AVERAGE (ARITHMETIC MEAN) – QUICK REVISION NOTES
(Designed for JKSSB – Social Forestry Worker & similar exams)
1. What is an Average?
- Definition – The arithmetic mean (average) of a set of numbers is the sum of the observations divided by the number of observations.
- Why it matters – It gives a single representative value that summarises a data set; used in everyday calculations (marks, wages, production, rainfall, etc.).
2. Basic Formula
| Symbol | Meaning |
|---|---|
| \( \bar{x} \) | Arithmetic mean |
| \( \sum x_i \) | Sum of all observations |
| \( n \) | Number of observations |
| \( \bar{x} = \dfrac{\sum x_i}{n} \) | Core formula |
3. Steps to Compute the Mean (Manual)
- List all data values.
- Add them together → \( \sum x_i \).
- Count the total number of values → \( n \).
- Divide the sum by the count → \( \bar{x} \).
Tip: Use a calculator for large data sets; keep track of units (e.g., ₹, kg, cm).
4. Mean of Grouped Data (Frequency Distribution)
When data are presented with frequencies \(f_i\) and class marks (mid‑points) \(x_i\):
\[
\bar{x}= \frac{\sum f_i x_i}{\sum f_i}
\]
Procedure
| Step | Action |
|---|---|
| 1 | Find the class mark (mid‑point) for each class: \(x_i = \frac{\text{Lower limit}+ \text{Upper limit}}{2}\). |
| 2 | Multiply each class mark by its frequency → \(f_i x_i\). |
| 3 | Sum all \(f_i x_i\) → numerator. |
| 4 | Sum all frequencies → denominator. |
| 5 | Divide numerator by denominator. |
5. Weighted Mean
Used when different observations carry different importance (weights).
\[
\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}
\]
- \(w_i\) = weight attached to observation \(x_i\).
- If all weights are equal, weighted mean reduces to simple arithmetic mean.
Common exam scenarios – average speed with different distances, average cost with different quantities, average marks with different subject credits.
6. Properties of the Arithmetic Mean
| Property | Explanation |
|---|---|
| 1. Sum of deviations = 0 | \(\sum (x_i – \bar{x}) = 0\) |
| 2. Minimises squared deviations | \(\sum (x_i – \bar{x})^2\) is smallest for the mean. |
| 3. Linear transformation | If each observation is changed to \(y_i = a x_i + b\), then \(\bar{y}= a\bar{x}+b\). |
| 4. Effect of adding a constant | Adding \(c\) to each observation increases the mean by \(c\). |
| 5. Effect of multiplying by a constant | Multiplying each observation by \(k\) multiplies the mean by \(k\). |
| 6. Sensitivity to outliers | Extreme values pull the mean toward them (unlike median). |
Mnemonic: S.M.L.A.M.E. – Sum zero, Minimise squares, Linear, Add constant, Multiply constant, Extreme sensitivity. —
7. Relationship with Median & Mode (Empirical Formula)
For a moderately skewed distribution:
\[
\text{Mode} \approx 3 \times \text{Median} – 2 \times \text{Mean}
\]
or rearranged
\[
\text{Mean} \approx \frac{3 \times \text{Median} – \text{Mode}}{2}
\]
Use: Quick check when two of the three measures are known. —
8. Short‑Cut Techniques (Useful for Time‑Bound Exams)
| Situation | Shortcut |
|---|---|
| Large numbers | Subtract a convenient base (assumed mean) from each value, work with smaller deviations, then add the base back. |
| Equal class widths | Use the step deviation method: \( \bar{x}= A + \frac{\sum f_i u_i}{\sum f_i} \times h\) where \(u_i = \frac{x_i – A}{h}\). |
