Is it correct to say: Using the upper or lower class limit as xᵢ when finding the mean.?

Use the class mark (mid-point) of each class: for 20–30, xᵢ = 25.

Is it correct to say: Taking cf as the cumulative frequency of the median class itself.?

In the median formula, cf is the cumulative frequency of the class BEFORE the median class.

Is it correct to say: Choosing the class with the highest frequency as the median class.?

That is the modal class. The median class is found from the cumulative frequencies and n/2.

Is it correct to say: Forgetting to multiply by h in the step-deviation method.?

x̄ = a + h × (Σfᵢuᵢ/Σfᵢ). Without h, the term added to a is off by a factor of the class size.

Is it correct to say: Using cumulative frequencies, or the wrong neighbouring classes, for f₀ and f₂.?

f₀ and f₂ are the ordinary frequencies of the classes immediately before and after the modal class.

Is it correct to say: Writing the empirical relation as 3 Mode = Median + 2 Mean.?

Correct form: 3 Median = Mode + 2 Mean. Equivalently, Mode = 3 Median − 2 Mean.

📄 Formula sheet (PDF)

Statistics

Mean of grouped data by the direct, assumed-mean and step-deviation methods, mode and median of grouped data, and the empirical relationship between them — NCERT Class 10 Maths Ch 13

📐 11 formulas✏️ 4 examples🎯 6 practice⚖️ Unit VII (Statistics and Probability): 11 marks (with Ch 14)🏫 CBSE📚 Class 10✓ 2025–26 syllabus

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Board Exam Tips

→Draw the full table (class, fᵢ, xᵢ, dᵢ or uᵢ, fᵢuᵢ, cf) neatly with column totals. Much of the working is in the table.
→Choose the assumed mean a as a class mark near the middle of the table. Step-deviation then gives small whole-number uᵢ.
→For the mode, write the modal class and then the values of l, h, f₁, f₀ and f₂ separately before substituting.
→For the median, find n/2 first, then the median class. The cf in the formula belongs to the class BEFORE the median class.
→If the classes are not continuous (e.g. 10–19, 20–29), make them continuous before using the median or mode formula.
→Use 3 Median = Mode + 2 Mean to find the third measure when two are given.

📐 Formulas(11)

Class Mark

x_{i} = \frac{upper class limit + lower class limit}{2}

Mean: Direct Method

x ˉ = \frac{\sum f _{i} x _{i}}{\sum f _{i}}

Symbol	Meaning
$x ˉ$	Mean of the data
$f_{i}$	Frequency of the i-th class
$x_{i}$	Class mark of the i-th class

Mean: Assumed Mean Method

x ˉ = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}, d_{i} = x_{i} - a

Symbol	Meaning
$a$	Assumed mean (one of the class marks)
$d_{i}$	Deviation of the class mark from a

Mean: Step-Deviation Method★ Board fav

x ˉ = a + h (\frac{\sum f _{i} u _{i}}{\sum f _{i}}), u_{i} = \frac{x _{i} - a}{h}

Symbol	Meaning
$h$	Class size (common factor of the dᵢ)
$u_{i}$	Step deviation of the i-th class

Modal Class

modal class = class with the maximum frequency

Mode of Grouped Data★ Board fav

Mode = l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h

Symbol	Meaning
$l$	Lower limit of the modal class
$h$	Class size
$f_{1}$	Frequency of the modal class
$f_{0}$	Frequency of the class preceding the modal class
$f_{2}$	Frequency of the class succeeding the modal class

Cumulative Frequency (Less Than Type)

c f_{k} = \sum_{i = 1}^{k} f_{i}

Median Class

median class = class whose c f is greater than and nearest to \frac{n}{2}

Median of Grouped Data★ Board fav

Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h

Symbol	Meaning
$l$	Lower limit of the median class
$n$	Total frequency, Σfᵢ
$c f$	Cumulative frequency of the class preceding the median class
$f$	Frequency of the median class
$h$	Class size

Empirical Relationship★ Board fav

3 Median = Mode + 2 Mean

Making Classes Continuous

adjustment = \frac{1}{2} [(lower limit of next class) - (upper limit of a class)]

✏️ Solved Examples

1Solved Exampleeasy3 steps

For a grouped distribution, the mode is 29 and the median is 26. Find the mean using the empirical relationship.

Write the empirical relationship.

