What is the formula for the mode of grouped data?

Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h, where l is the lower limit of the modal class (the class with the highest frequency), f₁ is its frequency, f₀ and f₂ are the frequencies of the classes before and after it, and h is the class width.

What is the formula for the median of grouped data?

Median = l + [(n/2 − cf)/f] × h, where l is the lower limit of the median class, n is the total frequency, cf is the cumulative frequency before the median class, f is the frequency of the median class, and h is the class width.

NCERT Solutions for Class 10 Maths Chapter 13 Statistics — Exercise Solutions

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Complete NCERT Solutions for CBSE Class 10 Maths Chapter 13 — Statistics: mean of grouped data (direct, assumed-mean, step-deviation), mode, and median.

Key Formulas (Quick Recap)

Mean (step deviation): x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h, where uᵢ = (xᵢ − a)/h.
Mode: l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h.
Median: l + [(n/2 − cf)/f] × h.
Empirical relation: Mode = 3 Median − 2 Mean.

Exercise 13.1 — Mean of Grouped Data

Q1. Find the mean number of plants per house from the data below.

Answer

Plants	0–2	2–4	4–6	6–8	8–10	10–12	12–14
Houses (fᵢ)	1	2	1	5	6	2	3

Midpoints xᵢ: 1, 3, 5, 7, 9, 11, 13. Σfᵢ = 20. Σfᵢxᵢ = 1(1) + 2(3) + 1(5) + 5(7) + 6(9) + 2(11) + 3(13) = 1 + 6 + 5 + 35 + 54 + 22 + 39 = 162. Mean = 162/20 = 8.1 plants per house.

Q2. (Step-deviation method) For class marks 17.5, 22.5, 27.5, 32.5, 37.5 with frequencies 12, 21, 20, 32, 15 (a = 27.5, h = 5): uᵢ = (xᵢ − 27.5)/5 = −2, −1, 0, 1, 2. Σfᵢ = 100. Σfᵢuᵢ = 12(−2) + 21(−1) + 20(0) + 32(1) + 15(2) = −24 − 21 + 0 + 32 + 30 = 17. Mean = 27.5 + (17/100)(5) = 27.5 + 0.85 = 28.35.

Exercise 13.2 — Mode

Q1. Find the mode of the data below.

Answer

Class	0–20	20–40	40–60	60–80	80–100	100–120
Frequency	10	35	52	61	38	29

Modal class = 60–80 (highest frequency 61). l = 60, f₁ = 61, f₀ = 52, f₂ = 38, h = 20. Mode = 60 + [(61 − 52)/(2·61 − 52 − 38)] × 20 = 60 + [9/(122 − 90)] × 20 = 60 + (9/32)(20) = 60 + 5.625 = 65.625.

Exercise 13.3 — Median

Q1. Find the median of the marks data below.

Answer

Marks	0–10	10–20	20–30	30–40	40–50
Students (f)	5	8	15	16	6
cf	5	13	28	44	50

n = 50 → n/2 = 25. The cumulative frequency just reaches 25 in the class 20–30 (cf before = 13, f = 15, l = 20, h = 10). Median = 20 + [(25 − 13)/15] × 10 = 20 + (12/15)(10) = 20 + 8 = 28.

Q2. Using the empirical relation, find the mode if Mean = 30 and Median = 28. Mode = 3 Median − 2 Mean = 3(28) − 2(30) = 84 − 60 = 24.

Common Mistakes

Using class limits instead of midpoints for the mean.
Picking the modal class by width instead of by highest frequency.
Forgetting to build the cumulative frequency column before finding the median class.

Exam Tips

Step-deviation method is fastest when class widths are equal — choose a (assumed mean) near the middle.
For median, the median class is where the cumulative frequency first reaches or exceeds n/2.
Use Mode = 3 Median − 2 Mean to find any one measure from the other two.

Frequently Asked Questions

Direct method: x̄ = Σfᵢxᵢ / Σfᵢ. Assumed-mean method: x̄ = a + Σfᵢdᵢ / Σfᵢ where dᵢ = xᵢ − a. Step-deviation method: x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h where uᵢ = (xᵢ − a)/h. All three give the same mean.