| Mean of two groups | If group 1 has \(n_1\) items with mean \(\bar{x}_1\) and group 2 has \(n_2\) items with mean \(\bar{x}_2\): \(\displaystyle \bar{x}= \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2}\). |
| Mean after adding/removing items | New mean = \(\displaystyle \frac{Old\ sum \pm value}{Old\ count \pm 1}\). |
| Mean of consecutive numbers | For an arithmetic progression, mean = \(\frac{\text{First} + \text{Last}}{2}\). (Same as median.) |
| Mean of first n natural numbers | \(\displaystyle \bar{x}= \frac{n+1}{2}\). |
| Mean of first n even numbers | \(\displaystyle \bar{x}= n+1\). |
| Mean of first n odd numbers | \(\displaystyle \bar{x}= n\). |
9. Common Problem Types & Solving Strategies
| Problem Type | Typical Data | Key Formula / Trick |
|---|---|---|
| Simple average of marks/scores | List of numbers | Direct \(\frac{\sum}{n}\). |
| Average speed | Different distances at different speeds | Use harmonic mean when distances equal: \( \text{Average speed}= \frac{2ab}{a+b}\) for two legs; otherwise total distance/total time. |
| Average weight/price with quantities | Weighted data | Weighted mean: \(\frac{\sum (quantity \times price)}{\sum quantity}\). |
| Average after inclusion/exclusion | One value added or removed | Adjust sum and count as shown in shortcut table. |
| Average of grouped data (frequency table) | Class intervals & frequencies | Use mid‑point method or step deviation. |
| Finding missing value when mean known | Mean, \(n-1\) values known | Missing value = \(n \times \text{mean} – \sum(\text{known values})\). |
| Combined average of two sections | Section sizes & means | Weighted mean using section sizes as weights. |
| Average of first n numbers (AP) | Sequence like 2,4,6,… | \(\frac{\text{first}+\text{last}}{2}\) or formula for AP. |
| Correcting erroneous entry | One value mis‑recorded | Correct sum = old sum – wrong value + right value; then recompute mean. |
10. Worked Examples (Illustrative)
Example 1: Simple Mean
Find the average of 12, 15, 9, 20, 14.
Solution
\[
\sum = 12+15+9+20+14 = 70,\quad n=5 \\
\bar{x}= \frac{70}{5}=14
\]
Example 2: Weighted Mean (Cost of Mixed Rice)
A shopkeeper mixes 5 kg rice @ ₹30/kg with 3 kg rice @ ₹45/kg. Find average price per kg.
Solution
\[
\text{Total cost}=5\times30+3\times45=150+135=285\\
\text{Total weight}=5+3=8\text{ kg}\\
\bar{x}= \frac{285}{8}= ₹35.625\text{ per kg}
\]
Example 3: Assumed Mean Method
Data: 102, 108, 115, 119, 121 (n=5). Assume mean \(A=110\).
| \(x_i\) | \(d_i = x_i-A\) |
|---|---|
| 102 | -8 |
| 108 | -2 |
| 115 | 5 |
| 119 | 9 |
|121|11|\[
\sum d_i = -8-2+5+9+11 = 15\\
\bar{x}= A + \frac{\sum d_i}{n}=110+\frac{15}{5}=110+3=113
\]
Example 4: Combined Average of Two Classes
Class A: 30 students, average marks = 68.
Class B: 45 students, average marks = 74.
Solution
\[
\text{Total marks}=30\times68 + 45\times74 = 2040 + 3330 = 5370\\
\text{Total students}=30+45=75\\
\bar{x}= \frac{5370}{75}=71.6
\]
Example 5: Finding Missing Observation
The mean of 5 numbers is 18. Four of them are 12, 20, 22, 15. Find the fifth.
Solution
\[
\text{Sum of 5 numbers}=5\times18=90\\
\text{Sum of known four}=12+20+22+15=69\\
\text{Fifth}=90-69=21\]
Example 6: Correcting a Mis‑recorded Value
Average of 8 observations was found to be 25. Later it was discovered that one observation 42 was wrongly taken as 24. Find the correct average.