3 Median = Mode + 2 Mean

2Solved Exampleboard4 steps

The daily pocket money (in ₹) of 50 students is given below. Find the mean pocket money by the step-deviation method. Classes: 100–120, 120–140, 140–160, 160–180, 180–200; number of students: 12, 14, 8, 6, 10.

Class marks xᵢ: 110, 130, 150, 170, 190. Take a = 150 and h = 20.

u_{i} = \frac{x _{i} - 150}{20} : - 2, - 1, 0, 1, 2

3Solved Exampleboard3 steps

The ages (in years) of 70 patients admitted to a hospital on one day are given below. Find the mode. Classes: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60; number of patients: 6, 10, 14, 20, 16, 4.

The maximum frequency is 20, so the modal class is 30–40.

l = 30, h = 10, f_{1} = 20, f_{0} = 14, f_{2} = 16

4Solved ExampleHOTS5 steps

The median of the following distribution is 32 and the total frequency is 50. Find the missing frequencies x and y. Classes: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60; frequencies: 4, x, 12, 15, y, 6.

Total frequency

4 + x + 12 + 15 + y + 6 = 50 \Rightarrow x + y = 13

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using the upper or lower class limit as xᵢ when finding the mean.
✓Use the class mark (mid-point) of each class: for 20–30, xᵢ = 25.
2
Taking cf as the cumulative frequency of the median class itself.
✓In the median formula, cf is the cumulative frequency of the class BEFORE the median class.
3
Choosing the class with the highest frequency as the median class.
✓That is the modal class. The median class is found from the cumulative frequencies and n/2.
4
Forgetting to multiply by h in the step-deviation method.
✓x̄ = a + h × (Σfᵢuᵢ/Σfᵢ). Without h, the term added to a is off by a factor of the class size.
5
Using cumulative frequencies, or the wrong neighbouring classes, for f₀ and f₂.
✓f₀ and f₂ are the ordinary frequencies of the classes immediately before and after the modal class.
6
Writing the empirical relation as 3 Mode = Median + 2 Mean.
✓Correct form: 3 Median = Mode + 2 Mean. Equivalently, Mode = 3 Median − 2 Mean.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the mean by the direct method. Classes: 0–4, 4–8, 8–12, 12–16; frequencies: 3, 5, 8, 4.
Q2
The mean of the distribution below is 31. Find p. Classes: 10–20, 20–30, 30–40, 40–50; frequencies: 5, p, 10, 5.
Q3
Find the mode. Classes: 10–20, 20–30, 30–40, 40–50, 50–60; frequencies: 3, 7, 12, 9, 5.
Q4
Find the median. Classes: 0–20, 20–40, 40–60, 60–80, 80–100; frequencies: 6, 9, 15, 12, 8.
Q5
For a grouped distribution the mean is 40 and the median is 42. Find the mode using the empirical relationship.
Q6
Make the classes 5–9, 10–14, 15–19 continuous and state the class size.

📝 Notes

Statistics

This chapter finds a single representative value (mean, mode or median) for data grouped into classes. Every question is a table followed by one formula.

Mean: three routes, one answer

The direct, assumed-mean and step-deviation methods always give the same mean; they differ only in how much arithmetic you do.

Direct: use when the class marks and frequencies are small.
Assumed mean: subtract a central class mark a from every xᵢ to get smaller dᵢ.
Step-deviation: divide the dᵢ by the common factor h as well, so the uᵢ are small integers such as −2, −1, 0, 1, 2.

The class mark is used for each class because we assume the values in a class are centred on its mid-point. If the question names a method, you must use that method.

Mode and median of grouped data

Mode: find the class with the highest frequency, then use $l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$ .
Median: add a cumulative frequency column, find n/2, locate the median class, then use $l + \frac{n/2 - cf}{f} \times h$ .

The two formulas look similar but use different columns. The mode uses the frequencies of the neighbouring classes. The median uses the cumulative frequency of the previous class. Classes must be continuous before you read off l.

The empirical relationship

For moderately skewed data, 3 Median = Mode + 2 Mean. It is a quick way to find a third measure from two known ones, but it is approximate: if the question asks you to compute the mode or median from a table, use the formula, not this relation.

Missing-frequency questions

When a frequency is unknown, write two equations: one from the total frequency, and one from the given mean or median. For a given median, first decide which class it lies in; that class is the median class.