
  • 3.1

CBSE 2026 · Set 31/4/3 · 1 mark

While calculating mean of a grouped frequency distribution using step deviation method x̄ = a + (h/n)Σfu, it was found that x̄ = 62, a = 47.5, h = 5. The value of ū is:

  • 3
  • 14.5
  • 2.9
  • 3.1

CBSE 2026 · Set 31/5/1 · 1 mark

Assertion (A): Median of a data is the value of N/2, where N represents sum of all frequencies. Reason (R): Median divides the whole distribution in two equal parts.

  • Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)
  • Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A)
  • Assertion (A) is true, but Reason (R) is false
  • Assertion (A) is false, but Reason (R) is true

CBSE 2026 · Set 31/5/1 · 5 marks

Find mean and mode of the following data: Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80, Frequency 5 4 10 13 12 10 6

CBSE 2026 · Set 31/5/2 · 5 marks

Find mean and mode of the following data: Class 0-15, 15-30, 30-45, 45-60, 60-75, 75-90, 90-105 with Frequency 4, 6, 8, 10, 12, 7, 3

CBSE 2026 · Set 31/5/3 · 5 marks

Find the missing frequencies p and q in the following frequency distribution, when sum of frequencies is 40 and mean is 19: Class 0-5: 2, 5-10: 5, 10-15: 6, 15-20: p, 20-25: 10, 25-30: q, 30-35: 4

CBSE 2026 · Set 31/1164/30 · 1 mark

If the mean of first n natural numbers is 6n/11, then n is

  • 11
  • 6
  • 12
  • 22

CBSE 2026 · Set 31/1164/30 · 1 mark

Assertion (A) : If the Mode and Mean of a data are 12k and 15k, then Median of the data is 14k. Reason (R) : The relation between the Mean, Mode and Median of a data is : Mean = 3 Median – 2 Mode.

  • Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  • Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 31/1164/30 · 3 marks

Find the mean of the following distribution: Class 30–40 40–50 50–60 60–70 70–80; Frequency 6 13 8 12 11

CBSE 2026 · Set 31/1167/430 · 1 mark

Assertion (A): If the value of mode and mean for a distribution is 50 and 56 respectively, then the value of median is 54. Reason (R): Median = 1/3 (Mode – 2 Mean)

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation of Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

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Class 10 Maths Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 10 Mathematics Previous Year Questions, Chapter-wise

All 1107 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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