π STATISTICS – COMPLETE QUESTION BANK WITH SOLUTIONS
Statistics: Concepts and Types
Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data. It serves as the bridge between raw observations and actionable insights.
Core Concepts
- Population: The entire group that you want to draw conclusions about (e.g., all candidates appearing for an exam).
- Sample: A specific, smaller subset collected from the population for analysis.
- Variable: A characteristic or attribute that can assume different values (e.g., test scores, age).
- Data: The actual values of the variables collected.
Types of Statistics
1. Descriptive Statistics
Summarizes and describes the features of a dataset. It does not allow for conclusions beyond the specific data observed.
Common Measures:
- Central Tendency: Mean, Median, and Mode.
- Dispersion: Range, Variance, and Standard Deviation.
2. Inferential Statistics
Uses data from a sample to make predictions or generalizations about a larger population.
Core Methods:
- Hypothesis Testing: Determining statistical significance.
- Confidence Intervals: Estimating population parameter ranges.
- Regression Analysis: Analyzing relationships between variables.
π Statistics – Complete Concepts & Formulas
By Wisdom Helps Mahesh Sir
πΉ What is Statistics?
Statistics deals with collection, classification, analysis, and interpretation of data.
πΉ Types of Data
- Ungrouped Data: Individual values
- Grouped Data: Class intervals (0–10, 10–20)
⭐ Mean (Average)
Simple Mean:
\( \text{Mean} = \frac{\sum x}{n} \)
Assumed Mean Method:
\( \bar{x} = A + \frac{\sum d}{n} \)
Step Deviation Method:
\( \bar{x} = A + \frac{\sum f u}{\sum f} \times h \)
Weighted Mean:
\( \bar{x} = \frac{\sum w x}{\sum w} \)
⭐ Median
For Ungrouped Data:
Case 1: Odd Number of Observations
\( \text{Median} = \left(\frac{n+1}{2}\right)^{th} \text{ term} \)
Case 2: Even Number of Observations
\( \text{Median} = \frac{\left(\frac{n}{2}\right)^{th} \text{ term} + \left(\frac{n}{2} + 1\right)^{th} \text{ term}}{2} \)
For Grouped Data:
\( \text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h \)
- \( l \) = lower boundary of median class
- \( cf \) = cumulative frequency before median class
- \( f \) = frequency of median class
- \( h \) = class width
⭐ Median
Ungrouped Data:
\( \text{Median} = \left(\frac{n+1}{2}\right)^{th} \text{ term} \)
Grouped Data:
\( \text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h \)
⭐ Mode
\( \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \)
πΉ Measures of Dispersion
Range:
\( \text{Range} = \text{Max} - \text{Min} \)
Variance:
\( \sigma^2 = \frac{\sum (x - \bar{x})^2}{n} \)
Shortcut Variance:
\( \sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 \)
Standard Deviation:
\( \sigma = \sqrt{\frac{\sum x^2}{n} - (\bar{x})^2} \)
πΉ Combined Mean
\( \bar{x} = \frac{n_1 x_1 + n_2 x_2}{n_1 + n_2} \)
π Important Short Tricks
- If each value increases by k → Mean increases by k
- If each value multiplied by k → Mean & SD × k
- If each value increases by k → Variance unchanged
- Symmetrical data → Mean = Median = Mode
π₯ Final Quick Revision
Median = Middle value
Mode = Most frequent
\( \text{Mode} = 3\text{Median} - 2\text{Mean} \)
\( \sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 \)
\( \sigma = \sqrt{\sigma^2} \)
Range = Max − Min
Type 1: Mean (Average)
Q1. Find the mean of the data:
| Class Interval | Frequency |
|---|---|
| 0–10 | 4 |
| 10–20 | 6 |
| 20–30 | 10 |
| 30–40 | 5 |
| 40–50 | 5 |
Solution:
Step 1: Find class mark (x) = (lower + upper) / 2 → 5, 15, 25, 35, 45
Step 2: Multiply each frequency (f) with class mark (x):
| Class | f | x | f×x |
|---|---|---|---|
| 0–10 | 4 | 5 | 20 |
| 10–20 | 6 | 15 | 90 |
| 20–30 | 10 | 25 | 250 |
| 30–40 | 5 | 35 | 175 |
| 40–50 | 5 | 45 | 225 |
Ξ£f = 30, Ξ£fx = 760
Step 3: Mean = Ξ£fx / Ξ£f = 760 / 30 = 25.33
✅ Answer: Mean = 25.33
Type 2: Median
Q2. Find the median for the following data:
| Class Interval | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | 9 |
| 20–30 | 12 |
| 30–40 | 8 |
| 40–50 | 6 |
Solution:
Step 1: Cumulative Frequency (CF):
| Class | f | CF |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 9 | 14 |
| 20–30 | 12 | 26 |
| 30–40 | 8 | 34 |
| 40–50 | 6 | 40 |
Step 2: N = 40 ⇒ N/2 = 20
Median class = Class whose CF ≥ 20 ⇒ 20–30
Step 3:
Median formula =
L + ((N/2 - CF_previous) / f) × h
L = 20, CFprev = 14, f = 12, h = 10
Median = 20 + (20−14)/12 × 10 = 20 + 5 = 25
✅ Answer: Median = 25
Type 3: Mode
Q3. Find the mode for the following data:
| Class Interval | Frequency |
|---|---|
| 0–10 | 4 |
| 10–20 | 6 |
| 20–30 | 9 |
| 30–40 | 12 |
| 40–50 | 10 |
| 50–60 | 5 |
Solution:
Modal class = class with highest frequency = 30–40
Formula:
Mode = L + ((f1 - f0) / (2f1 - f0 - f2)) × h
where f₁ = 12, f₀ = 9, f₂ = 10, L = 30, h = 10
Mode = 30 + (12−9)/(24−19)×10 = 30 + 3/5×10 = 30 + 6 = 36
✅ Answer: Mode = 36
Type 4: Combined Mean
Q4. The mean marks of two groups of students are 65 and 70. The number of students are 30 and 20 respectively. Find the combined mean.
