18
15
1.0
5
Correct answer is C
Mean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6
Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\)
\(\frac{31}{2}\) = 15.5
\(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065
1.0(1 d.p)
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