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Given:

Option 1: Arithmetic mean

Option 2: Harmonic mean

Option 3: Geometric mean

Option 4: Median

Formula Used:

Median is the middle value of a dataset when arranged in ascending or descending order.

It is not influenced by extreme values, unlike mean-based measures.

Calculation:

⇒ Arithmetic mean, Harmonic mean, and Geometric mean are all influenced by extreme observations.

⇒ Median remains unaffected by extreme observations, as it solely depends on the middle value(s).

∴ The correct answer is Median.

Given:

To find the average ratio like price/unit, work done/hour, kilometre/hour under certain conditions.

Arithmetic mean

Geometric mean

Harmonic mean

Mode

Correct Option: 1

Formula Used:

Harmonic Mean (HM) = n / (Σ(1/x))

Arithmetic Mean (AM) = Σx / n

Geometric Mean (GM) = (Πx)^1/n

Calculation:

⇒ Average ratio calculations for price/unit, work done/hour, kilometre/hour use arithmetic mean for simplicity and direct averaging.

∴ The correct answer is arithmetic mean.

Given:

Marks more than 10: 100 candidates

Marks more than 20: 75 candidates

Marks more than 30: 60 candidates

Marks more than 40: 40 candidates

Formula Used:

Class Frequency = Number of candidates in a specific range

Average Marks = (Σ(Midpoint × Frequency)) / Total Frequency

Calculation:

Class intervals: 10-20, 20-30, 30-40, 40 and above

Frequencies: 100 - 75 = 25, 75 - 60 = 15, 60 - 40 = 20, 40 (last class)

Midpoints: (10 + 20)/2 = 15, (20 + 30)/2 = 25, (30 + 40)/2 = 35, 40 (open-ended class, assume midpoint = 45)

⇒ Σ(Midpoint × Frequency) = (15 × 25) + (25 × 15) + (35 × 20) + (45 × 40)

⇒ Σ(Midpoint × Frequency) = 375 + 375 + 700 + 1800

⇒ Σ(Midpoint × Frequency) = 3250

⇒ Total Frequency = 100

⇒ Average Marks = Σ(Midpoint × Frequency) / Total Frequency

⇒ Average Marks = 3250 / 100

⇒ Average Marks = 32.5

∴ The correct answer is 32.5.

Given:

Arithmetic mean of 200 observations = 60

Each observation is multiplied by 5

Formula Used:

New Arithmetic Mean = Old Arithmetic Mean × Multiplication Factor

Calculation:

⇒ New Arithmetic Mean = 60 × 5

⇒ New Arithmetic Mean = 300

∴ The correct answer is 300.

Given:

Frequencies = 45, 40, 55

Means = 2, 2.5, 2

Formula Used:

Combined Mean = (Σ(Frequency × Mean)) / Σ(Frequency)

Calculation:

⇒ Σ(Frequency × Mean) = (45 × 2) + (40 × 2.5) + (55 × 2)

⇒ Σ(Frequency × Mean) = 90 + 100 + 110

⇒ Σ(Frequency × Mean) = 300

⇒ Σ(Frequency) = 45 + 40 + 55

⇒ Σ(Frequency) = 140

⇒ Combined Mean = 300 / 140

⇒ Combined Mean = 2.14

∴ The correct answer is 2.14.

Given:

Option 1: Arithmetic mean

Option 2: Median

Option 3: Mode

Option 4: Geometric mean

Option 5: Not provided

Correct option: 2

Formula Used:

Positional average is determined by the position of data values in a sorted dataset, such as Median.

Calculation:

⇒ Median is the middle value of a sorted dataset, making it a positional average.

∴ The correct answer is Median.

Given:

Two frequency distributions are given.

Median of the first group = 20.8 m

Mean of the second group = 17.3 m

Unknown frequencies:

First group: u, v

Second group: 3u, 2v

Use median condition for the first group

Class intervals (width h = 5)

Median value 20.8 lies in the class 20–25

Lower limit L = 20

Frequency of median class f = 75

Cumulative frequency before median class:

cf = u + v + 11 + 52 = u + v + 63

Total frequency N = u + v + 11 + 52 + 75 + 22

⇒ N = u + v + 160

Median formula:

Median = L + [ (N/2 − cf) ÷ f ] × h

Substitute values:

⇒ 20.8 = 20 + [ ((u + v + 160)/2 − (u + v + 63)) ÷ 75 ] × 5

⇒ 0.8 = [ (u + v + 160 − 2u − 2v − 126) ÷ 150 ] × 5

⇒ 0.8 = [ (34 − (u + v)) ÷ 150 ] × 5

⇒ 0.8 = (34 − u − v) ÷ 30

⇒ 34 − u − v = 24

⇒ u + v = 10 ……… (1)

Use mean condition for the second group

Midpoints x: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5

Second group frequencies:

3u, 2v, 40, 50, 30, 28

Σf = 3u + 2v + 148

Σfx = (3u × 2.5) + (2v × 7.5) + (40 × 12.5) + (50 × 17.5) + (30 × 22.5) + (28 × 27.5)

⇒ Σfx = 7.5u + 15v + 500 + 875 + 675 + 770

⇒ Σfx = 7.5u + 15v + 2820

Mean formula:

Mean = Σfx ÷ Σf

Substitute mean = 17.3

⇒ (7.5u + 15v + 2820) ÷ (3u + 2v + 148) = 17.3

⇒ 7.5u + 15v + 2820 = 17.3(3u + 2v + 148)

⇒ 7.5u + 15v + 2820 = 51.9u + 34.6v + 2560.4

⇒ 44.4u + 19.6v = 259.6

⇒ Divide by 2

⇒ 22.2u + 9.8v = 129.8 ……… (2)

From (1): v = 10 − u

Substitute in (2):

⇒ 22.2u + 9.8(10 − u) = 129.8

⇒ 22.2u + 98 − 9.8u = 129.8

⇒ 12.4u = 31.8

⇒ u ≈ 2.56 ≈ 3 (nearest integer)

∴ The value of u is 3.

Given:

Two frequency distributions are given.

Median of the first group = 20.8 m

Mean of the second group = 17.3 m

Unknown frequencies:

First group: u, v

Second group: 3u, 2v

Use median condition for the first group

Class intervals (width h = 5)

Median value 20.8 lies in the class 20–25

Lower limit L = 20

Frequency of median class f = 75

Cumulative frequency before median class:

cf = u + v + 11 + 52 = u + v + 63

Total frequency N = u + v + 11 + 52 + 75 + 22

⇒ N = u + v + 160

Median formula:

Median = L + [ (N/2 − cf) ÷ f ] × h

Substitute values:

⇒ 20.8 = 20 + [ ((u + v + 160)/2 − (u + v + 63)) ÷ 75 ] × 5

⇒ 0.8 = [ (u + v + 160 − 2u − 2v − 126) ÷ 150 ] × 5

⇒ 0.8 = [ (34 − (u + v)) ÷ 150 ] × 5

⇒ 0.8 = (34 − u − v) ÷ 30

⇒ 34 − u − v = 24

⇒ u + v = 10 ……… (1)

Use mean condition for the second group

Midpoints x: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5

Second group frequencies:

3u, 2v, 40, 50, 30, 28

Σf = 3u + 2v + 148

Σfx = (3u × 2.5) + (2v × 7.5) + (40 × 12.5) + (50 × 17.5) + (30 × 22.5) + (28 × 27.5)

⇒ Σfx = 7.5u + 15v + 500 + 875 + 675 + 770

⇒ Σfx = 7.5u + 15v + 2820

Mean formula:

Mean = Σfx ÷ Σf

Substitute mean = 17.3

⇒ (7.5u + 15v + 2820) ÷ (3u + 2v + 148) = 17.3

⇒ 7.5u + 15v + 2820 = 17.3(3u + 2v + 148)

⇒ 7.5u + 15v + 2820 = 51.9u + 34.6v + 2560.4

⇒ 44.4u + 19.6v = 259.6

⇒ Divide by 2

⇒ 22.2u + 9.8v = 129.8 ……… (2)

From (1): v = 10 − u

Substitute in (2):

⇒ 22.2u + 9.8(10 − u) = 129.8

⇒ 22.2u + 98 − 9.8u = 129.8

⇒ 12.4u = 31.8

⇒ u ≈ 2.56 ≈ 3 (nearest integer)

u + v = 10

⇒ v = 10 − u

⇒ v = 10 − 3

⇒ v = 7

∴ The value of v is 7.

Given:

Median = 24

Formula used:

Median = L + [( (N/2 − cf) / f ) × h]

Calculation:

Total number of students, N = 55 + k

⇒ N/2 = (55 + k)/2

Median value 24 lies in the class 20–30

Lower limit (L) = 20

Class width (h) = 10

Cumulative frequency before median class (cf) = 30

Frequency of median class (f) = (30 + k) − 30 = k

Substitute in median formula:

⇒ 24 = 20 + [ ( (55 + k)/2 − 30 ) ÷ k ] × 10

⇒ 4 = [ (55 + k − 60) ÷ (2k) ] × 10

⇒ 4 = ( (k − 5) ÷ (2k) ) × 10

⇒ 4 = (5(k − 5)) ÷ k

⇒ 4k = 5k − 25

⇒ k = 25

∴ The value of k is 25.

Calculation:

Median = L + [( (N/2 − cf) / f ) × h]

Calculation:

Total number of students, N = 55 + k

⇒ N/2 = (55 + k)/2

Median value 24 lies in the class 20–30

Lower limit (L) = 20

Class width (h) = 10

Cumulative frequency before median class (cf) = 30

Frequency of median class (f) = (30 + k) − 30 = k

Substitute in median formula:

⇒ 24 = 20 + [ ( (55 + k)/2 − 30 ) ÷ k ] × 10

⇒ 4 = [ (55 + k − 60) ÷ (2k) ] × 10

⇒ 4 = ( (k − 5) ÷ (2k) ) × 10

⇒ 4 = (5(k − 5)) ÷ k

⇒ 4k = 5k − 25

⇒ k = 25

Convert to class frequencies

Σf = 80

Σ(f × x) = 25 + 375 + 625 + 630 + 315 = 1970

Mean = Σ(f × x) ÷ Σf

⇒ Mean = 1970 ÷ 80

⇒ Mean = 24.625

∴ The mean of the distribution is 24.625.