Solution
\[\text{Incorrect sum}=8\times25=200\\
\text{Correct sum}=200-24+42=218\\
\text{Correct mean}= \frac{218}{8}=27.25
\]
Example 7: Step Deviation (Grouped Data)
| Class | Frequency (f) | Mid‑point (x) |
|---|---|---|
| 0‑10 | 4 | 5 |
| 10‑20 | 6 | 15 |
| 20‑30 | 10 | 25 |
| 30‑40 | 8 | 35 |
| 40‑50 | 2 | 45 |
Assume \(A=25\), class width \(h=10\).
Compute \(u_i = \frac{x_i-A}{h}\) and \(f_i u_i\):
| x | u = (x-25)/10 | f | f·u |
|---|---|---|---|
| 5 | -2 | 4 | -8 |
| 15 | -1 | 6 | -6 |
| 25 | 0 | 10 | 0 |
| 35 | 1 | 8 | 8 |
| 45 | 2 | 2 | 4 |
\[\sum f = 4+6+10+8+2 = 30\\
\sum f u = -8-6+0+8+4 = -2\\
\bar{x}= A + \frac{\sum f u}{\sum f}\times h = 25 + \frac{-2}{30}\times10 = 25 – \frac{20}{30}=25-0.6667\approx 24.33
\]
11. Mnemonics & Memory Aids
| Concept | Mnemonic |
|---|---|
| Mean formula | “SUM over N” – Sum Up M divided by N. |
| Properties | S.M.L.A.M.E. (see section 6). |
| Weighted mean | “W” for weight – think of a see‑saw: heavier weight pulls the average toward it. |
| Assumed mean | “A” for Anchor – pick an anchor, compute deviations, then shift back. |
| Combined average | “Size × Mean” – multiply each group’s size by its mean, add, divide by total size. |
| Correcting error | “Swap & Re‑sum” – subtract wrong, add right, then recompute. |
| Empirical relation | “3M‑2M = Mode” – 3×Median – 2×Mean ≈ Mode. |
12. Quick Reference Tables #### 12.1 Formulas at a Glance
| Situation | Formula |
|---|---|
| Simple mean | \(\displaystyle \bar{x}= \frac{\sum x_i}{n}\) |
| Weighted mean | \(\displaystyle \bar{x}_w= \frac{\sum w_i x_i}{\sum w_i}\) |
| Assumed mean | \(\displaystyle \bar{x}= A+ \frac{\sum f_i d_i}{\sum f_i}\) |
| Step deviation | \(\displaystyle \bar{x}= A+ \frac{\sum f_i u_i}{\sum f_i}\times h\) |
| Combined mean (2 groups) | \(\displaystyle \bar{x}= \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}\) |
| Mean after adding value \(v\) | \(\displaystyle \bar{x}_{new}= \frac{n\bar{x}+v}{n+1}\) |
| Mean after removing value \(v\) | \(\displaystyle \bar{x}_{new}= \frac{n\bar{x}-v}{n-1}\) |
| Mean of first n natural numbers | \(\displaystyle \frac{n+1}{2}\) |
| Mean of first n even numbers | \(\displaystyle n+1\) |
| Mean of first n odd numbers | \(\displaystyle n\) |
| Harmonic mean (two speeds) | \(\displaystyle \frac{2ab}{a+b}\) |
| Empirical relation (Mode) | \(\displaystyle \text{Mode}\approx 3\text{Median}-2\text{Mean}\) |
12.2 Common Pitfalls & How to Avoid Them
| Pitfall | Why it Happens | Remedy |
|---|---|---|
| Forgetting to include all observations (especially zeroes) | Overlook a value or treat zero as “nothing”. | Write the full list; zero still contributes to sum and count. |
| Using arithmetic mean for rates (speed, density) when time/distance varies | Mean of rates is not appropriate unless weighted by time/distance. | Use harmonic mean for equal distances, or compute total distance/total time. |
| Mis‑applying weighted mean with wrong weights | Using frequencies instead of actual quantities or vice‑versa. | Identify what each weight represents (quantity, credit, time). |
| Incorrect class mark in grouped data | Using lower limit instead of midpoint. | Always compute \((\text{lower}+\text{upper})/2\). |
| Sign errors in assumed/deviation methods | Adding instead of subtracting the assumed mean. | Keep a column for deviations and double‑check signs. |
| Rounding too early in intermediate steps | Leads to cumulative error. | Keep full precision (or at least 2‑3 extra decimals) until final answer. |
| Confusing mean with median/mode when data skewed | Assuming symmetry. | Check skewness; if large, recall median is more robust. |
13. Exam‑Specific Tips (JKSSB – Social Forestry Worker)
- Time Management – Expect 2–3 direct average questions; allocate ~2 minutes each.
