Allied & ICAR Exam Bank • ALLIED_2284 NAT • 2 Marks

Section 8: General Aptitude

Quantitative Aptitude

Subtopic: Data Interpretation

Question Statement:

The frequency table of grain sizes has intervals: 1-3 mm (freq 10), 3-5 mm (freq 25), 5-7 mm (freq 15). The estimated mean grain size in mm using class midpoints is:

Answer Key:

4.2

Step-by-Step Solution & Explanation

Midpoints: 2 mm (10), 4 mm (25), 6 mm (15). $\sum fx = 2(10) + 4(25) + 6(15) = 20 + 100 + 90 = 210$. Total $N = 50$. Mean $= 210 / 50 = 4.2$ mm.
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