Arithmetic Mean – Complete Numerical Problems & Properties.

Arithmetic Mean – అంకగణిత సగటు | Formulas, Properties & Numerical Problems

APPSC Statistics – Measures of Central Tendency | Part 2

ఈ Topicలో:

✓ Meaning of Arithmetic Mean
✓ Individual Series
✓ Discrete Series
✓ Continuous Series
✓ Direct Method
✓ Assumed Mean Method
✓ Step-Deviation Method
✓ Weighted Arithmetic Mean
✓ Combined Mean
✓ Corrected Mean
✓ Missing Value Problems
✓ Properties of Arithmetic Mean
✓ Effect of Change of Origin & Scale
✓ APPSC Numerical Shortcuts
✓ Practice MCQs & Advanced Problems

1. Arithmetic Mean అంటే ఏమిటి?

Arithmetic Mean (AM) అనేది Measures of Central Tendencyలో అత్యంత సాధారణంగా ఉపయోగించే average.

Observations అన్నింటి మొత్తాన్ని వాటి సంఖ్యతో భాగిస్తే వచ్చే valueను Arithmetic Mean అంటారు.

Arithmetic Mean = Sum of all Observations ÷ Number of Observations

X̄ = ΣX / N

Important:

Arithmetic Mean calculationలో ప్రతి observation ఉపయోగించబడుతుంది.

2. Arithmetic Mean – Symbols

Symbol	Meaning
X̄	Arithmetic Mean
X	Observation / Variable Value
Σ	Summation
f	Frequency
N	Total number of observations / Σf
A	Assumed Mean
d	Deviation from Assumed Mean
i or h	Common class interval

3. Arithmetic Mean – Individual Series

Frequency ఇవ్వకుండా observations మాత్రమే ఇచ్చినప్పుడు దానిని Individual Seriesగా తీసుకోవచ్చు.

X̄ = ΣX / N

Example 1:

Find the Arithmetic Mean of:

10, 20, 30, 40, 50

ΣX = 10 + 20 + 30 + 40 + 50 = 150

N = 5

X̄ = 150 / 5 = 30

Example 2:

The heights of five persons are:
160, 137, 149, 153 and 161 cm.

Find the mean height.

ΣX = 160 + 137 + 149 + 153 + 161
= 760

N = 5

Mean Height = 760 / 5 = 152 cm

4. Discrete Frequency Distribution

ఒక value ఎన్ని సార్లు occur అయిందో frequency (f)తో ఇచ్చినప్పుడు:

X̄ = ΣfX / Σf

Example 3:

Find Arithmetic Mean:

X	f	fX
10	2	20
20	3	60
30	4	120
40	1	40
Total	Σf = 10	ΣfX = 240

X̄ = 240 / 10 = 24

Common Mistake:

Discrete seriesలో denominator number of X-values కాదు.

Denominator = Σf

5. Continuous Frequency Distribution

Continuous class intervals ఇచ్చినప్పుడు మొదట ప్రతి class యొక్క Class Midpoint (m) కనుగొనాలి.

Midpoint = (Lower Limit + Upper Limit) / 2

X̄ = Σfm / Σf

Example 4:

Find the Arithmetic Mean:

Class	f	Midpoint (m)	fm
0–10	5	5	25
10–20	8	15	120
20–30	12	25	300
30–40	8	35	280
40–50	7	45	315
Total	40	—	1040

X̄ = 1040 / 40 = 26

6. Three Methods of Calculating Arithmetic Mean

Method	Formula	Best Situation
Direct Method	X̄ = ΣfX / Σf	Small/simple values
Assumed Mean Method	X̄ = A + Σfd / Σf	Large values
Step-Deviation Method	X̄ = A + (Σfu / Σf) × i	Equal class intervals

7. Assumed Mean Method

Values పెద్దవిగా ఉన్నప్పుడు ఒక convenient valueను Assumed Mean (A) గా తీసుకొని deviations calculate చేస్తాం.

d = X − A

X̄ = A + Σfd / Σf

Example 5:

X = 10, 20, 30, 40, 50
f = 2, 3, 5, 3, 2

Assume A = 30.

X	f	d = X − 30	fd
10	2	−20	−40
20	3	−10	−30
30	5	0	0
40	3	10	30
50	2	20	40
Total	15	—	0

X̄ = 30 + 0/15

X̄ = 30

Shortcut:

If Σfd = 0,

Mean = Assumed Mean

8. Step-Deviation Method

Equal class intervals ఉన్నప్పుడు calculationsను మరింత simplify చేయడానికి Step-Deviation Method ఉపయోగించవచ్చు.

u = (X − A) / i

X̄ = A + (Σfu / Σf) × i

Example 6:

Class	f	m	u	fu
0–10	2	5	−2	−4
10–20	3	15	−1	−3
20–30	5	25	0	0
30–40	3	35	1	3
40–50	2	45	2	4
Total	15	—	—	0

A = 25 and i = 10

X̄ = 25 + (0/15) × 10

X̄ = 25

9. Weighted Arithmetic Mean

అన్ని observationsకు equal importance లేకపోతే simple Arithmetic Mean సరిపోదు. ప్రతి observationకు ఇచ్చిన importanceను weight (W) అంటారు.

Weighted Mean = ΣWX / ΣW

Example 7:

A student obtains the following marks:

Subject	Marks (X)	Weight (W)	WX
A	80	3	240
B	70	2	140
C	60	1	60
Total	—	6	440

Weighted Mean = 440 / 6
= 73.33

Exam Clue:

Different importance / credits / quantities / weights

→ Use Weighted Arithmetic Mean

10. Combined Arithmetic Mean

రెండు లేదా అంతకంటే ఎక్కువ groups యొక్క individual means మరియు group sizes ఇచ్చినప్పుడు మొత్తం group meanను Combined Mean అంటారు.

Combined Mean = (N₁X̄₁ + N₂X̄₂) / (N₁ + N₂)

Example 8:

Group A: N₁ = 40, Mean = 60
Group B: N₂ = 60, Mean = 70

Find combined mean.

Combined Total = (40 × 60) + (60 × 70)
= 2400 + 4200
= 6600

Total N = 40 + 60 = 100

Combined Mean = 6600 / 100 = 66

Very Important Trap:

(60 + 70) / 2 = 65 చేయకూడదు.

ఎందుకంటే రెండు groups sizes equal కావు.

11. Equal-Sized Groups Shortcut

Group sizes equal అయితే మాత్రమే group means యొక్క simple average తీసుకోవచ్చు.

Group A: 50 students, Mean = 60
Group B: 50 students, Mean = 70

Combined Mean = (60 + 70)/2 = 65

12. Finding Missing Value Using Mean

Example 9:

Mean of 5 observations is 20.
Four observations are 10, 15, 20 and 25.
Find the fifth observation.

Mean = Total / N

Therefore,
Total = Mean × N
= 20 × 5
= 100

Known Total = 10 + 15 + 20 + 25 = 70

Missing Value = 100 − 70 = 30

Fast Formula:

Required Total = Mean × Number of Observations

13. Mean after Adding a New Observation

Example 10:

Mean of 10 observations = 20.
A new observation 31 is added.
Find the new mean.

Old Total = 10 × 20 = 200

New Total = 200 + 31 = 231

New N = 11

New Mean = 231 / 11 = 21

14. Mean after Removing an Observation

Example 11:

Mean of 10 observations = 25.
One observation 34 is removed.
Find the new mean.

Original Total = 10 × 25 = 250

New Total = 250 − 34 = 216

New N = 9

New Mean = 216 / 9 = 24

15. Corrected Arithmetic Mean

ఒక observationను తప్పుగా record చేసినప్పుడు old totalను correct చేసి new mean calculate చేయాలి.

Correct Total = Wrong Total − Wrong Value + Correct Value

Correct Mean = Correct Total / N

Example 12:

Mean of 20 observations was calculated as 30.
Later it was found that 45 was wrongly recorded as 25.
Find the correct mean.

Wrong Total = 20 × 30 = 600

Correct Total = 600 − 25 + 45
= 620

Correct Mean = 620 / 20 = 31

Fast Correction Formula:

Correct Mean = Wrong Mean + (Correct Value − Wrong Value)/N

Correct Mean = 30 + (45 − 25)/20 = 31

16. Missing Frequency Problem

Example 13:

Mean of the following distribution is 25. Find missing frequency k.

X	f	fX
10	2	20
20	3	60
30	k	30k
40	1	40

Σf = 6 + k

ΣfX = 120 + 30k

25 = (120 + 30k)/(6 + k)

150 + 25k = 120 + 30k

30 = 5k

k = 6

17. Important Property – Sum of Deviations from Mean

Arithmetic Mean యొక్క అత్యంత ముఖ్యమైన mathematical property:

Σ(X − X̄) = 0

Frequency distributionలో:

Σf(X − X̄) = 0

Example 14:

Values = 10, 20, 30
Mean = 20

X	X − X̄
10	−10
20	0
30	+10
Total	0

Golden Property:

Algebraic sum of deviations from Arithmetic Mean = Zero.

18. Least Squares Property of Arithmetic Mean

Arithmetic Mean నుంచి తీసుకున్న squared deviations మొత్తం, మరే ఇతర value నుంచి తీసుకున్న squared deviations మొత్తంకంటే తక్కువ లేదా సమానం.

Σ(X − X̄)² is Minimum

APPSC Concept:

Sum of deviations → Zero

Sum of squared deviations → Minimum

ఈ రెండింటిని confuse చేయకండి.

19. Effect of Adding a Constant

ప్రతి observationకు ఒకే constant k add చేస్తే mean కూడా kతో పెరుగుతుంది.

If Y = X + k, then Ȳ = X̄ + k

Mean of X = 25.
Every observation is increased by 7.
Find new mean.

New Mean = 25 + 7 = 32

20. Effect of Subtracting a Constant

If Y = X − k, then Ȳ = X̄ − k

Mean = 50.
Every observation is reduced by 8.

New Mean = 50 − 8 = 42

21. Effect of Multiplication

If Y = kX, then Ȳ = kX̄

Mean = 15.
Every observation is multiplied by 4.

New Mean = 15 × 4 = 60

22. Effect of Division

If Y = X/k, then Ȳ = X̄/k

Mean = 60.
Every observation is divided by 3.

New Mean = 60/3 = 20

23. General Linear Transformation

If every observation is transformed as:

Y = aX + b

then:

Ȳ = aX̄ + b

Example 15:

Mean of X = 10.
Y = 3X + 5.
Find mean of Y.

Ȳ = 3(10) + 5
= 35

24. Reverse Transformation Problem

Mean of Y = 35 and Y = 3X + 5.
Find mean of X.

35 = 3X̄ + 5

30 = 3X̄

X̄ = 10

25. Finding One Group Mean from Combined Mean

Example 16:

Mean of 100 students = 60.
Mean of 40 students = 50.
Find mean of remaining 60 students.

Total marks of 100 students = 100 × 60 = 6000

Total marks of first 40 = 40 × 50 = 2000

Remaining total = 6000 − 2000 = 4000

Mean of remaining 60 students = 4000/60
= 66.67

26. Average Replacement Problem

Example 17:

Average age of 10 persons is 30 years. One person aged 40 years leaves the group and another person joins. The new average becomes 29 years.

Find the age of the new person.

Old Total = 10 × 30 = 300

New Total = 10 × 29 = 290

After 40-year-old leaves:
300 − 40 = 260

New Person's Age = 290 − 260 = 30 years

27. Important Properties of Arithmetic Mean

Property	Result
Based on all observations	Yes
Rigidly defined	Yes
Algebraic treatment	Possible
Sum of deviations from mean	Zero
Sum of squared deviations from mean	Minimum
Add k to every value	Mean increases by k
Subtract k from every value	Mean decreases by k
Multiply every value by k	Mean multiplied by k
Divide every value by k	Mean divided by k
Effect of extreme values	Mean is affected

28. Merits of Arithmetic Mean

✓ Simple to understand

✓ Easy to calculate

✓ Rigidly defined

✓ Based on all observations

✓ Suitable for algebraic treatment

✓ Useful for further statistical analysis

✓ Relatively stable from sample to sample

29. Limitations of Arithmetic Mean

✓ Extreme values can strongly influence the mean.

