
Shortcut Trick
Identify the class with the highest frequency: 20-30 has the maximum frequency of 35.
Use the grouped mode formula: Mode = L + [(f1 − f0) ÷ (2f1 − f0 − f2)] × h
Substitute L = 20, f1 = 35, f0 = 29, f2 = 25, h = 10.
Mode = 20 + [(35 − 29) ÷ (70 − 54)] × 10 = 20 + 3.75 = 23.75 years.
∴ The correct answer is 23.75 years.

Alternate Method
Given:
Frequency distribution of ages of patients.
Maximum frequency is 35, which corresponds to the class interval 20-30.
Formula Used:
Mode of grouped data is given by:
Mode = L + [ (f1 − f0) ÷ (2f1 − f0 − f2) ] × h
Where:
L = Lower limit of the modal class
f1 = Frequency of the modal class
f0 = Frequency of the class preceding the modal class
f2 = Frequency of the class succeeding the modal class
h = Width of the class interval

Calculations:
From the given data, the modal class is 20-30 because it has the highest frequency of 35.
⇒ L = 20
⇒ f1 = 35
⇒ f0 = 29
⇒ f2 = 25
⇒ h = 30 − 20 = 10
Substituting these values into the formula:
⇒ Mode = 20 + [ (35 − 29) ÷ (2 × 35 − 29 − 25) ] × 10
⇒ Mode = 20 + [ 6 ÷ (70 − 54) ] × 10
⇒ Mode = 20 + [ 6 ÷ 16 ] × 10
⇒ Mode = 20 + 0.375 × 10
⇒ Mode = 20 + 3.75
⇒ Mode = 23.75 years
Option Analysis:
⇒ Option (1) 39.44 is incorrect because the mode must lie within the modal class interval of 20-30.
⇒ Option (2) 6.56 is incorrect as it lies completely outside the range of the given data classes.
⇒ Option (4) 22.58 is incorrect due to calculation errors in the frequency differences.
∴ The correct answer is 23.75 years.

Additional Information
Empirical Relationship between Mean, Median, and Mode
For a moderately skewed distribution, the relationship is given by: Mode = 3 × Median − 2 × Mean.
Median of Grouped Data
Median = L + [ (N/2 − CF) ÷ f ] × h, where L is the lower limit of the median class, CF is the cumulative frequency of the preceding class, f is the frequency of the median class, and h is the class width.
Mean of Grouped Data (Direct Method)
Mean = Σ(fi × xi) ÷ Σfi, where xi is the class mark (midpoint) of each class interval and fi is the corresponding frequency.