Correct Answer - Option 2 : 77.5
Concept:
Mode of the given frequency table
Mode = L + \(\left [\frac{f_1~-~f_o}{2f_1~-~f_o~-~f_2} \right ] × h\)
Where,
L = lowest class of highest frequency, f1 = Highest frequency, f2 = second highest frequency, f0 = third highest frequency,
h = Class difference.
Given:
L = 70, f1 = 16, f2 = 14, f3 = 10, h = 70 - 80 = 10.
Calculation:
Mode = L + \(\left [\frac{f_1~-~f_o}{2f_1~-~f_o~-~f_2} \right ] × h\)
= 70 + \(\left [\frac{16~-~10}{2~×16~-~10~-~14} \right ] × 10\)
= 70 + \(\frac{6}{8}\) × 10
= 70 + 7.5
= 77.5