✓ It may not represent a highly skewed distribution satisfactorily.

✓ The calculated mean need not be an actual observation in the dataset.

✓ It is not appropriate for purely qualitative categories.

✓ Open-ended distributions can create difficulties when class midpoints are not known.

30. APPSC Numerical Shortcuts

Shortcut 1
Mean × Number = Total

Shortcut 2
Missing Value = Required Total − Known Total

Shortcut 3
Correct Total = Wrong Total − Wrong Item + Correct Item

Shortcut 4
Combined Mean = Combined Total / Combined Number

Shortcut 5
Every value + k → Mean + k

Shortcut 6
Every value × k → Mean × k

Shortcut 7
Σ(X − X̄) = 0

Shortcut 8
Σ(X − X̄)² = Minimum

31. APPSC Practice MCQs

Q1. Arithmetic Mean of 10, 20, 30, 40 and 50 is:

A) 20
B) 25
C) 30
D) 35

Answer: C) 30

Q2. If mean of 8 observations is 15, their total is:

A) 100
B) 120
C) 125
D) 150

Answer: B) 120
15 × 8 = 120.

Q3. Mean of 5 numbers is 20. If four numbers are 10, 15, 20 and 25, the fifth number is:

A) 25
B) 30
C) 35
D) 40

Answer: B) 30

Q4. The algebraic sum of deviations from Arithmetic Mean is:

A) 1
B) Maximum
C) Minimum
D) Zero

Answer: D) Zero

Q5. The sum of squared deviations is minimum when deviations are measured from:

A) Mode
B) Median
C) Arithmetic Mean
D) Range

Answer: C) Arithmetic Mean

Q6. If 5 is added to every observation, Arithmetic Mean:

A) remains unchanged
B) increases by 5
C) decreases by 5
D) becomes five times

Answer: B) increases by 5

Q7. Mean of a series is 12. If every value is multiplied by 3, new mean is:

A) 4
B) 12
C) 15
D) 36

Answer: D) 36

Q8. Mean of 20 observations was calculated as 30. A value 25 should actually have been 45. Correct mean is:

A) 29
B) 30
C) 31
D) 32

Answer: C) 31

Q9. Mean of 40 students is 60 and mean of 60 students is 70. Combined mean is:

A) 64
B) 65
C) 66
D) 68

Answer: C) 66

Q10. Which formula is used for Weighted Arithmetic Mean?

A) ΣX/N
B) ΣWX/ΣW
C) Σf/ΣX
D) ΣW/ΣWX

Answer: B) ΣWX/ΣW

32. APPSC Advanced Numerical Problems

Q11.

Average of 20 observations is 15. If each observation is increased by 4 and then multiplied by 2, what is the new average?

Original Mean = 15

After adding 4:
Mean = 19

After multiplying by 2:
Mean = 38

Answer = 38

Q12.

Mean of 50 observations is 40. Mean of 30 of them is 35. Find mean of remaining 20 observations.

Total of 50 = 50 × 40 = 2000

Total of 30 = 30 × 35 = 1050

Remaining Total = 950

Remaining Mean = 950 / 20 = 47.5

Q13.

Mean of 10 observations is 18. If one observation 12 is replaced by 32, find the new mean.

Old Total = 10 × 18 = 180

New Total = 180 − 12 + 32 = 200

New Mean = 200/10 = 20

Q14.

Mean of X is 20. If Y = 5X − 8, find mean of Y.

Ȳ = 5(20) − 8
= 100 − 8
= 92

Q15.

If the mean of 25 observations is 16 and the sum of deviations of 24 observations from 16 is +7, what is the deviation of the remaining observation from the mean?

Σ(X − X̄) = 0

Required deviation = −7

33. Statement-Based Questions

Q16. Consider the following statements:

1. Arithmetic Mean is based on all observations.
2. Sum of deviations from Arithmetic Mean is zero.
3. Arithmetic Mean is unaffected by extreme values.
4. Arithmetic Mean permits algebraic treatment.

Which are correct?

A) 1 and 2 only
B) 1, 2 and 4 only
C) 2, 3 and 4 only
D) All four

Answer: B) 1, 2 and 4 only

Statement 3 is false because Arithmetic Mean is affected by extreme values.

Q17. Which statements are correct?

1. Adding 10 to every value increases mean by 10.
2. Multiplying every value by 2 doubles the mean.
3. Sum of squared deviations from mean is minimum.
4. Mean must always be one of the observed values.

A) 1 and 2 only
B) 1, 2 and 3 only
C) 2, 3 and 4 only
D) All four

Answer: B) 1, 2 and 3 only

34. Most Important Exam Traps

Trap 1:
Mean × N = Total — not Mean + N.

Trap 2:
Frequency distributionలో N = Σf.

Trap 3:
Continuous seriesలో class limitsను directly Xగా తీసుకోకూడదు; midpoint ఉపయోగించాలి.

Trap 4:
Different group sizes ఉన్నప్పుడు group meansను directly average చేయకూడదు.

Trap 5:
Σ(X − X̄) = 0.

Trap 6:
Σ(X − X̄)² is minimum, not zero.

Trap 7:
Wrong value correctionలో wrong itemను subtract చేసి correct itemను add చేయాలి.

Trap 8:
Every observation changes అయితే mean కూడా exactly corresponding linear transformationను follow చేస్తుంది.

Trap 9:
Arithmetic Mean extreme values వల్ల affected అవుతుంది.

Trap 10:
Mean datasetలో actual observation కావాల్సిన అవసరం లేదు.

35. Formula Revision Table

Situation	Formula
Individual Series	X̄ = ΣX / N
Discrete Series	X̄ = ΣfX / Σf
Continuous Series	X̄ = Σfm / Σf
Assumed Mean	X̄ = A + Σfd / Σf
Step Deviation	X̄ = A + (Σfu / Σf) × i
Weighted Mean	X̄w = ΣWX / ΣW
Combined Mean	(N₁X̄₁ + N₂X̄₂)/(N₁ + N₂)
Corrected Total	Wrong Total − Wrong Value + Correct Value
Linear Transformation	If Y = aX + b, then Ȳ = aX̄ + b
Deviation Property	Σ(X − X̄) = 0
Least Squares Property	Σ(X − X̄)² = Minimum

36. One-Minute Final Revision

Arithmetic Mean
= Sum of Observations / Number of Observations

Individual Series
X̄ = ΣX/N

Frequency Series
X̄ = ΣfX/Σf

Continuous Series
First find Midpoints → then Σfm/Σf

Assumed Mean
X̄ = A + Σfd/Σf

Step Deviation
X̄ = A + (Σfu/Σf) × i

Weighted Mean
ΣWX/ΣW

Combined Mean
Combined Total / Combined Number

Missing Value
Mean × N − Known Total

Correction
Wrong Total − Wrong Value + Correct Value

Golden Property 1
Σ(X − X̄) = 0

Golden Property 2
Σ(X − X̄)² = Minimum

Transformation
Y = aX + b
⇒ Ȳ = aX̄ + b

Major Limitation
Arithmetic Mean is affected by extreme values.

Measures of Central Tendency – Concept

Measures of Central Tendency – Meaning, Objectives & Characteristics of a Good Average

APPSC Statistics – Module 2 | Part 1

ఈ Topicలో ముఖ్యంగా:

✓ Meaning of Central Tendency
✓ Meaning of Average
✓ Objectives
✓ Characteristics of a Good Average
✓ Types of Averages
✓ Simple vs Special Averages
✓ Basic applications
✓ Selection of suitable average
✓ APPSC Objective Traps & MCQs

1. Measures of Central Tendency – Meaning

ఒక datasetలో చాలా observations ఉన్నప్పుడు ప్రతి observationను విడివిడిగా పరిశీలించడం కంటే, మొత్తం data యొక్క ముఖ్యమైన లక్షణాన్ని ఒకే representative value ద్వారా తెలియజేయడం సులభం.

ఈ representative లేదా central valueను కొలిచే statistical measureను Measure of Central Tendency అంటారు.

Central Tendency:

ఒక group of observationsను represent చేసే central, typical or representative value.

Large Set of Observations

↓

Summarisation

↓

One Representative Value

↓

Measure of Central Tendency

2. Why is it called Central Tendency?

ఒక distributionలో observations సాధారణంగా ఒక central లేదా typical value చుట్టూ concentrate అయ్యే tendencyను చూపుతాయి.

ఆ central positionను సూచించే measure కాబట్టి దీనిని Measure of Central Tendency అంటారు.

Example:

Marks = 40, 50, 60, 70, 80

ఈ observations యొక్క central representative value = 60.

3. Average అంటే ఏమిటి?

Statisticsలో Average అనే పదాన్ని Measure of Central Tendencyకు సాధారణంగా ఉపయోగిస్తారు.

ఒక పెద్ద group of observations యొక్క ముఖ్యమైన లక్షణాన్ని ఒకే value ద్వారా represent చేసే విలువను Average అంటారు.

APPSC Important:

Average అనగానే Arithmetic Mean మాత్రమే అని భావించకూడదు.

Mean, Median, Mode, Geometric Mean, Harmonic Mean మొదలైనవి కూడా Measures of Central Tendency.

4. Average యొక్క అవసరం – Simple Example

ఒక classలో 50 మంది విద్యార్థుల marks ఉన్నాయి.

ప్రతి విద్యార్థి markను విడిగా చెప్పడం కంటే:

“Average marks of the class = 68”

అని చెప్పడం ద్వారా మొత్తం class performance గురించి ఒక general idea వస్తుంది.

అంటే average పెద్ద మొత్తంలోని dataను condense చేసి సులభంగా అర్థమయ్యే ఒక representative figureగా మారుస్తుంది.

5. Objectives of Measures of Central Tendency

1. Dataను ఒకే Representative Valueలో చూపించడం

పెద్ద సంఖ్యలో observationsను ఒక single value ద్వారా represent చేయడం ప్రధాన objective.

1000 Observations → One Representative Average

2. Complex Dataను Simplify చేయడం

Raw data చాలా పెద్దదిగా లేదా complicatedగా ఉన్నప్పుడు average దానిని సులభంగా అర్థం చేసుకునే summaryగా మారుస్తుంది.

3. Comparisonను సులభం చేయడం

రెండు లేదా అంతకంటే ఎక్కువ groupsను compare చేయడానికి averages ఉపయోగపడతాయి.

Example:

Class A Average Marks = 72
Class B Average Marks = 64

Individual marks అన్నింటినీ compare చేయకుండా రెండు classes యొక్క general performanceను averages ద్వారా compare చేయవచ్చు.

4. General Level of Dataను తెలుసుకోవడం

Income, wages, prices, marks, production, sales, expenditure వంటి variables యొక్క general లేదా typical levelను తెలుసుకోవచ్చు.

5. Decision Makingకు సహాయం

Economics, Business, Government Planning, Administration మరియు Researchలో averages ఆధారంగా decisions తీసుకోవచ్చు.

6. Further Statistical Analysisకు Basis

Central values అనేక statistical calculations మరియు comparisonsకు foundationగా ఉపయోగపడతాయి.

6. Types of Averages

మన syllabusలో ఐదు ప్రధాన averages ఉన్నాయి.

