Method of central tendency

Reference link : https://jmp.sh/s/oZN1QLomLJ9YzxbFxuzG

To-Do question:

q3,4,20,21

1. Arithmetic Mean (A.M.)

🔹 For Ungrouped Data:

Formula:

$$\text{A.M.} = \frac{x_1 + x_2 + \cdots + x_n}{n}$$

Example:

Marks: 10, 15, 20, 25, 30

$$\text{A.M.} = \frac{10 + 15 + 20 + 25 + 30}{5} = \frac{100}{5} = 20$$

🟢 Use it when you have raw individual values.

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For Grouped Data:

Formula:

$$\text{A.M.} = \frac{\sum fx}{\sum f}$$

Where $x$ is the midpoint of each class.

Example from Q3:

$$\text{A.M.} = \frac{2375000}{910} \approx 2615.38$$

🟢 Use it when data is in class intervals with frequencies.

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2. Median

🔹 For Ungrouped Data:

• Sort values

• If $n$ is odd, median = middle value

• If $n$ is even, median = average of two middle values

Example:

Data: 3, 8, 5, 7, 9 → Sorted: 3, 5, 7, 8, 9 Median = 7 (middle value)

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For Grouped Data:

Formula:

$$\text{Median} = L + \left( \frac{N/2 - CF}{f} \right) \cdot h$$

Example from Q3:

• Total N = 910 → N/2 = 455

• Median class: 2000–3000 → L = 2000, f = 225, CF = 320, h = 1000

$$\text{Median} = 2000 + \left( \frac{455 - 320}{225} \right) \cdot 1000 = 2611.11$$

🟢 Use it when you're finding a central value in grouped data.

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✅ 3. Mode

🔹 For Ungrouped Data:

• Most frequent value

Example from Q23:

Data: 15, 12, 5, 13, 12, 15, 8, 8, 9, 9, 10, 15 → Mode = 15 (appears 3 times)

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🔹 For Grouped Data:

Formula:

$$\text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \cdot h$$

Example from Q20:

$$\text{Mode} = 2000 + \left( \frac{75 - 50}{2 \cdot 75 - 50 - 68} \right) \cdot 500 = 2000 + \left( \frac{25}{32} \right) \cdot 500 = 2390.63$$

🟢 Use it when you want the most common value in grouped data.

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✅ 4. Harmonic Mean (H.M.)

Formula:

$$\text{H.M.} = \frac{n}{\sum \frac{1}{x_i}}$$

Example from Q9:

Data: 6, 12, 24

$$\text{H.M.} = \frac{3}{\frac{1}{6} + \frac{1}{12} + \frac{1}{24}} = \frac{3}{\frac{7}{24}} = \frac{72}{7} \approx 10.29$$

🟢 Use it when data involves rates or ratios (like speed).

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✅ 5. Geometric Mean (G.M.)

Formula:

$$\text{G.M.} = \sqrt[n]{x_1 \cdot x_2 \cdot \cdots \cdot x_n}$$

Example from Q10:

Data: 3, 12, 48

$$\text{G.M.} = \sqrt[3]{3 \cdot 12 \cdot 48} = \sqrt[3]{1728} = 12$$

🟢 Use it when comparing relative growth rates or percentages.

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✅ 6. Relation: Mean, Median, Mode

Formula:

$$\text{Mode} = 3 \cdot \text{Median} - 2 \cdot \text{Mean}$$

Example:

Mean = 25, Median = 30

$$\text{Mode} = 3 \cdot 30 - 2 \cdot 25 = 90 - 50 = 40$$

🟢 Use it when one measure is missing or data is skewed.

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✅ 7. Combined Mean

Formula:

$$\text{Combined Mean} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$

Example from Q12:

• Group 1: 30 students, mean = 20

• Group 2: 30 students, mean = 25

$$\text{Mean} = \frac{30 \cdot 20 + 30 \cdot 25}{60} = \frac{1350}{60} = 22.5$$

🟢 Use it when combining two datasets.

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✅ 8. Corrected Mean

Formula:

$$\text{Correct Mean} = \text{Wrong Mean} + \frac{\text{Correct} - \text{Wrong}}{n}$$

Example from Q13:

• Wrong Mean = 40

• One wrong value = 50 (should be 40)

• $n = 100$

$$\text{Correct Mean} = 40 + \frac{40 - 50}{100} = 40 - 0.1 = 39.9$$

🟢 Use it when you’ve fixed a mistake in one value.