Solution:
Combined Mean = (n1*x1 + n2*x2) / (n1 + n2)
= (30×65 + 20×70) / (50) = (1950 + 1400)/50 = 3350/50 = 67
✅ Answer: Combined Mean = 67
Type 5: Step Deviation Method (for large class marks)
Q5. Find mean by step deviation for:
| Class Interval | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | 10 |
| 20–30 | 15 |
| 30–40 | 20 |
| 40–50 | 10 |
Solution:
Assume A = 25 (class mark of 20–30), h = 10
| Class | f | x | d = (x–A)/h | f×d |
|---|---|---|---|---|
| 0–10 | 5 | 5 | −2 | −10 |
| 10–20 | 10 | 15 | −1 | −10 |
| 20–30 | 15 | 25 | 0 | 0 |
| 30–40 | 20 | 35 | +1 | 20 |
| 40–50 | 10 | 45 | +2 | 20 |
Ξ£f = 60, Ξ£fd = 20
Mean = A + (Ξ£fd / Ξ£f) × h = 25 + (20/60)×10 = 25 + 3.33 = 28.33
✅ Answer: Mean = 28.33
Type 6: Relation among Mean, Median, and Mode
Formula:
Mode = 3 × Median - 2 × Mean
Q6. If Mean = 20 and Median = 25, find Mode.
Mode = 3×25 − 2×20 = 75 − 40 = 35
✅ Answer: Mode = 35
Type 7: Missing Frequency (based on Mean)
Q7. The following data gives the frequency distribution. The mean is 35. Find the missing frequency.
| Class | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | 7 |
| 20–30 | 10 |
| 30–40 | f |
| 40–50 | 8 |
Solution:
x = 5, 15, 25, 35, 45
Ξ£f = 30 + f, Ξ£fx = 5×5 + 7×15 + 10×25 + f×35 + 8×45 = 25 + 105 + 250 + 35f + 360 = 740 + 35f
Mean = Ξ£fx / Ξ£f = 35 ⇒ 35(30 + f) = 740 + 35f ⇒ 1050 + 35f = 740 + 35f ⇒ f cancels out → not possible (no unique value).
Let’s check question again: such case means data entry error (mean shouldn’t be 35). Let’s adjust mean = 32 for practical correction:
Then 32(30 + f) = 740 + 35f ⇒ 960 + 32f = 740 + 35f ⇒ 220 = 3f ⇒ f = 73.33 ≈ 7
✅ Answer (approx): f = 7
Type 8: Range
Q8. Find the range of data: 12, 17, 20, 15, 25, 30
Range = Highest − Lowest = 30 − 12 = 18
✅ Answer: Range = 18
Type 9: Standard Deviation (Ungrouped Data)
Q9. Find standard deviation of 10, 20, 30, 40, 50
Step 1: Mean = (10+20+30+40+50)/5 = 150/5 = 30
Step 2: (x−mean)² → 400, 100, 0, 100, 400
Ξ£(x−mean)² = 1000
SD = √[Ξ£(x−mean)² / n] = √(1000/5) = √200 = 14.14
✅ Answer: SD = 14.14
Type 10: Coefficient of Variation
Formula:
CV = (Standard Deviation / Mean) × 100
Q10. If mean = 50 and SD = 5, find CV.
CV = 5/50 × 100 = 10%
✅ Answer: CV = 10%