Averages

↓

Simple Averages

 

Special Averages

Category	Averages
Simple Averages	Arithmetic Mean Median Mode
Special Averages	Geometric Mean Harmonic Mean

Very Important Classification:

Mean + Median + Mode → Simple Averages

Geometric Mean + Harmonic Mean → Special Averages

7. Other Measures of Location

Medianతో పాటు dataలో specific positionsను గుర్తించడానికి:

Measure	Division of Data	Dividing Values
Quartiles	4 equal parts	Q₁, Q₂, Q₃
Deciles	10 equal parts	D₁ to D₉
Percentiles	100 equal parts	P₁ to P₉₉

Q₂ = D₅ = P₅₀ = Median

8. Characteristics of a Good Measure of Central Tendency

అన్ని averages ప్రతి situationలో equally suitable కావు. ఒక satisfactory లేదా good averageకు కొన్ని desirable characteristics ఉండాలి.

1. It should be rigidly defined

Averageకు clear and definite definition ఉండాలి.

ఒకే dataకు ఒకే method ఉపయోగిస్తే different investigatorsకు different answers రాకూడదు.

Rigidly Defined = Definite and Unique Value

2. It should be easy to understand

Average యొక్క meaning సులభంగా అర్థమయ్యే విధంగా ఉండాలి.

3. It should be easy to calculate

Calculation complicated లేదా tediousగా ఉండకూడదు.

4. It should be based on all observations

Ideal average సాధ్యమైనంత వరకు datasetలోని observations అన్నింటినీ consider చేయాలి.

Important:

Arithmetic Mean → అన్ని observationsను ఉపయోగిస్తుంది.

Median → ప్రధానంగా positionపై ఆధారపడుతుంది.

Mode → frequencyపై ఆధారపడుతుంది.

5. It should not be unduly affected by extreme values

Very high లేదా very low observations వల్ల average ఎక్కువగా మారిపోకపోవడం desirable.

Exam Trap:

Arithmetic Mean extreme values వల్ల ప్రభావితమవుతుంది.

Median extreme values వల్ల comparatively తక్కువగా ప్రభావితమవుతుంది.

6. It should be representative

Average dataset యొక్క overall natureను సాధ్యమైనంత బాగా represent చేయాలి.

7. It should be capable of further algebraic treatment

Averageను further mathematical calculationsలో ఉపయోగించగలగాలి.

ఈ విషయంలో Arithmetic Meanకు ప్రత్యేక advantage ఉంది.

8. It should be stable with regard to sampling

Different samples తీసుకున్నప్పుడు averageలో చాలా పెద్ద fluctuations రావకుండా relatively stableగా ఉండటం desirable.

9. Good Average – One-Minute Table

Characteristic	Meaning
Rigidly Defined	Definite and unique
Easy to Understand	Simple interpretation
Easy to Calculate	No unnecessary complexity
Based on Observations	Should adequately use available data
Not Unduly Affected by Extremes	Extreme values should not distort it excessively
Representative	Should represent the whole group
Algebraically Treatable	Useful for further mathematical analysis
Stable in Sampling	Should show relatively small sampling fluctuations

10. Basic Idea of Arithmetic Mean

Arithmetic Mean అనేది అత్యంత commonly used measure of central tendency.

Arithmetic Mean = Sum of Observations / Number of Observations

X̄ = ΣX / N

Example:

10, 20, 30, 40, 50

Sum = 150
N = 5

Mean = 150 / 5 = 30

Keyword: All observations → Think Arithmetic Mean

11. Basic Idea of Median

Observationsను ascending లేదా descending orderలో arrange చేసినప్పుడు middle positionలో ఉండే value Median.

Data:

10, 20, 30, 40, 50

Median = 30

Median distributionను రెండు equal positional partsగా divide చేస్తుంది.

Keywords:

Middle Value
Positional Average
Extreme Values
Skewed Distribution

→ Think Median

12. Basic Idea of Mode

Distributionలో అత్యధిక frequencyతో occur అయ్యే valueను Mode అంటారు.

Data:

2, 3, 3, 3, 5, 6, 8

3 occurs most frequently.

Mode = 3

Common Trap:

Mode = Highest Frequency కలిగిన value.

Mode = Highest numerical value కాదు.

13. Basic Idea of Geometric Mean

Geometric Mean observations యొక్క product ఆధారంగా లెక్కించబడే special average.

GM = (X₁ × X₂ × ... × Xₙ)^1/n

Example:

Values = 1, 10, 100

GM = (1 × 10 × 100)^1/3
= (1000)^1/3
= 10

Application Clue:

Growth Rates
Compound Growth
Multiplicative Changes

→ Think Geometric Mean

14. Basic Idea of Harmonic Mean

Harmonic Mean reciprocals ఆధారంగా లెక్కించబడే special average.

HM = n / Σ(1/X)

Application Clue:

Rates / Ratios / Certain Average-Speed Problems

→ Think Harmonic Mean

15. Which Average Should Be Used?

ఏ ఒక్క average కూడా ప్రతి situationకు best కాదు. Purpose మరియు nature of dataను బట్టి suitable averageను select చేయాలి.

Situation	Suitable Measure
General numerical average	Arithmetic Mean
All observations should be used	Arithmetic Mean
Extreme values / highly skewed numerical data	Median
Middle positional value	Median
Most frequently occurring value	Mode
Most popular item/category	Mode
Growth or compound changes	Geometric Mean
Certain rate and average-speed situations	Harmonic Mean

16. Example – Choosing the Correct Average

Problem 1:

A shoe company wants to know which shoe size is demanded most frequently. Which average should be used?

Answer: Mode

Reason: Company needs the most frequently demanded size.

Problem 2:

Monthly incomes of five persons are:

₹20,000, ₹22,000, ₹23,000, ₹25,000, ₹5,00,000

Which measure may better describe the typical central position when the very high income is an outlier?

Answer: Median

₹5,00,000 is an extreme value. Arithmetic Mean would be strongly influenced by it.

Problem 3:

A researcher wants an average suitable for further algebraic calculations. Which measure is generally preferred?

Answer: Arithmetic Mean

17. APPSC High-Yield Traps

Trap 1:
Average ≠ Arithmetic Mean only.

Trap 2:
Mean, Median and Mode → Simple Averages.

Trap 3:
GM and HM → Special Averages.

Trap 4:
Arithmetic Mean uses all observations.

Trap 5:
Median is a positional measure.

Trap 6:
Mode depends on frequency.

Trap 7:
Mode is not necessarily the largest value.

Trap 8:
Quartiles divide data into 4 parts but there are only 3 quartile dividing points.

Trap 9:
Deciles divide data into 10 parts but there are 9 decile points.

Trap 10:
Percentiles divide data into 100 parts but there are 99 percentile points.

Trap 11:
Q₂ = D₅ = P₅₀ = Median.

Trap 12:
No single average is suitable for every situation.

18. APPSC Practice MCQs

Q1. A single value representing the entire group of observations is generally called:

A) Dispersion
B) Average
C) Correlation
D) Index Number

Answer: B) Average

Q2. Which of the following is a simple average?

A) Geometric Mean
B) Harmonic Mean
C) Median
D) None

Answer: C) Median

Q3. Which pair consists of special averages?

A) Mean and Median
B) Median and Mode
C) Geometric Mean and Harmonic Mean
D) Mean and Mode

Answer: C) Geometric Mean and Harmonic Mean

Q4. Which characteristic requires an average to have a definite and unique value?

A) Simplicity
B) Rigid definition
C) Sampling stability
D) Representativeness

Answer: B) Rigid definition

Q5. Which average is based on all observations?

A) Arithmetic Mean
B) Median only
C) Mode only
D) Quartile

Answer: A) Arithmetic Mean

Q6. Which measure is comparatively less affected by extreme observations?

A) Arithmetic Mean
B) Median
C) Geometric Mean
D) Weighted Mean

Answer: B) Median

Q7. The value occurring with the greatest frequency is:

A) Mean
B) Median
C) Mode
D) Quartile

Answer: C) Mode

Q8. Which measure is particularly useful for compound growth?

A) Median
B) Mode
C) Geometric Mean
D) Quartile

Answer: C) Geometric Mean

Q9. Harmonic Mean is especially associated with:

A) Frequencies only
B) Certain rates and speeds
C) Qualitative categories only
D) Cumulative frequencies

Answer: B) Certain rates and speeds

Q10. Quartiles divide an ordered distribution into:

A) 3 equal parts
B) 4 equal parts
C) 10 equal parts
D) 100 equal parts

Answer: B) 4 equal parts

Q11. Number of quartile dividing values is:

A) 2
B) 3
C) 4
D) 5

Answer: B) 3

Q12. Number of decile dividing values is:

A) 9
B) 10
C) 11
D) 100

Answer: A) 9

Q13. Which of the following equals Median?

A) Q₁
B) D₁
C) P₂₅
D) P₅₀

Answer: D) P₅₀

Q14. Which relation is correct?

A) Q₁ = D₅ = P₅₀
B) Q₂ = D₅ = P₅₀
C) Q₃ = D₅ = P₂₅
D) Q₂ = D₁ = P₁₀

Answer: B) Q₂ = D₅ = P₅₀

Q15. A manufacturer wants to know the most popular shirt size. The most appropriate measure is:

A) Mean
B) Median
C) Mode
D) Geometric Mean

Answer: C) Mode

19. Statement-Based Questions

Q16. Consider the following statements:

1. Arithmetic Mean uses all observations.
2. Median is a positional average.
3. Mode depends upon frequency.
4. Quartiles divide data into ten equal parts.

Which are correct?

A) 1 and 2 only
B) 1, 2 and 3 only
C) 2, 3 and 4 only
D) All four

Answer: B) 1, 2 and 3 only

Quartiles divide data into four equal parts.

Q17. Which of the following are desirable characteristics of a good average?

1. Rigidly defined
2. Easy to calculate
3. Capable of algebraic treatment
4. Stable with regard to sampling

A) 1 and 2 only
B) 1, 2 and 3 only
C) 2, 3 and 4 only
D) 1, 2, 3 and 4

Answer: D) 1, 2, 3 and 4

20. Assertion–Reason

Q18.

Assertion (A): Median is often preferred when extreme observations distort the arithmetic mean.

Reason (R): Median is determined primarily by the position of observations rather than their magnitudes.

Answer: Both A and R are true, and R correctly explains A.

Q19.

Assertion (A): Mode can be useful for identifying the most popular item.

Reason (R): Mode represents the value or category having the greatest frequency.

Answer: Both A and R are true, and R correctly explains A.

Q20.

Assertion (A): One particular average is best for every statistical problem.

Reason (R): Selection of an average depends upon the nature of data and purpose of analysis.

Answer: Assertion is false; Reason is true.

21. 10-Second Identification Table

Question Keyword	Think Immediately
All observations	Arithmetic Mean
General numerical average	Arithmetic Mean
Extreme values	Median
Middle position	Median
Most frequent	Mode
Most popular category	Mode
Compound growth	Geometric Mean
Rates / certain speeds	Harmonic Mean
4 equal parts	Quartiles
10 equal parts	Deciles
100 equal parts	Percentiles
50th percentile	Median

22. One-Minute Master Table

Measure	Basic Principle	Main Clue
Arithmetic Mean	ΣX / N	All observations
Median	Middle position	Extreme values / position
Mode	Maximum frequency	Most popular
Geometric Mean	nth root of product	Growth
Harmonic Mean	Reciprocal-based average	Rates
Quartiles	4 equal parts	Q₁, Q₂, Q₃
Deciles	10 equal parts	D₁–D₉
Percentiles	100 equal parts	P₁–P₉₉

చివరి నిమిషం పునశ్చరణ

Central Tendency
→ Large dataను one representative valueతో summarise చేస్తుంది.

Simple Averages
→ Mean + Median + Mode

Special Averages
→ Geometric Mean + Harmonic Mean

Arithmetic Mean
→ All observations
→ Algebraic treatmentకు suitable
→ Extreme values ప్రభావం ఉంటుంది.

