Important Formulas (Statistics)

• 1. Mean (x̄):
  • Direct Method: x̄ = Σ(f_i x_i) / Σf_i
  • Assumed Mean Method: x̄ = a + [Σ(f_i d_i) / Σf_i], where d_i = x_i - a
  • Step-Deviation Method: x̄ = a + [Σ(f_i u_i) / Σf_i] × h, where u_i = (x_i - a) / h
• 2. Mode:
  Mode = l + [ (f₁ - f₀) / (2f₁ - f₀ - f₂) ] × h
  (where l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of preceding class, f₂ = frequency of succeeding class, h = class size)
• 3. Median:
  Median = l + [ (N/2 - cf) / f ] × h
  (where l = lower limit of median class, N = total frequency, cf = cumulative frequency of preceding class, f = frequency of median class, h = class size)
• 4. Empirical Relationship:
  3 × Median = Mode + 2 × Mean

Question 1 (Direct Method for Mean)

A survey was conducted by a group of students to find the number of plants in 20 houses. Find the mean number of plants per house.

Solution:

We first calculate the class mark x_i = (Upper limit + Lower limit) / 2 for each interval:

• For 0-2: x_i = 1, f_i x_i = 1 × 1 = 1
• For 2-4: x_i = 3, f_i x_i = 2 × 3 = 6
• For 4-6: x_i = 5, f_i x_i = 1 × 5 = 5
• For 6-8: x_i = 7, f_i x_i = 5 × 7 = 35
• For 8-10: x_i = 9, f_i x_i = 6 × 9 = 54
• For 10-12: x_i = 11, f_i x_i = 2 × 11 = 22
• For 12-14: x_i = 13, f_i x_i = 3 × 13 = 39

Σf_i = 1 + 2 + 1 + 5 + 6 + 2 + 3 = 20
Σ(f_i x_i) = 1 + 6 + 5 + 35 + 54 + 22 + 39 = 162

Mean (x̄) = Σ(f_i x_i) / Σf_i = 162 / 20 = 8.1 plants.

Question 2 (Assumed Mean Method)

Find the mean daily wages of 50 workers of a factory using an appropriate method.

Solution:

Let Assumed Mean (a) = 550 (Class mark of 540-560).

• For 500-520: x_i = 510, d_i = 510 - 550 = -40 ⇒ f_id_i = 12 × (-40) = -480
• For 520-540: x_i = 530, d_i = 530 - 550 = -20 ⇒ f_id_i = 14 × (-20) = -280
• For 540-560: x_i = 550, d_i = 550 - 550 = 0 ⇒ f_id_i = 8 × 0 = 0
• For 560-580: x_i = 570, d_i = 570 - 550 = 20 ⇒ f_id_i = 6 × 20 = 120
• For 580-600: x_i = 590, d_i = 590 - 550 = 40 ⇒ f_id_i = 10 × 40 = 400

Σf_i = 50
Σ(f_i d_i) = -480 - 280 + 0 + 120 + 400 = -240

Mean (x̄) = a + [Σ(f_i d_i) / Σf_i] = 550 + (-240 / 50) = 550 - 4.8 = ₹ 545.20.

Question 3 (Finding Missing Frequency)

The mean pocket allowance is ₹ 18. Find the missing frequency 'f'.

Solution:

Class marks (x_i): 12, 14, 16, 18, 20, 22, 24.

• Σf_i = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f
• Σ(f_i x_i) = (7×12) + (6×14) + (9×16) + (13×18) + (f×20) + (5×22) + (4×24)
• Σ(f_i x_i) = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f

Given Mean = 18:
Mean = Σ(f_i x_i) / Σf_i
18 = (752 + 20f) / (44 + f)
18 × (44 + f) = 752 + 20f
792 + 18f = 752 + 20f
792 - 752 = 20f - 18f
40 = 2f
f = 20.

Question 4 (Calculating Mode)

Find the mode of the ages of patients admitted to a hospital.

Solution:

The maximum frequency is 23, which belongs to the class interval 35-45. Thus, the modal class is 35-45.

• Lower limit of modal class (l) = 35
• Frequency of modal class (f₁) = 23
• Frequency of preceding class (f₀) = 21
• Frequency of succeeding class (f₂) = 14
• Class size (h) = 10

Mode = l + [ (f₁ - f₀) / (2f₁ - f₀ - f₂) ] × h
Mode = 35 + [ (23 - 21) / (2(23) - 21 - 14) ] × 10
Mode = 35 + [ 2 / (46 - 35) ] × 10
Mode = 35 + (20 / 11)
Mode = 35 + 1.82 = 36.82 years.

Question 5 (Calculating Median)

Calculate the median for the following monthly electricity consumption data of 68 consumers.

Solution:

Construct Cumulative Frequency (cf) Table:

• 65-85: f = 4, cf = 4
• 85-105: f = 5, cf = 9
• 105-125: f = 13, cf = 22
• 125-145: f = 20, cf = 42
• 145-165: f = 14, cf = 56
• 165-185: f = 8, cf = 64
• 185-205: f = 4, cf = 68

Total Frequency (N) = 68 ⇒ N/2 = 34.
The cumulative frequency just greater than 34 is 42, which belongs to the class interval 125-145. Hence, the median class is 125-145.

• Lower limit (l) = 125
• Cumulative frequency of preceding class (cf) = 22
• Frequency of median class (f) = 20
• Class width (h) = 20

Median = l + [ (N/2 - cf) / f ] × h
Median = 125 + [ (34 - 22) / 20 ] × 20
Median = 125 + 12 = 137 units.