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All MCQ Answers + Tricks

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Q1. Relation between mean, median, mode

🟩 Answer: (a) Mean – Mode = 3(Mean – Median)
🧠 Trick: Standard empirical formula
→ Mode = 3Median – 2Mean ⇒ Rearranged: Mean – Mode = 3(Mean – Median)

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Q2. A.M. of 1, 2, 3, ..., n

🟩 Answer: (b) (n+1)/2
🧠 Trick: A.M. of first ( n ) natural numbers = (n+1)/2

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Q6. Median of: 12, 5, 7, 10, 4, 9, 8, 11, 3, 6

🟩 Answer: (a) 7
🧠 Trick: Sort and take average of 5th & 6th term (even ( n = 10 ))

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Q7. A.M. of 16, 24, 8, 54, 26, 54

🟩 Answer: (c) 21
🧠 Trick: Add all and divide by 6
→ ( \frac{182}{6} = 30.33 ) ❌ wait! Recheck: (sum = 182, mean ≈ 30.33)
🟥 There's likely a mismatch in answer key; the correct mean is 30.33, not 21.
PDF might have wrong option.

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Q8. Mode of: 13, 6, 8, 11, 5, 10, 15, 3, 2

🟩 Answer: (d) None of these
🧠 Trick: All values occur once ⇒ No mode

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Q9. H.M. of 6, 12, 24

🟩 Answer: (a) 3/(1/6+1/12+1/24+1/72) = 72/7
🧠 Trick: Use H.M. = n / (1/n1+1/n2+1/n3)

n= number of element

n1,n2,n3= elements

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Q10. G.M. of 3, 12, 48

🟩 Answer: (a) 12
🧠 Trick: Multiply: see formula of GM

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Q11. Highest point of frequency curve

🟩 Answer: (c) Both (a) and (b)
🧠 Trick: In normal distribution, mean = median = mode = peak point

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Q16. G.M. of 3, 12, 48

🟩 Answer: (a) 12
🧠 Trick: Repetition of Q10

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Q17. Median of: 12, 5, 7, 10, 4, 9, 15, 14, 2

🟩 Answer: (a) 7
🧠 Trick: Sort and take middle (odd ( n = 9 )) → 7 is 5th

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Q18. Which of the following is false?

🟩 Answer: (b) A.M × H.M = (G.M)^4 / G.M
🧠 Trick: Valid relation: A.M × H.M = (G.M)^2
→ So this one is false

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Q22. Relation between A.M., G.M., H.M.

🟩 Answer: (a) A.M ≥ G.M ≥ H.M
🧠 Trick: Always: AM ≥ GM ≥ HM

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Q23. Mode of: 15, 12, 5, 13, 12, 15, 8, 8, 9, 9, 10, 15

🟩 Answer: (a) 15
🧠 Trick: Count frequency → 15 occurs most (3 times)

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Q27. G.M. of 3, 12, 48

🟩 Answer: (a) 12
🧠 Trick: Again same Q repeated

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Q28. A.M. of 1 to m

🟩 Answer: (b)
🧠 Trick: Same as Q2, formula-based

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Q31. Relation between mean, median, mode

🟩 Answer: (a) Mean – Mode = 3(Mean – Median)
🧠 Trick: Standard identity

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Q36. Highest point of frequency curve

🟩 Answer: (c) Mode
🧠 Trick: Peak of frequency curve = Mode

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Q37. A.M. = 25, G.M. = 15 ⇒ H.M. = ?

🟩 Answer: (d) 10
🧠 Trick: Use: AM × HM = (GM)^2
🟥 Correction: Actually HM = 9, not 10
→ Looks like the PDF key is wrong again.

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Q38. A.M. of 7, x–2, x+3 is 9. Find x

🟩 Answer: (b) 10
🧠 **Trick:(**7+x-2+x+3)/3=9 solve for x

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Q43. G.M. of 3, 12, 48

🟩 Answer: (a) 12
🧠 Trick: Again a repetition

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Q44. H.M. of 6, 12, 14

🟩 Answer: (a) ( \frac{72}{7} )
🧠 Trick: Use H.M. = ( \frac{3}{\frac{1}{6} + \frac{1}{12} + \frac{1}{14}} )

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Q45. If 2x = 7y, G.M. of y = 1 ⇒ G.M. of x = ?

🟩 Answer: (c) 7/2
🧠 Trick: If 2x=7×1

\=> 2x=7 =>x=7/2 (ans)

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Q46. If 3x + 5y = 15 and median of x = 2 ⇒ median of y?

🟩 Answer: (a) 1.8

explain: x=2, 3×2+5y=15 → 5y=15-6 →5y=9 → y=1.8

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Q47. A.M. of 1, 2, ..., n

🟩 Answer: (b) (n+1)/2 (rem)
🧠 Trick: Same as Q2/Q28 — memorize this

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