Median
→ Positional average
→ Middle value
→ Extreme values ఉన్నప్పుడు useful.

Mode
→ Maximum frequency
→ Most popular value/category.

GM
→ Growth / compound changes.

HM
→ Rates / certain speed problems.

Quartiles → 4 parts → 3 dividing points.

Deciles → 10 parts → 9 dividing points.

Percentiles → 100 parts → 99 dividing points.

Golden Relation:
Q₂ = D₅ = P₅₀ = Median

Andhra Pradesh Socio-Economic Survey 2025–26 rankings + NITI Aayog indices.

ఆంధ్రప్రదేశ్ ర్యాంకులు 2025–26 – Socio Economic Survey & NITI Aayog Indices

APPSC Special Study Material

ఆంధ్రప్రదేశ్ Socio Economic Survey 2025–26 మరియు NITI Aayog సూచీలలో ఆంధ్రప్రదేశ్ సాధించిన ముఖ్యమైన ర్యాంకులు, స్కోర్లు, ఉత్పత్తి గణాంకాలు మరియు Exam-Oriented One-Liners.

PART–A: ఆంధ్రప్రదేశ్ Socio Economic Survey 2025–26

1. భౌగోళిక & జనాభా గణాంకాలు

అంశం	ఆంధ్రప్రదేశ్ ర్యాంకు	ముఖ్య గణాంకం
వైశాల్యం (Area)	8వ స్థానం	1,62,970 చ.కి.మీ.
జనాభా (Population)	10వ స్థానం	4.96 కోట్లు
తీరరేఖ పొడవు	3వ స్థానం	1,053 కి.మీ.

APPSC Memory Trick:

Area → 8
Population → 10
Coastline → 3

2. ఆర్థిక రంగం & ఎగుమతులు

National Exports Share: ఎగుమతులు చేసే పెద్ద రాష్ట్రాల జాబితాలో ఆంధ్రప్రదేశ్ 6వ స్థానంలో ఉంది.

3. ఉద్యానవన రంగం – Horticulture

ఆంధ్రప్రదేశ్ వ్యవసాయ అనుబంధ రంగాల్లో ముఖ్యంగా ఉద్యానవన పంటల ఉత్పత్తి మరియు ఉత్పాదకతలో అనేక అగ్రస్థానాలను కలిగి ఉంది.

ఉత్పత్తి పరంగా ర్యాంకులు

పంట / అంశం	AP Rank	ముఖ్య సమాచారం
మొత్తం పండ్ల ఉత్పత్తి	1వ స్థానం	జాతీయ GVA వాటా 15.6%
అరటిపండు ఉత్పత్తి	1వ స్థానం	దేశంలో అగ్రస్థానం
Oil Palm – Area & Production	1వ స్థానం	వైశాల్యం మరియు ఉత్పత్తి రెండింటిలో
మొత్తం Horticulture GVA	3వ స్థానం	జాతీయ స్థాయి
కొబ్బరి ఉత్పత్తి	4వ స్థానం	—
కాఫీ ఉత్పత్తి	4వ స్థానం	—

ఉత్పాదకత (Productivity)లో AP No.1

1వ స్థానం సాధించిన పంటలు:

✓ Oil Palm
✓ Papaya – బొప్పాయి
✓ Lime – నిమ్మ
✓ Cocoa – కోకో
✓ Tomato – టమోటా
✓ Coconut – కొబ్బరి

మిరప ఉత్పాదకత (Chilli Productivity) → 2వ స్థానం

Important Trap:

కొబ్బరి ఉత్పత్తిలో → 4వ స్థానం
కొబ్బరి ఉత్పాదకతలో → 1వ స్థానం

4. సాగునీటి రంగం – Micro Irrigation

Micro Irrigation Implementation → ఆంధ్రప్రదేశ్ 1వ స్థానం

దేశంలో అత్యధిక Micro Irrigation జరుగుతున్న Top-10 జిల్లాల్లో ఆంధ్రప్రదేశ్‌కు చెందిన 6 జిల్లాలు ఉన్నాయి.

ఆ 6 జిల్లాలు:

1. అనంతపురం
2. వైఎస్సార్ కడప
3. అన్నమయ్య
4. శ్రీ సత్యసాయి
5. ప్రకాశం
6. చిత్తూరు

5. పశుసంవర్ధక & మత్స్య రంగాలు

రంగం	AP Rank	ముఖ్య గణాంకం
గుడ్ల ఉత్పత్తి	1వ స్థానం	2,739.03 కోట్ల గుడ్లు
Fisheries Sector	1వ స్థానం	జాతీయ చేపల ఉత్పత్తిలో 29% వాటా
Shrimp Exports	అగ్రస్థానం	75% వాటా
Tassar Silk	1వ స్థానం	టస్సార్ పట్టు ఉత్పత్తి
Mulberry & Total Silk	2వ స్థానం	కర్ణాటక మొదటి స్థానం
Poultry Population	2వ స్థానం	—
Sheep Population	2వ స్థానం	—
Meat Production	4వ స్థానం	—
Buffalo Population	6వ స్థానం	—
Milk Production	7వ స్థానం	139.46 లక్షల మెట్రిక్ టన్నులు
Goat Population	11వ స్థానం	—
Total Cattle	14వ స్థానం	—

Most Important APPSC Sequence:

Eggs → 1
Poultry Population → 2
Sheep → 2
Meat → 4
Buffalo → 6
Milk → 7
Goat → 11
Total Cattle → 14

6. ప్రాంతీయ ఉత్పత్తుల ప్రత్యేకతలు

ప్రాంతం	ప్రత్యేకత
రాయలసీమ	Horticulture ఉత్పత్తిలో రాష్ట్రంలో 1వ స్థానం – 52% వాటా
కోనసీమ	రాష్ట్ర కొబ్బరి ఉత్పత్తిలో 40% పైగా వాటా
Visakhapatnam Economic Region	అరటిపండు ఉత్పత్తిలో 34% వాటా
తూర్పు గోదావరి & కాకినాడ	Oil Palm ఉత్పత్తిలో రాష్ట్రంలో మొదటి స్థానం

7. ప్రభుత్వ పథకాలు & పురస్కారాలు

iGOT Karmayogi App
వినియోగం మరియు పనితీరులో → 1వ స్థానం

MGNREGS
అమలు తీరు మరియు ఎక్కువ పనిదినాల కల్పనలో → 3వ స్థానం

మహిళల భాగస్వామ్యం → 60% పైగా

Gopal Ratna Award:
ఆంధ్రప్రదేశ్‌కు చెందిన శ్రీమతి అనురాధ → 3వ బహుమతి / ర్యాంకు

PART–B: NITI Aayog Indicesలో ఆంధ్రప్రదేశ్

8. SDG India Index

Report	AP Score
2020–21	72
2023–24	74

2020–21 నుంచి 2023–24 వరకు AP SDG Score

72 → 74

అంటే 2 పాయింట్ల మెరుగుదల.

AP Best Performing SDG:

SDG 7 – Affordable & Clean Energy
Score → 100/100
Rank → 1వ స్థానం

AP Weakest Goal:

SDG 9 – Industry, Innovation & Infrastructure
Score → 52 నుంచి 49కు తగ్గింది
Rank → 21వ స్థానం
Category → Aspirant

9. AP Rank in 16 SDG Goals

SDG	Goal	AP Rank
SDG 1	No Poverty	3
SDG 2	Zero Hunger	10
SDG 3	Good Health & Well-being	11
SDG 4	Quality Education	20
SDG 5	Gender Equality	14
SDG 6	Clean Water & Sanitation	11
SDG 7	Affordable & Clean Energy	1
SDG 8	Decent Work & Economic Growth	16
SDG 9	Industry, Innovation & Infrastructure	21
SDG 10	Reduced Inequalities	17
SDG 11	Sustainable Cities & Communities	9
SDG 12	Responsible Consumption & Production	4
SDG 13	Climate Action	12
SDG 14	Life Below Water	2
SDG 15	Life on Land	14
SDG 16	Peace, Justice & Strong Institutions	12

High-Yield SDG Ranks:

SDG 7 → 1st
SDG 14 → 2nd
SDG 1 → 3rd
SDG 12 → 4th

Weakest ranking:
SDG 9 → 21st

10. Multidimensional Poverty Index – MPI

Indicator	AP Status
National Rank	9వ స్థానం
NFHS-4 Poverty	11.77%
NFHS-5 Poverty	6.06%
National Average	14.96%
Reduction	5.71 percentage points

AP Multidimensional Poverty:

11.77% → 6.06%

దేశంలో అత్యంత తక్కువ పేదరికంతో కేరళ మొదటి స్థానంలో ఉన్నట్లు ఇచ్చిన నోట్స్ పేర్కొంటున్నాయి.

11. India Innovation Index

Indicator	AP Performance
Overall – Major States	9వ స్థానం
Score	13.32
Enablers Rank	8వ స్థానం
Performance Rank	14వ స్థానం
Business Environment Score	37.06
National Average	28.13

Innovation Weaknesses:

Knowledge Workers Score → 4.04
Investments Score → 4.48

12. Energy Efficiency & Climate Readiness

State Energy Efficiency Index

AP Score → 79.3%
Rank → 1వ స్థానం
Category → Front Runner

విద్యుత్ పొదుపు మరియు సమర్థవంతమైన వినియోగంలో కూడా ఆంధ్రప్రదేశ్‌కు ప్రథమ స్థానం గుర్తింపు లభించినట్లు నోట్స్ పేర్కొంటున్నాయి.

Climate Readiness Index

నిర్దిష్ట వాతావరణ లక్ష్యాల సాధన మరియు nodal agencies నిర్వహణలో AP → 1వ స్థానం

13. School Education Quality

Indicator	AP Data
Upper Primary GER	81.36% → 101%
Private School Students	50% కంటే ఎక్కువ
Secondary Dropout Rate	15.5%
Primary-only Schools	38,212
Classes 1–12 at same institution	557 schools

14. Investment Friendliness Index

Major States Rank: 6వ స్థానం
Overall Rank: 8వ స్థానం
Score: 48.7%

Key Strengths:

Per-capita Digital Transactions → 18
Industrial Power Supply → 23.71 hours/day
STEM Students Share → 48%

Key Concern:

GSDPలో liabilities → 35%

15. Export Preparedness Index

Indicator	AP Performance
Major States Rank	5వ స్థానం
Score	60.65
Exports Value	₹1.6 లక్షల కోట్లు
Exports Value National Rank	6వ స్థానం

Major AP Exports:

✓ Shrimp
✓ Pharma
✓ Rice
✓ Tobacco
✓ Chilli
✓ Automobiles
✓ Petroleum Products

భారత రొయ్యల ఎగుమతుల్లో సుమారు 2/3 వంతు AP నుంచి వెళ్తుందని ఇచ్చిన NITI notes పేర్కొంటున్నాయి.

16. Fiscal Health Index

AP Rank: 18 ప్రధాన రాష్ట్రాల్లో 17వ స్థానం
Score: 23.11
Category: Aspirational

Fiscal Indicator	Value
CAGR	12.7%
Debt-to-GSDP Ratio – 2023–24	33%
Other Major States Average	28.3%
Interest Payments / Revenue	16.2%

ఇచ్చిన నోట్స్ ప్రకారం AP Fiscal Health Rank 2014–15లో 3వ స్థానం నుంచి 17వ స్థానానికి తగ్గింది.

