Statistics Class 10: Notes, Formulas and Important Questions

This guide covers Statistics Class 10, chapter 13 of NCERT Class 10 Mathematics, which is about the mean, mode and median of grouped data. You will learn three ways to find the mean, the mode and median formulas, how cumulative frequency works and how to fix class intervals with gaps. Worked tables, important questions and common mistakes are included.

Grouped data and class marks

Grouped data puts observations into class intervals, which makes large data easier to study. The three measures of central tendency are:

• Mean: the average.
• Mode: the most frequent value.
• Median: the middle-most value.

Grouping loses some detail, so values from grouped data are estimates, not the exact values of the raw data.

We assume each class's frequency is centred at its class mark (mid-point):

Class mark = (Upper class limit + Lower class limit)/2

For 25–40, the class mark is (25 + 40)/2 = 32.5.

Mean: the direct method

x̄ = Σfᵢxᵢ / Σfᵢ, where xᵢ is the class mark and fᵢ the frequency. It is best when the numbers are small and easy to multiply.

x̄ = 1860/30 = 62.

Mean: assumed mean and step-deviation methods

Assumed mean method: choose one class mark as the assumed mean a (usually a middle one). Find dᵢ = xᵢ − a. Then

x̄ = a + Σfᵢdᵢ / Σfᵢ

Step-deviation method: if all dᵢ share the class size h, use uᵢ = (xᵢ − a)/h. Then

x̄ = a + h × (Σfᵢuᵢ / Σfᵢ)

Same data, with a = 47.5 and h = 15:

• Assumed mean: x̄ = 47.5 + 435/30 = 47.5 + 14.5 = 62.
• Step-deviation: x̄ = 47.5 + 15 × 29/30 = 47.5 + 14.5 = 62.

All three methods give the same mean. The choice of a does not change the answer; a central value just keeps the numbers small.

Mode of grouped data

The modal class is the class with the highest frequency. Then

Mode = l + (f₁ − f₀) / (2f₁ − f₀ − f₂) × h

• l = lower limit of the modal class
• f₁ = frequency of the modal class
• f₀ = frequency of the class before it
• f₂ = frequency of the class after it
• h = class size

Example: classes 10–20, 20–30, 30–40 with frequencies 2, 5, 3. The modal class is 20–30, so l = 20, f₁ = 5, f₀ = 2, f₂ = 3, h = 10. Mode = 20 + 3/(10 − 2 − 3) × 10 = 20 + (3/5) × 10 = 26.

The modal class is an interval; the mode is a value inside it.

Median and cumulative frequency

Less-than cumulative frequency is a running total. For frequencies 2, 5, 3 it is 2, 7, 10. The more-than form counts from the top: 10, 8, 3.

The median class is the first class whose cumulative frequency exceeds n/2. Then

Median = l + (n/2 − cf) / f × h

• l = lower limit of the median class
• n = total frequency
• cf = cumulative frequency of the class before the median class
• f = frequency of the median class
• h = class size

Example (heights of 51 girls): n/2 = 25.5. The median class is 145–150, with l = 145, cf = 11, f = 18, h = 5. Median = 145 + (25.5 − 11)/18 × 5 = 145 + 72.5/18 ≈ 145 + 4.03 = 149.03 cm. So about half the girls are shorter than 149.03 cm and half are taller.

If you are given less-than data, get each class frequency by subtracting: if less than 140 is 4 and less than 145 is 11, the class 140–145 has frequency 11 − 4 = 7.

Class boundaries and choosing a measure

Fix gaps first. Classes like 118–126, 127–135, 136–144 (measured to the nearest millimetre) have gaps. Subtract 0.5 from each lower limit and add 0.5 to each upper limit: 117.5–126.5, 126.5–135.5, 135.5–144.5. Now each width is 126.5 − 117.5 = 9 mm, not 8 mm. The half-unit adjustment depends on how precisely the data was measured; it is not always 0.5.

In continuous classes like 10–20 and 20–30, the value 20 belongs to the second class, so nothing is counted twice.

Empirical relationship: 3 × Median ≈ Mode + 2 × Mean. It is a rule of thumb for suitable distributions, not something every data set obeys.

Remember this

• Class mark = (upper limit + lower limit)/2.
• Mean: x̄ = Σfᵢxᵢ/Σfᵢ, or a + Σfᵢdᵢ/Σfᵢ, or a + h × Σfᵢuᵢ/Σfᵢ.
• Mode = l + (f₁ − f₀)/(2f₁ − f₀ − f₂) × h.
• Median = l + (n/2 − cf)/f × h, where cf is for the class before.
• Make classes continuous before using the median or mode formula.

Important questions with answers

1. Find the class mark of 25–40. (25 + 40)/2 = 32.5.

2. The assumed mean is 50 and Σfᵢdᵢ/Σfᵢ = −2.5. Find the mean. x̄ = 50 + (−2.5) = 47.5.

3. Does the choice of assumed mean change the mean? No. Any class mark gives the same answer; a central one keeps the arithmetic simple.

4. The class 30–40 has the highest frequency, 15. Classes 20–30 and 40–50 have 7 and 10. Find the mode. l = 30, f₁ = 15, f₀ = 7, f₂ = 10, h = 10. Mode = 30 + (15 − 7)/(30 − 7 − 10) × 10 = 30 + (8/13) × 10 = 30 + 80/13 ≈ 36.15.

5. Find the median of: 0–10 (5), 10–20 (8), 20–30 (4), 30–40 (3). n = 20, n/2 = 10. Cumulative frequencies: 5, 13, 17, 20. The median class is 10–20, with l = 10, cf = 5, f = 8, h = 10. Median = 10 + (10 − 5)/8 × 10 = 10 + 6.25 = 16.25.

6. If n = 50, which class is the median class? You cannot tell from n alone. Find n/2 = 25 and pick the first class whose cumulative frequency exceeds 25.

7. Lengths are measured to the nearest millimetre. What are the boundaries and width of the class 127–135 mm? 126.5 mm and 135.5 mm; width = 9 mm. Use l = 126.5 in the formulas.

8. Using the empirical relationship, estimate the mode if the mean is 20 and the median is 22. Mode ≈ 3 × 22 − 2 × 20 = 66 − 40 = 26.

Common mistakes to avoid

• Using the cumulative frequency of the median class itself as cf. It must be the class before.
• Using the modal class's upper limit instead of its lower limit for l.
• Forgetting to make classes continuous, which gives a wrong l and h.
• Choosing the median class by the highest frequency. That finds the modal class, not the median class.
• Calling a grouped-data estimate the exact value of the raw data.

Build each table neatly, column by column, and study this chapter with Joy whenever you want to practise more data.