\(\frac{31}{15}\)
\(\frac{31}{5}\)
\(\frac{129}{5}\)
\(\frac{43}{5}\)
\(\frac{16}{5}\)
Correct answer is D
F = 1, 2, 3, 4, 5
x = 1, 2, 4, 8, 16
fx = 1, 4, 12, 32, 80, 3f = 15
(average size) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{129}{15}\)
= \(\frac{43}{5}\)
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