17. AP No.1 – అత్యంత ముఖ్యమైన జాబితా

ఆంధ్రప్రదేశ్ 1వ స్థానంలో ఉన్న ముఖ్య అంశాలు:

✓ మొత్తం పండ్ల ఉత్పత్తి
✓ అరటిపండు ఉత్పత్తి
✓ Oil Palm Area & Production
✓ Oil Palm Productivity
✓ Papaya Productivity
✓ Lime Productivity
✓ Cocoa Productivity
✓ Tomato Productivity
✓ Coconut Productivity
✓ Micro Irrigation Implementation
✓ Egg Production
✓ Fisheries Sector
✓ Tassar Silk
✓ iGOT Karmayogi App Usage
✓ SDG 7 – Affordable & Clean Energy
✓ State Energy Efficiency Index
✓ Climate Readiness

18. Rank-Wise Quick Revision

Rank	Important AP Indicators
1st	Fruits, Banana, Oil Palm, Eggs, Fisheries, Tassar Silk, Micro Irrigation, SDG-7, Energy Efficiency
2nd	Mulberry/Total Silk, Poultry Population, Sheep Population, Chilli Productivity, SDG-14
3rd	Coastline, Horticulture GVA, MGNREGS, SDG-1
4th	Coconut Production, Coffee Production, Meat Production, SDG-12
5th	Export Preparedness – Major States
6th	National Export Share, Buffalo Population, Investment Friendliness – Major States
7th	Milk Production
8th	Area; Investment Friendliness Overall
9th	MPI; India Innovation Index – Major States; SDG-11
10th	Population; SDG-2
17th	Fiscal Health Index; SDG-10
21st	SDG-9

19. APPSC Exam Traps

Trap 1:
AP Area Rank → 8; Population Rank → 10.

Trap 2:
Coastline → 3rd, not 1st.

Trap 3:
Coconut Production → 4th; Coconut Productivity → 1st.

Trap 4:
Egg Production → 1st; Poultry Population → 2nd.

Trap 5:
Milk Production → 7th; Meat Production → 4th.

Trap 6:
SDG Overall Score 2023–24 → 74.

Trap 7:
SDG 7 → 100/100 → 1st Rank.

Trap 8:
SDG 9 is the weak point → 21st Rank.

Trap 9:
MPI Poverty → 6.06%; National Average → 14.96%.

Trap 10:
Innovation Overall Rank → 9th among major states.

Trap 11:
Export Preparedness → 5th among major states; Export Value → 6th nationally.

Trap 12:
Fiscal Health Index → 17th among 18 major states.

20. APPSC Practice MCQs

Q1. వైశాల్యం పరంగా ఆంధ్రప్రదేశ్ స్థానం?

A) 6వ
B) 7వ
C) 8వ
D) 10వ

Answer: C) 8వ స్థానం

Q2. ఆంధ్రప్రదేశ్ తీరరేఖ పొడవు ఎంత?

A) 974 km
B) 1,053 km
C) 1,250 km
D) 1,620 km

Answer: B) 1,053 km

Q3. మొత్తం పండ్ల ఉత్పత్తిలో AP స్థానం?

A) 1వ
B) 2వ
C) 3వ
D) 4వ

Answer: A) 1వ స్థానం

Q4. కొబ్బరి ఉత్పాదకతలో AP స్థానం?

A) 1వ
B) 2వ
C) 3వ
D) 4వ

Answer: A) 1వ స్థానం

Q5. Micro Irrigation Implementationలో AP స్థానం?

A) 1వ
B) 2వ
C) 3వ
D) 5వ

Answer: A) 1వ స్థానం

Q6. గుడ్ల ఉత్పత్తిలో AP స్థానం?

A) 1వ
B) 2వ
C) 3వ
D) 4వ

Answer: A) 1వ స్థానం

Q7. పాల ఉత్పత్తిలో AP స్థానం?

A) 4వ
B) 5వ
C) 6వ
D) 7వ

Answer: D) 7వ స్థానం

Q8. 2023–24 SDG India Indexలో AP Score?

A) 70
B) 72
C) 74
D) 79

Answer: C) 74

Q9. ఏ SDGలో AP 100/100 సాధించింది?

A) SDG 1
B) SDG 7
C) SDG 9
D) SDG 14

Answer: B) SDG 7

Q10. SDG 14 – Life Below Waterలో AP Rank?

A) 1వ
B) 2వ
C) 3వ
D) 4వ

Answer: B) 2వ స్థానం

Q11. AP Multidimensional Poverty శాతం?

A) 5.71%
B) 6.06%
C) 11.77%
D) 14.96%

Answer: B) 6.06%

Q12. India Innovation Indexలో AP Rank among major states?

A) 6వ
B) 8వ
C) 9వ
D) 14వ

Answer: C) 9వ స్థానం

Q13. State Energy Efficiency Indexలో AP Score?

A) 60.65%
B) 74%
C) 79.3%
D) 81.36%

Answer: C) 79.3%

Q14. Export Preparedness Indexలో పెద్ద రాష్ట్రాల మధ్య AP Rank?

A) 3వ
B) 4వ
C) 5వ
D) 6వ

Answer: C) 5వ స్థానం

Q15. Fiscal Health Indexలో 18 ప్రధాన రాష్ట్రాల్లో AP Rank?

A) 15వ
B) 16వ
C) 17వ
D) 18వ

Answer: C) 17వ స్థానం

21. Statement-Based Questions

Q16. Consider the following statements:

1. AP ranks first in Egg Production.
2. AP ranks second in Poultry Population.
3. AP ranks first in Milk Production.
4. AP ranks fourth in Meat Production.

Which are correct?

A) 1 and 2 only
B) 1, 2 and 4 only
C) 2, 3 and 4 only
D) All four

Answer: B) 1, 2 and 4 only

Milk Productionలో AP → 7వ స్థానం.

Q17. Consider the following:

1. SDG 7 – AP 1st
2. SDG 14 – AP 2nd
3. SDG 1 – AP 3rd
4. SDG 12 – AP 4th

A) 1 and 2 only
B) 1, 2 and 3 only
C) 2, 3 and 4 only
D) All four

Answer: D) All four

Q18. Which pair is incorrectly matched?

A) SDG Score 2023–24 – 74
B) MPI Poverty – 6.06%
C) Energy Efficiency – 79.3%
D) Fiscal Health Rank – 7th

Answer: D)
Fiscal Health Rank → 17th.

22. సంఖ్యలను గుర్తుంచుకోవడానికి Super Revision

1st Rank:
Fruits – Banana – Oil Palm – Eggs – Fisheries – Tassar – Micro Irrigation – SDG 7 – Energy Efficiency

2nd Rank:
Poultry – Sheep – Silk – Chilli Productivity – SDG 14

3rd Rank:
Coastline – Horticulture GVA – MGNREGS – SDG 1

4th Rank:
Coconut Production – Coffee – Meat – SDG 12

Important Scores:
SDG → 74
SDG 7 → 100/100
MPI → 6.06%
Innovation → 13.32
Energy Efficiency → 79.3%
Investment Friendliness → 48.7%
Export Preparedness → 60.65
Fiscal Health → 23.11

23. చివరి నిమిషం పునశ్చరణ

AP Area → 8th

AP Population → 10th

Coastline → 3rd → 1,053 km

National Export Share → 6th

Fruit Production → 1st

Banana Production → 1st

Oil Palm → 1st

Micro Irrigation → 1st

Egg Production → 1st

Fisheries → 1st

Milk Production → 7th

SDG Score → 74

SDG 7 → 100/100 → 1st

SDG 14 → 2nd

SDG 9 → 21st

MPI Poverty → 6.06%

Innovation Index → 9th among major states

Energy Efficiency → 79.3% → 1st

Export Preparedness → 5th among major states

Fiscal Health → 17th among 18 major states

Top Daily Current Affairs Telugu 19.09.2026

కరెంట్ అఫైర్స్ 2026 – అవార్డులు, క్రీడలు, సైన్స్, అంతర్జాతీయ సంబంధాలు & ఆంధ్రప్రదేశ్ ముఖ్యాంశాలు

Competitive Exams Focus:

APPSC • TSPSC • Groups • Police • DSC • SSC • Banking • Railway మరియు ఇతర పోటీ పరీక్షలకు ఉపయోగపడే ముఖ్యమైన Current Affairs.

1. సినిమా మరియు జాతీయ పురస్కారాలు

దాదాసాహెబ్ ఫాల్కే పురస్కారం 2024

పురస్కార గ్రహీత: అనంత్ నాగ్
రంగం: సినిమా
బహుమతి: ₹15 లక్షలు + స్వర్ణ కమలం
పురస్కార కార్యక్రమం: 72వ జాతీయ చలనచిత్ర పురస్కారాలు
తేదీ: సెప్టెంబర్ 22, 2026

ప్రముఖ కన్నడ నటుడు అనంత్ నాగ్ 2024 సంవత్సరానికి దాదాసాహెబ్ ఫాల్కే పురస్కారానికి ఎంపికయ్యారు. కన్నడ, హిందీతో పాటు పలు భాషల్లో ఆయన 300కు పైగా చిత్రాల్లో నటించారు.

Static GK:

• దాదాసాహెబ్ ఫాల్కే పురస్కారం ప్రారంభం – 1969
• తొలి గ్రహీత – దేవికారాణి
• భారతీయ చలనచిత్ర పితామహుడు – ధుండిరాజ్ గోవింద్ ఫాల్కే
• భారతీయ సినిమాలో ఇచ్చే అత్యున్నత పురస్కారం – దాదాసాహెబ్ ఫాల్కే పురస్కారం

ఆస్కార్స్ 2027 – భారత అధికారిక ఎంట్రీ

చిత్రం: గోందల్ (Gondhal)
భాష: మరాఠీ
దర్శకుడు & రచయిత: సంతోష్ ధావకర్
సంగీతం: ఇళయరాజా
విభాగం: Best International Feature Film
ఎంపిక సంస్థ: Film Federation of India (FFI)

99వ అకాడమీ అవార్డ్స్‌కు Best International Feature Film విభాగంలో భారత అధికారిక ఎంట్రీగా మరాఠీ సైకలాజికల్ థ్రిల్లర్ గోందల్ ఎంపికైంది.

2. క్రీడలు – 20వ ఆసియా క్రీడలు 2026

అంశం	వివరం
Edition	20వ Asian Games
వేదిక	Aichi-Nagoya, Japan
తేదీలు	19 సెప్టెంబర్ – 4 అక్టోబర్ 2026
భారత పతాకదారులు	మనూ భాకర్ & తేజేందర్ పాల్ సింగ్ తూర్
అధికారిక మస్కట్	Honohan
నిర్వహణ సంస్థ	Olympic Council of Asia (OCA)
భారత బృందం	503 మంది అథ్లెట్లు – 37 క్రీడాంశాలు

Static GK:

Olympic Council of Asia (OCA)
• ఏర్పాటు – 1982
• ప్రధాన కార్యాలయం – కువైట్ సిటీ

3. పర్యావరణం, సైన్స్ & టెక్నాలజీ

A. భారత్ వన్ గ్రీన్ (Bharat Vana Green)

న్యూఢిల్లీలోని లోధి కాలనీ పార్క్‌లో కేంద్ర ఆర్థిక మంత్రి నిర్మల సీతారామన్ ఈ కార్యక్రమాన్ని ప్రారంభించారు.

లక్ష్యం: పట్టణ ప్రాంతాల్లో దట్టమైన Micro Forests అభివృద్ధి.

పద్ధతి: Miyawaki Method

అమలు: New Delhi Municipal Council (NDMC)

Miyawaki Methodను జపనీస్ వృక్షశాస్త్రవేత్త అకిరా మియావాకి 1970లలో అభివృద్ధి చేశారు. స్థానిక మొక్కల రకాలను ఉపయోగించి తక్కువ కాలంలో దట్టమైన అడవులను అభివృద్ధి చేయడం దీని ప్రత్యేకత.