- Unit Consistency – Convert all quantities to the same unit before summing (e.g., convert hectares to square meters if needed).
- Read the Word Problem – Identify whether the question asks for average price per unit, average yield per hectare, average number of trees per plot, etc.
- Look for Clues – Words like “mixed”, “combined”, “overall”, “per hectare”, “per worker” often signal weighted or combined average.
- Use Shortcut – If numbers are large or close to each other, apply assumed mean to reduce arithmetic load.
- Verify – After computing, do a quick sanity check: the mean should lie between the smallest and largest observations (unless extreme outlier).
- Practice – Solve at least 10 varied problems (simple, weighted, grouped, missing value, correction) before the exam.
14. Practice Question Bank (Select 10)
| No. | Question | Answer (for self‑check) |
|---|---|---|
| 1 | Find the mean of 7, 9, 12, 15, 18. | 12.2 |
| 2 | The average weight of 5 bags is 24 kg. Four bags weigh 22, 26, 23, 25 kg. Find the weight of the fifth bag. | 20 kg |
| 3 | A farmer mixes 8 kg of seed @ ₹120/kg with 12 kg of seed @ ₹150/kg. What is the average cost per kg? | ₹136/kg |
| 4 | The mean of 8 numbers is 45. If one number, 60, is wrongly recorded as 30, find the correct mean. | 48.75 |
| 5 | Class intervals and frequencies: 0‑10 (5), 10‑20 (12), 20‑30 (20), 30‑40 (10), 40‑50 (3). Find the mean. | 23.5 |
| 6 | The average marks of 30 students in Section A is 62. The average marks of 20 students in Section B is 71. Find the overall average. | 65.2 |
| 7 | Find the mean of the first 50 odd numbers. | 50 |
| 8 | The average speed of a car for a journey of 240 km is 60 km/h. If the first half was travelled at 50 km/h, what was the speed for the second half? | 75 km/h |
| 9 | Using assumed mean \(A=40\), find the mean of the data: 35, 38, 42, 45, 48. | 41.6 |
| 10 | The mean of 7 observations is 18. If each observation is increased by 4, what is the new mean? | 22 |
(Solve each on your own; compare with the answer column.)
15. Final Recap (Bullet‑Point Cheat Sheet) – Mean = Sum ÷ Count – Weighted Mean → weight matters (quantity, credit, time).
- Assumed/Step Deviation → reduce large numbers, work with deviations.
- Combined Mean → \(\frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}\). – Corrections → Adjust sum, then divide by count.
- Properties → Sum of deviations = 0; linear transformation; sensitive to outliers.
- Empirical Relation → Mode ≈ 3·Median – 2·Mean.
- Shortcuts → AP mean = (first+last)/2; first n naturals = (n+1)/2.
- Watch Out → Units, zero values, rates (use harmonic), sign errors, premature rounding.
You are now equipped with a solid, exam‑ready grasp of the arithmetic mean.
Revise the tables, practice the shortcuts, and solve the practice set – you’ll be ready to tackle any average‑related question in the JKSSB Social Forestry Worker (or similar) paper with confidence.
Good luck!
