Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Mean of Grouped Data
The class mark xᵢ = (lower limit + upper limit) / 2
Method 1 — Direct Method
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$
Worked example:
| Class | Frequency (f) | Class mark (x) | fx |
|---|---|---|---|
| 10–20 | 3 | 15 | 45 |
| 20–30 | 7 | 25 | 175 |
| 30–40 | 10 | 35 | 350 |
| 40–50 | 5 | 45 | 225 |
| Total | 25 | 795 |
Mean = 795/25 = 31.8
Method 2 — Assumed Mean Method (Shortcut)
Choose assumed mean A (usually middle class mark). Compute dᵢ = xᵢ − A.
$$\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$$
| Class | f | x | d = x − 35 | fd |
|---|---|---|---|---|
| 10–20 | 3 | 15 | −20 | −60 |
| 20–30 | 7 | 25 | −10 | −70 |
| 30–40 | 10 | 35 | 0 | 0 |
| 40–50 | 5 | 45 | 10 | 50 |
| Total | 25 | −80 |
Mean = 35 + (−80/25) = 35 − 3.2 = 31.8 ✓
Method 3 — Step Deviation Method
Compute uᵢ = (xᵢ − A) / h where h = class width.
$$\bar{x} = A + h \times \frac{\sum f_i u_i}{\sum f_i}$$
Particularly useful when class widths are equal and dᵢ values are large multiples of h.
Mode of Grouped Data
The modal class is the class with the highest frequency.
$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$
Where:
- l = lower class boundary of modal class
- f₁ = frequency of modal class
- f₀ = frequency of class preceding the modal class
- f₂ = frequency of class following the modal class
- h = class width
Worked example:
| Class | Frequency |
|---|---|
| 10–20 | 5 |
| 20–30 | 8 |
| 30–40 | 12 (highest → modal class) |
| 40–50 | 7 |
| 50–60 | 3 |
Modal class = 30–40; l = 30, f₁ = 12, f₀ = 8, f₂ = 7, h = 10
Mode = 30 + [(12−8)/(2×12 − 8 − 7)] × 10 = 30 + [4/9] × 10 = 30 + 4.44 = 34.44
Median of Grouped Data
Step 1: Calculate cumulative frequencies (cf) Step 2: Find n/2 Step 3: Find the class where cumulative frequency first ≥ n/2 → this is the median class
$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$
Where:
- l = lower class boundary of median class
- cf = cumulative frequency of class before the median class
- f = frequency of the median class
- h = class width
Worked example:
| Class | f | Cumulative f (cf) |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 15 | 20 |
| 20–30 | 20 | 40 ← median class (n/2 = 30, first cf ≥ 30) |
| 30–40 | 8 | 48 |
| 40–50 | 2 | 50 |
n = 50; n/2 = 25; Median class = 20–30 l = 20, cf = 20, f = 20, h = 10
Median = 20 + [(25−20)/20] × 10 = 20 + 2.5 = 22.5
Empirical Relationship
When data is not very symmetrical (moderately skewed):
$$\text{Mode} \approx 3 \times \text{Median} - 2 \times \text{Mean}$$
This can be used to find one measure if the other two are known.
Ogive (Cumulative Frequency Graph)
An ogive is a graph of cumulative frequency against the upper class boundaries.
Less Than Ogive
Plot cumulative frequency against upper class boundary. Curve goes upward from left to right.
More Than Ogive
Plot (n − cumulative frequency) against lower class boundary. Curve goes downward.
Finding Median from Ogive
1. Locate n/2 on y-axis 2. Draw horizontal line to the less-than ogive curve 3. Drop vertical to x-axis → read the median value
Alternative: Draw both 'less than' and 'more than' ogives on the same graph. Their intersection point gives the median.
Board Exam Tips
- 6–8 marks from this chapter — very high marks, mostly calculation
- Always make a table with all columns before substituting in formulas
- Modal class = class with highest frequency (not highest cumulative frequency)
- For median, use cumulative frequency — the median class is where cf first ≥ n/2
- Empirical relation (Mode = 3Median − 2Mean) is asked in "find the missing value" type problems
- Ogive graph: plot points carefully; mark axes with class boundaries (not midpoints) for cumulative frequency
- Class mark = midpoint = (lower + upper)/2 — always compute this column
📚 RD Sharma Mathematics Class 10 — Statistics Chapter
📝 Oswaal CBSE Maths Question Bank — Statistics Board Questions