B. ఆంధ్రప్రదేశ్‌లో కొత్త మొక్క జాతి

Scientific Name: Caralluma umbellata
స్థానిక పేరు: కుందేటి కొమ్ములు
కనుగొన్నవారు: యోగి వేమన విశ్వవిద్యాలయ పరిశోధకులు
ప్రాంతం: పోరుమామిళ్ల అటవీ ప్రాంతం, వైఎస్సార్ కడప జిల్లా

ఈ మొక్కకు నక్షత్ర ఆకారంలో ముదురు ఎరుపు రంగు పువ్వులు ఉంటాయి. రక్తంలో చక్కెర స్థాయిల నియంత్రణ కోసం ఆయుర్వేదంలో దీనిని ఉపయోగిస్తారు.

C. పర్యావరణ మరియు వాతావరణ సదస్సు

అంశం	వివరం
వేదిక	న్యూఢిల్లీ
తేదీలు	సెప్టెంబర్ 19–20, 2026
ప్రారంభించినవారు	ప్రధాని నరేంద్ర మోదీ
నిర్వాహకులు	National Green Tribunal (NGT)

NGT – Static GK

• NGT Act – 2010
• ఏర్పాటు – 18 అక్టోబర్ 2010
• ప్రధాన కార్యాలయం – న్యూఢిల్లీ
• చైర్‌పర్సన్ – జస్టిస్ ప్రకాష్ శ్రీవాస్తవ్

D. అత్యంత పిన్న వయస్సు గ్రహం – Elias 224b

గ్రహం: Elias 224b
వయస్సు: 10 లక్షల సంవత్సరాల కంటే తక్కువ
దూరం: భూమికి సుమారు 450 కాంతి సంవత్సరాలు
రకం: బృహస్పతి పరిమాణంలోని వాయు గ్రహం

4. అంతర్జాతీయ సంబంధాలు

A. భారత్ – న్యూజిలాండ్ స్వేచ్ఛా వాణిజ్య ఒప్పందం

FTA – Key Points

• భారత్ నుంచి వెళ్లే అన్ని ఎగుమతులకు 100% Duty-Free ప్రవేశం.
• 15 సంవత్సరాల్లో న్యూజిలాండ్ భారత్‌లో $20 బిలియన్ పెట్టుబడి.
• 5,000 మంది భారతీయ IT, Healthcare, Engineering నిపుణులకు 3 సంవత్సరాల Work Visas.
• Dairy, Agriculture, Sugar, Artificial Honey వంటి రంగాలకు మినహాయింపులు.

B. EU Associate Membership

European Commission President Ursula von der Leyen “Alliance for the Future” కింద Canadaకు తొలి Associate Membership హోదాను ప్రతిపాదించారు.

EU Quick Facts:

• సభ్యదేశాలు – 27
• ప్రధాన కేంద్రం – Brussels, Belgium
• Eurozone – 21 దేశాలు
• Bulgaria – 1 January 2026న 21వ Eurozone దేశంగా చేరింది.

C. Djibouti – Artemis Accords

Djibouti Artemis Accordsపై సంతకం చేసిన 72వ దేశం మరియు 8వ ఆఫ్రికా దేశం.

Artemis Accordsను NASA నేతృత్వంలో శాంతియుత అంతరిక్ష పరిశోధన కోసం 2020లో ప్రారంభించారు.

D. India–Japan: Year of Shared Horizons 2027

భారత్–జపాన్ దౌత్య సంబంధాలకు 75 సంవత్సరాలు పూర్తవుతున్న సందర్భంగా 2027ను “Year of Shared Horizons”గా జరుపుకోవాలని నిర్ణయించారు.

5. రక్షణ, ఆర్థిక రంగం & సంస్థలు

A. Shahzad Bhatti Network – UAPA

పాకిస్తాన్ కేంద్రంగా పనిచేస్తున్న Shahzad Bhatti Networkను కేంద్ర హోం మంత్రిత్వ శాఖ UAPA, 1967 కింద ఉగ్రవాద సంస్థగా ప్రకటించింది.

B. 43వ Indian Coast Guard Commanders' Conference

అంశం	వివరం
Edition	43వ
వేదిక	ICG Headquarters, New Delhi
ప్రారంభించినవారు	సంజయ్ సేథ్
ICG ఏర్పాటు	1 February 1977
నినాదం	వయం రక్షామః – “We Protect”
Director General	పరమేష్ శివమణి

C. World Bank Group Chief Economist

నియామకం: Michael Kremer
బాధ్యతలు: 1 October 2026 నుంచి
ప్రత్యేకత: 2019 Economics Nobel Laureate

Michael Kremer 2019లో Abhijit Banerjee మరియు Esther Dufloతో కలిసి ఆర్థిక శాస్త్రంలో నోబెల్ బహుమతి పొందారు.

D. 69వ Commonwealth Parliamentary Conference

వేదిక: Cape Town, South Africa
భారత బృందం నేత: హరివంశ్ నారాయణ్ సింగ్ – రాజ్యసభ డిప్యూటీ ఛైర్మన్

6. ఆంధ్రప్రదేశ్ & ప్రాంతీయ కరెంట్ అఫైర్స్

A. SARIKA AI – Passenger Assistance

South Coast Railway విశాఖపట్నం రైల్వే స్టేషన్‌లో ప్రయాణికులకు సహాయం చేయడానికి SARIKA AI ఆధారిత Voice/Text సేవలను pilot projectగా ప్రారంభించింది.

B. దుగరాజపట్నం Ship Building Cluster

సంస్థ: Mazagon Dock Shipbuilders Limited (MDL)
ప్రాంతం: దుగరాజపట్నం, తిరుపతి జిల్లా
ప్రతిపాదిత పెట్టుబడి: ₹15,000 కోట్లు

C. గురజాడ అప్పారావు జయంతి – రాష్ట్ర పండుగ

మహాకవి, సాంఘిక సంస్కర్త గురజాడ అప్పారావు జయంతి అయిన సెప్టెంబర్ 21ను అధికారికంగా రాష్ట్ర పండుగ (State Panduga)గా జరపాలని ఆంధ్రప్రదేశ్ ప్రభుత్వం నిర్ణయించింది.

Remember:

గురజాడ అప్పారావు → కన్యాశుల్కం
గురజాడ అప్పారావు → “దేశమును ప్రేమించమన్నా”
జయంతి → సెప్టెంబర్ 21

D. ప్రపంచంలోనే అత్యంత ఎత్తైన Twin Flower Fields

ప్రాంతం: Leh District, Ladakh
ప్రదేశాలు: Choglamsar & Stakna
ఎత్తు: 10,600 అడుగులు
ప్రారంభించినవారు: లడఖ్ లెఫ్టినెంట్ గవర్నర్ వినయ్ కుమార్ సక్సేనా

7. One-Minute Current Affairs Master Table

Keyword	Answer
Dadasaheb Phalke Award 2024	అనంత్ నాగ్
India's Oscars 2027 Entry	Gondhal
Asian Games 2026	Aichi-Nagoya, Japan
Asian Games Mascot	Honohan
Bharat Vana Green	Miyawaki Method
Caralluma umbellata	కుందేటి కొమ్ములు
New plant location	పోరుమామిళ్ల, YSR Kadapa
Young Planet	Elias 224b
Djibouti	72nd Artemis Accords signatory
India–Japan 2027	Year of Shared Horizons
World Bank Chief Economist	Michael Kremer
SARIKA AI	Visakhapatnam Railway Station
MDL Investment	₹15,000 crore
Gurajada Jayanti	September 21 – State Panduga
Twin Flower Fields	Ladakh – 10,600 ft

8. Current Affairs Practice MCQs

Q1. 2024 దాదాసాహెబ్ ఫాల్కే పురస్కారానికి ఎంపికైన ప్రముఖ నటుడు ఎవరు?

A) రజనీకాంత్
B) అనంత్ నాగ్
C) కమల్ హాసన్
D) మిథున్ చక్రవర్తి

Answer: B) అనంత్ నాగ్

Q2. ఆస్కార్స్ 2027 Best International Feature Film విభాగానికి భారత అధికారిక ఎంట్రీ ఏది?

A) కాంతార
B) లాపతా లేడీస్
C) గోందల్
D) ఆల్ వి ఇమాజిన్ యాజ్ లైట్

Answer: C) గోందల్

Q3. 20వ ఆసియా క్రీడల్లో భారత పతాకదారులు ఎవరు?

A) నీరజ్ చోప్రా & పి.వి. సింధు
B) మనూ భాకర్ & తేజేందర్ పాల్ సింగ్ తూర్
C) హర్మన్‌ప్రీత్ సింగ్ & లోవ్లినా బోర్గోహెయిన్
D) సాత్విక్‌సాయిరాజ్ & చిరాగ్ శెట్టి

Answer: B) మనూ భాకర్ & తేజేందర్ పాల్ సింగ్ తూర్

Q4. Shahzad Bhatti Networkను ఏ చట్టం కింద ఉగ్రవాద సంస్థగా ప్రకటించారు?

A) NIA Act, 2008
B) POTA, 2002
C) UAPA, 1967
D) AFSPA, 1958

Answer: C) UAPA, 1967

Q5. ‘కుందేటి కొమ్ములు’ మొక్కను ఏ జిల్లాలో గుర్తించారు?

A) విశాఖపట్నం
B) వైఎస్సార్ కడప
C) తిరుపతి
D) అనకాపల్లి

Answer: B) వైఎస్సార్ కడప

Q6. Bharat Vana Green కార్యక్రమంలో ఉపయోగించే పద్ధతి?

A) Permaculture
B) Miyawaki
C) Social Forestry
D) Agroforestry

Answer: B) Miyawaki Method

Q7. World Bank Group Chief Economistగా బాధ్యతలు స్వీకరించనున్న Nobel Laureate?

A) Abhijit Banerjee
B) Amartya Sen
C) Michael Kremer
D) Esther Duflo

Answer: C) Michael Kremer

Q8. Artemis Accordsలో 72వ దేశంగా చేరిన దేశం?

A) Kenya
B) Nigeria
C) Djibouti
D) Egypt

Answer: C) Djibouti

Q9. SARIKA AI సేవను ఏ రైల్వే స్టేషన్‌లో ప్రారంభించారు?

A) విజయవాడ
B) విశాఖపట్నం
C) సికింద్రాబాద్
D) తిరుపతి

Answer: B) విశాఖపట్నం

Q10. Elias 224b అంచనా వయస్సు?

A) 1 కోటి సంవత్సరాల కంటే తక్కువ
B) 10 లక్షల సంవత్సరాల కంటే తక్కువ
C) 50 లక్షల సంవత్సరాల కంటే తక్కువ
D) 1 లక్ష సంవత్సరాల కంటే తక్కువ

Answer: B) 10 లక్షల సంవత్సరాల కంటే తక్కువ

9. Fill in the Blanks – Quick Revision

Q11. గురజాడ అప్పారావు జయంతి సెప్టెంబర్ 21ను ఆంధ్రప్రదేశ్ ప్రభుత్వం __________గా ప్రకటించింది.

Answer: రాష్ట్ర పండుగ (State Panduga)

Q12. దుగరాజపట్నం Ship Building Clusterలో MDL ప్రతిపాదించిన పెట్టుబడి __________.

Answer: ₹15,000 కోట్లు

Q13. లడఖ్‌లో 10,600 అడుగుల ఎత్తులో ఏర్పాటు చేసినవి ప్రపంచంలోనే అత్యంత ఎత్తైన __________.

Answer: Twin Flower Fields

Q14. India–New Zealand FTA ప్రకారం 15 సంవత్సరాల్లో ప్రతిపాదిత పెట్టుబడి __________.

Answer: $20 billion

Q15. National Green Tribunal ఏర్పాటు చేయబడిన సంవత్సరం __________.

