a. X F FX 3 4 12 4 6 24 5 4 20 6 5 30 7 6 42 8 2 16 Σf =27 Σfx = 144 Median position = \(\frac{∑F + 1}{2}\) (since ∑F is odd so we added 1) \(\frac{27 + 1}{2} = \frac{28}{2}\) = 14th position. counting down the frequency column, the median position occurs at 4. Therefore, the median height is 5m. b i. mean(x̄) = \(\frac{Σfx}{Σf}\) = \(\frac{144}{27}\) ∴ x̄ = 5.3m ( to 1 d.p) ii. x f fx \(fx^2\) 3 4 12 (3 x 3)4 =36 4 6 24 (4 x 4)6 =96 5 4 20 ( 5 x 5)4 =100 6 5 30 (6 x 6)5 =180 7 6 42 (7 x 7)6 =294 8 2 16 (8 x 8)2 =128 Σf =27 Σfx = 144 Σ\(fx^2\) = 834 Standard deviation = \(\sqrt{vaiance} = \sqrt{\frac{Σfx^2}{Σf} - (\frac{Σfx}{Σf})^2}\) S D = \(\sqrt{\frac{834}{27} - (\frac{144}{27})^2}\) S.D = \(\sqrt{30.888 - (5.333)^2}\) = \(\sqrt{30.888 - 28.441}\) = \(\sqrt{2.447}\) = 1.5738 ≈ 1.6 ( to 1 d.p)