Answer: 2010

10. చివరి నిమిషం పునశ్చరణ

అనంత్ నాగ్ → దాదాసాహెబ్ ఫాల్కే పురస్కారం 2024

దేవికారాణి → తొలి దాదాసాహెబ్ ఫాల్కే పురస్కార గ్రహీత

Gondhal → Oscars 2027 India Entry

Aichi-Nagoya → Asian Games 2026

Honohan → Asian Games Mascot

Miyawaki → Bharat Vana Green

Caralluma umbellata → కుందేటి కొమ్ములు → YSR Kadapa

NGT → 2010 → New Delhi

Elias 224b → Young Planet → < 1 million years

Djibouti → 72nd Artemis Accords country

2027 → India–Japan “Year of Shared Horizons”

UAPA 1967 → Shahzad Bhatti Network

Michael Kremer → World Bank Group Chief Economist

SARIKA AI → Visakhapatnam Railway Station

₹15,000 crore → Duggarajapatnam Ship Building Cluster

September 21 → Gurajada Apparao Jayanti → AP State Panduga

APPSC Statistics – Advanced Sampling & Probability PYQ Problems, Numerical Shortcuts and Exam Traps

APPSC Statistics – Advanced Sampling & Probability PYQ Problems, Numerical Shortcuts and Exam Traps

APPSC Statistics questionsలో Sampling మరియు Probability నుంచి కేవలం definitions మాత్రమే కాకుండా, statement-based questions, sample identification, probability inequalities, with/without replacement, systematic sampling sequences, conditional probability, sampling errors వంటి tricky patterns కూడా ముఖ్యమైనవి.

ఈ Advanced Practice Postలో:

✓ Circular Systematic Sampling
✓ SRSWR & SRSWOR
✓ Finite Population Correction
✓ Standard Error of Sample Mean
✓ Probability Bounds
✓ Conditional Probability Bounds
✓ Sampling Frame & Selection Error
✓ Sampling vs Non-Sampling Error
✓ Two-Way Table Probability
✓ Independence from a Contingency Table
✓ Bayes' Theorem
✓ Repeated Attempts
✓ Sampling Method Identification
✓ APPSC Statement Traps
✓ Advanced Numerical MCQs

Important: Standard Error and Finite Population Correction are treated here as PYQ-oriented extension material. They should not be confused with the basic syllabus wording “Methods of Sampling (Random, Non-Random)”.

PART–1: Circular Systematic Sampling

Ordinary systematic samplingలో population unitsను ఒక ordered listలో arrange చేసి, మొదట random startను select చేసి, తరువాత ప్రతి kth unitను select చేస్తాము.

Sampling Interval k = N/n

Circular systematic samplingలో end of the population list చేరిన తరువాత selection తిరిగి beginning నుంచి కొనసాగుతుంది.

Main Idea:

If the next selected number exceeds N,

New Position = Obtained Position − N

Example 1 – Circular Selection

A population contains 12 units numbered 1 to 12. A sample of 4 units is required by circular systematic sampling. Take sampling interval k = 3 and random start = 11. Find the sample.

Start with 11.

11 → 14 → 17 → 20

But the population contains only 12 units. Therefore wrap around the list.

14 − 12 = 2

17 − 12 = 5

20 − 12 = 8

Selected Sample = 11, 2, 5, 8

⚡ Circular Shortcut:

Keep adding k.
Whenever value > N, subtract N.

Example 2 – Identify the Correct Circular Sample

Population consists of:

A, B, C, D, E, F, G, H, I, J, K, L

Sample size = 4 and interval = 3.
If the random start is K, identify the sample.

Positions:

K → B → E → H

Answer: K, B, E, H

Example 3 – Which Sequence is Impossible?

Suppose the interval in a circular systematic sample is fixed at 3. Which pattern cannot represent the selected units?

A) A, D, G, J
B) C, F, I, L
C) K, B, E, H
D) A, E, H, K

Check the movement between successive units.

A → D → G → J = +3 each time

C → F → I → L = +3 each time

K → B → E → H = +3 circularly

But:

A → E = +4

Answer: D

APPSC Shortcut:

For an “impossible sample” question, do not reconstruct the whole procedure. Check whether the circular gap between consecutive selected units remains constant.

PART–2: Simple Random Sampling With and Without Replacement

Feature	SRSWR	SRSWOR
Meaning	With Replacement	Without Replacement
Selected unit returned?	Yes	No
Population available for next draw	Same	Reduced
Successive draws	Independent under random selection	Generally dependent
Finite Population Correction	Not required	Applicable

Example 4 – Probability With Replacement

A population contains 8 units, of which 3 possess characteristic A. Two units are selected independently with replacement. Find the probability that both possess A.

P(A on first draw) = 3/8

P(A on second draw) = 3/8

P(Both) = 3/8 × 3/8 = 9/64

Answer: 9/64

Example 5 – Same Population Without Replacement

From the same population of 8 units, of which 3 possess A, two units are selected without replacement. Find the probability that both possess A.

P = 3/8 × 2/7

Answer: 3/28

Exam Trap:

With replacement → 3/8 × 3/8

Without replacement → 3/8 × 2/7

PART–3: Standard Error – SRSWR

For a simple random sample of size n selected with replacement from a population having standard deviation σ:

SE(X̄) = σ / √n

Example 6

Population standard deviation σ = 12 and sample size n = 36. Find the standard error of the sample mean under SRSWR.

SE(X̄) = 12/√36

= 12/6 = 2

Answer: 2

Example 7 – Reverse Problem

Population standard deviation is 20. If SE(X̄)=2 under sampling with replacement, find sample size.

2 = 20/√n

√n = 10

n = 100

Answer: 100

⚡ Important Relationship:

SE ∝ 1/√n

If sample size becomes 4 times,
SE becomes 1/2.

If sample size becomes 9 times,
SE becomes 1/3.

PART–4: SRSWOR and Finite Population Correction

When sampling is without replacement from a finite population, the standard error is reduced by the Finite Population Correction (FPC).

FPC = √[(N − n)/(N − 1)]

SE(X̄) = (σ/√n) × √[(N − n)/(N − 1)]

Example 8 – PYQ-Type Standard Error Problem

A population contains N = 101 units. Population standard deviation σ = 12. A sample of n = 20 units is selected.

Find the standard error of the sample mean:
(i) With replacement
(ii) Without replacement

(i) With Replacement

SE = 12/√20

Since √20 = 2√5:

SE = 12/(2√5) = 6/√5

Rationalising:

6/√5 = 6√5/5 = 1.2√5

SRSWR: SE = 1.2√5

(ii) Without Replacement

SE = 1.2√5 × √[(101−20)/(101−1)]

= 1.2√5 × √(81/100)

= 1.2√5 × 9/10

= 1.08√5

SRSWOR: SE = 1.08√5

Very Important APPSC Trap:

SE without replacement is smaller than SE with replacement because:

0 ≤ FPC ≤ 1

Example 9 – Find FPC Directly

N = 100 and n = 20. Find the finite population correction.

FPC = √[(100−20)/(100−1)]

= √(80/99)

Answer: √(80/99) ≈ 0.899

Example 10 – Census as an Extreme Case

What happens to the sampling standard error under SRSWOR when n=N?

FPC = √[(N−N)/(N−1)] = 0

SE = 0

Answer: Sampling error due to sample-to-sample variation disappears when the entire population is enumerated.

PART–5: Probability Bounds – Very Important Objective Pattern

For any two events A and B:

max[0, P(A)+P(B)−1] ≤ P(A∩B) ≤ min[P(A),P(B)]

Example 11 – Minimum Intersection

P(A)=0.8 and P(B)=0.6. Find the minimum possible value of P(A∩B).

Minimum = max[0, 0.8+0.6−1]

= 0.4

Answer: 0.4

Example 12 – Maximum Intersection

P(A)=0.8 and P(B)=0.6. Find the maximum possible value of P(A∩B).

Maximum = min(0.8,0.6)

Answer: 0.6

Example 13 – Range of Intersection

If P(A)=0.7 and P(B)=0.5, determine the possible range of P(A∩B).

Minimum:

0.7 + 0.5 − 1 = 0.2

Maximum:

min(0.7,0.5)=0.5

0.2 ≤ P(A∩B) ≤ 0.5

⚡ Memorise:

Minimum Intersection
= P(A)+P(B)−1, but never below 0.

Maximum Intersection
= Smaller of P(A), P(B).

PART–6: Conditional Probability Bounds

P(A|B) = P(A∩B)/P(B)

If only P(A) and P(B) are known, first determine possible bounds for P(A∩B).

Example 14

P(A)=0.7 and P(B)=0.5. Find the minimum possible value of P(A|B).

First find minimum intersection:

P(A∩B) ≥ 0.7+0.5−1 = 0.2

Therefore:

P(A|B) ≥ 0.2/0.5

Answer: Minimum P(A|B)=0.4

Example 15

P(A)=0.6 and P(B)=0.8. Find the maximum possible value of P(A|B).

Maximum possible intersection:

P(A∩B) ≤ min(0.6,0.8)=0.6

P(A|B) ≤ 0.6/0.8

Answer: Maximum = 0.75

PART–7: Sampling Frame and Selection Error

A Sampling Frame is the list of population members from which the sample is actually selected.

Target Population
All subjects about whom information is required.

↓

Sampling Frame
Operational list used for selecting the sample.

↓

Sample
Units actually selected.

Example 16 – Identify the Error

A researcher wants information about all households in a city but selects households only from a telephone directory. Some households are not listed in the directory. What is the main problem?

Answer: Selection Error due to an incomplete/unrepresentative Sampling Frame.

Example 17

A college wants to estimate the average GPA of all students but a professor uses only students attending her own Statistics class. Identify the likely sampling method.

Answer: Convenience Sampling – a Non-Random Sampling method.

PART–8: Sampling Error vs Non-Sampling Error

Feature	Sampling Error	Non-Sampling Error
Main cause	Chance variation from observing a sample	Human/procedural errors
Occurs in sample survey?	Yes	Yes
Occurs in census?	No sampling error	Yes
Examples	Sample result differs from census result	Selection, nonresponse, response, recording errors

Example 18 – Sampling Error

True population mean = 80.
Sample mean = 83.
Assuming no other errors, find the signed sampling error.

83 − 80 = +3

Answer: +3

Example 19 – Nonresponse Error

A random sample of 1000 households is selected. Only 400 households answer the survey, and the nonrespondents systematically differ from respondents. Which error should be suspected?

Answer: Nonresponse Error.

Example 20 – Response Error

A respondent intentionally reports an incorrect income. What type of error is this?

Answer: Response Error.

Example 21 – Voluntary Response Error

A newspaper publishes an online question and asks readers to voluntarily vote. Which error/bias is particularly likely?

Answer: Voluntary Response Error/Bias.

Most Important Trap:

Increasing sample size may reduce random sampling variability, but it does not automatically remove systematic bias or non-sampling errors.

PART–9: Two-Way Table + Probability

500 employees are classified as follows:

	Retirement Benefit – Yes	Retirement Benefit – No	Total
Men	225	75	300
Women	150	50	200
Total	375	125	500

Example 22 – Marginal Probability

One employee is selected randomly. Find P(Woman).

P(Woman)=200/500

Answer: 0.4

Example 23 – Joint Probability

Find P(Woman ∩ Retirement Benefit).

P = 150/500

Answer: 0.30

Example 24 – Conditional Probability

Given that the selected employee is a woman, find the probability that she has retirement benefits.

P(Benefit | Woman)=150/200

Answer: 0.75

Example 25 – Reverse Conditional Probability

Given that an employee has retirement benefits, find the probability that the employee is a woman.

P(Woman | Benefit)=150/375

Answer: 0.40

APPSC Trap:

P(Benefit | Woman) = 150/200

but

P(Woman | Benefit) = 150/375

Conditional probability changes the denominator.

PART–10: Independence from a Two-Way Table

Example 26

Using the previous employee table:

P(Woman)=200/500=0.4

P(Benefit)=375/500=0.75

P(Woman∩Benefit)=150/500=0.30

Now:

P(Woman)P(Benefit)=0.4×0.75=0.30

Since:

P(Woman∩Benefit)=P(Woman)P(Benefit)

For this table, the two events are independent.

⚡ Independence Test:

Calculate:

P(A∩B)

and

P(A) × P(B)

Equal → Independent
Not Equal → Dependent

PART–11: Mutually Exclusive vs Independent – Major Trap

Mutually Exclusive	Independent
Cannot occur together	Occurrence of one does not change probability of other
P(A∩B)=0	P(A∩B)=P(A)P(B)
For non-zero probabilities, generally dependent	Can occur together

Example 27

A and B are mutually exclusive events with P(A)=0.4 and P(B)=0.3. Are they independent?

Since mutually exclusive:

P(A∩B)=0

But if independent:

P(A)P(B)=0.4×0.3=0.12

Since:

0 ≠ 0.12

Answer: No. They are dependent.

PART–12: Bayes' Theorem – Source Identification Problems

Example 28 – Factory Defect Problem

Factories F₁, F₂ and F₃ produce 30%, 30% and 40% of total production. Their defective rates are 4%, 3% and 2% respectively. A randomly selected product is defective. Find the probability that it came from F₃.

Factory	Production Share	Defect Rate	Joint Contribution
F₁	0.30	0.04	0.012
F₂	0.30	0.03	0.009
F₃	0.40	0.02	0.008

P(D)=0.012+0.009+0.008=0.029

P(F₃|D)=0.008/0.029

Answer: 8/29

Example 29 – Another Bayes Pattern

Machines A and B produce 60% and 40% of a factory's output. Their defect rates are 2% and 5% respectively. A defective product is selected. Find P(B | Defective).

A contribution = 0.60×0.02 = 0.012

B contribution = 0.40×0.05 = 0.020

Total Defective = 0.032

P(B|D)=0.020/0.032

Answer: 5/8 = 0.625

⚡ Bayes Table Shortcut:

Step 1: Source Probability
Step 2: Multiply by conditional outcome probability
Step 3: Add all joint contributions
Step 4: Required contribution ÷ Total contribution

PART–13: Repeated Independent Attempts

Let:

P(Success)=p     and     P(Failure)=q=1−p

Example 30 – No Success

Probability of success in one independent attempt is 0.7. Find the probability of no success in 4 attempts.

q=0.3

P(No Success)=0.3⁴=0.0081

Answer: 0.0081

Example 31 – At Least One Success

P(At Least One)=1−q⁴

=1−0.0081

Answer: 0.9919

Example 32 – First Success on 4th Attempt

Required sequence:

Failure → Failure → Failure → Success

P=q³p

=(0.3)³(0.7)

Answer: 0.0189

Example 33 – Exactly One Success in Four Attempts

p=0.7, q=0.3. Find probability of exactly one success in four independent attempts.

The one success can occupy any one of four positions.

P = 4 × (0.7)(0.3)³

= 4 × 0.0189

Answer: 0.0756

Do Not Confuse:

Success on 4th attempt → wording must be interpreted carefully.

First success on 4th attempt → q³p

At least one success in 4 attempts → 1−q⁴

Exactly one success in 4 attempts → 4pq³

PART–14: Identify the Sampling Method

Example 34

A university gives every student a number and randomly selects 100 numbers.

Simple Random Sampling

Example 35

After a random start, every 50th household is selected.

Systematic Random Sampling

Example 36

Population is divided into rural and urban groups and random samples are taken from both groups.

Stratified Random Sampling

Example 37

A state is divided into geographical clusters. A few clusters are randomly chosen for the survey.

Cluster Sampling

Example 38

A researcher interviews people who are easiest to contact.

Convenience Sampling

Example 39

An expert personally selects units considered most suitable for the study.

Judgment Sampling

Example 40

A population consists of 48% men and 52% women. The researcher intentionally chooses a sample containing exactly the same percentages.

Quota Sampling

PART–15: Stratified vs Quota – Favourite Trap

Feature	Stratified Random Sampling	Quota Sampling
Main category	Random/Probability Sampling	Non-Random Sampling
Population divided into groups?	Yes	Yes
Representation of groups	Sample drawn from each stratum	Quota fixed for groups
Random selection within groups	Yes, in stratified random sampling	Not required

Exam Trap:

“Groups are proportionately represented” alone does not prove that the method is stratified random sampling.

Check whether random selection is actually used.

PART–16: Stratified vs Cluster Sampling

Stratified Sampling	Cluster Sampling
Population → Strata	Population → Clusters
Sample from every stratum	One or a few clusters selected
Strata internally similar for stratification characteristic	Clusters ideally represent population
Example: income groups	Example: geographical areas

PART–17: Advanced APPSC Practice MCQs

Question 41

If P(A)=0.65 and P(B)=0.55, the minimum possible value of P(A∩B) is:

A) 0.10
B) 0.20
C) 0.55
D) 0.65

0.65+0.55−1=0.20

Answer: B – 0.20

Question 42

If P(A)=0.35 and P(B)=0.60, the maximum possible value of P(A∩B) is:

A) 0
B) 0.21
C) 0.35
D) 0.60

Answer: C – 0.35

Question 43

For N=101, n=20 and σ=12, SE(X̄) under SRSWR is:

A) 1.08√5
B) 1.2√5
C) 2.4√5
D) 12√5

Answer: B – 1.2√5

Question 44

For the same data, SE(X̄) under SRSWOR is:

A) 1.08√5
B) 1.2√5
C) 0
D) 6√5

Answer: A – 1.08√5

Question 45

Which error can occur only because a sample rather than the entire population is observed?

A) Response Error
B) Nonresponse Error
C) Sampling Error
D) Recording Error

Answer: C – Sampling Error

Question 46

Which can occur even in a complete census?

A) Sampling Error only
B) Non-Sampling Error
C) Finite Population Correction
D) Sample-to-sample variation

Answer: B – Non-Sampling Error

Question 47

A sample is selected from every income group after dividing a population into low, middle and high income groups. The method is:

A) Cluster Sampling
B) Convenience Sampling
C) Stratified Random Sampling
D) Judgment Sampling

Answer: C – Stratified Random Sampling

Question 48

A sample includes people who voluntarily respond to a newspaper poll. The principal concern is:

A) Sampling interval
B) Voluntary Response Bias
C) Finite Population Correction
D) Stratification

Answer: B – Voluntary Response Bias

Question 49

If A and B are mutually exclusive with P(A)>0 and P(B)>0, then:

A) They must be independent
B) P(A∩B)=P(A)P(B)
C) They are not independent
D) P(A∪B)=0

Answer: C – They are not independent

Question 50

If P(A)=0.4, P(B)=0.5 and P(A∩B)=0.20, A and B are:

A) Mutually exclusive
B) Independent
C) Impossible events
D) Complementary events

P(A)P(B)=0.4×0.5=0.20

Answer: B – Independent

PART–18: Assertion–Reason Practice

Question 51

Assertion (A): Sampling error does not occur in a census.

Reason (R): A census includes every element of the population.

Answer: Both A and R are true, and R explains A.

Question 52

Assertion (A): Non-sampling errors can occur in a census.

Reason (R): Recording and response errors are not eliminated merely by observing every population unit.

Answer: Both A and R are true, and R explains A.

Question 53

Assertion (A): A random sample must always exactly reproduce all population characteristics.

Reason (R): Random selection eliminates every possible sample-to-sample difference.

Answer: Both A and R are false.

Question 54

Assertion (A): Under SRSWOR, the standard error is generally smaller than under SRSWR for the same N, n and σ.

Reason (R): The finite population correction is at most 1.

Answer: Both A and R are true, and R explains A.

PART–19: APPSC 10-Second Shortcuts

Question Clue	Immediate Thought
Every kth unit	Systematic Sampling
List ends and selection continues from beginning	Circular Systematic Sampling
Each same-size sample equally likely	Simple Random Sampling
Sample from every subgroup	Stratified Sampling
One/few geographical groups selected	Cluster Sampling
Easiest people selected	Convenience Sampling
Expert chooses	Judgment Sampling
Exact population proportions imposed without random selection	Quota Sampling
Incomplete list used to select sample	Selection Error / Sampling Frame problem
Selected people do not reply	Nonresponse Error
Incorrect answer supplied	Response Error
People choose themselves to respond	Voluntary Response Error
At least one	1 − None
First success on nth trial	qⁿ⁻¹p
Minimum intersection	max[0,P(A)+P(B)−1]
Maximum intersection	min[P(A),P(B)]
Without replacement SE	Apply FPC

PART–20: One-Minute Formula Sheet

Systematic Sampling Interval
k = N/n

Sampling Fraction
f = n/N

SRSWR Standard Error
SE(X̄)=σ/√n

Finite Population Correction
FPC=√[(N−n)/(N−1)]

SRSWOR Standard Error
SE(X̄)=(σ/√n)√[(N−n)/(N−1)]

Conditional Probability
P(A|B)=P(A∩B)/P(B)

Independent Events
P(A∩B)=P(A)P(B)

Addition Rule
P(A∪B)=P(A)+P(B)−P(A∩B)

Minimum Intersection
max[0,P(A)+P(B)−1]

Maximum Intersection
min[P(A),P(B)]

At Least One Success
1−qⁿ

First Success on nth Attempt
qⁿ⁻¹p

Exactly One Success in n Independent Trials
npqⁿ⁻¹

PART–21: Final APPSC Exam Traps

Trap 1:
Random sample ≠ guaranteed perfectly representative sample.

Trap 2:
Random sampling and non-random sampling are not distinguished merely by sample size.

Trap 3:
Stratified Sampling ≠ Quota Sampling.

Trap 4:
Stratified Sampling ≠ Cluster Sampling.

Trap 5:
Sampling Error can occur in sample surveys, not as sampling error in a complete census.

Trap 6:
Non-Sampling Errors can occur in both sample surveys and censuses.

Trap 7:
Larger sample size does not automatically remove response or selection bias.

Trap 8:
Mutually Exclusive ≠ Independent.

Trap 9:
P(A|B) ≠ P(B|A) in general.

Trap 10:
With Replacement ≠ Without Replacement.

Trap 11:
Without replacement from a finite population → check whether FPC is required.

Trap 12:
“At least one” → complement is usually fastest.

Trap 13:
“First success on nth attempt” requires all previous attempts to fail.

Trap 14:
In probability-bound questions, P(A∩B) can never exceed the smaller marginal probability.

Trap 15:
In circular systematic sampling, crossing the last unit does not end selection; continue from the beginning.

చివరి నిమిషం పునశ్చరణ

Circular Systematic Sampling
→ Fixed interval + wrap around population list

SRSWR
→ With Replacement
→ SE = σ/√n

SRSWOR
→ Without Replacement
→ Apply Finite Population Correction

FPC
→ √[(N−n)/(N−1)]

Minimum P(A∩B)
→ max[0,P(A)+P(B)−1]

Maximum P(A∩B)
→ min[P(A),P(B)]

Sampling Frame Problem
→ Selection Error

No Reply
→ Nonresponse Error

Incorrect Reply
→ Response Error

Self-selected Respondents
→ Voluntary Response Error

Sampling Error
→ Sample Survey only

Non-Sampling Error
→ Sample Survey + Census

Two-Way Table
→ Joint + Marginal + Conditional Probability

Independence Check
→ P(A∩B)=P(A)P(B)

Bayes
→ Observed Outcome → Find Source

At Least One
→ 1−qⁿ

First Success on nth Attempt
→ qⁿ⁻¹p