a. The table shows the height of teak trees harvested by a farmer: Find the median height.
| Height(m) | 3 | 4 | 5 | 6 | 7 | 8 |
| number of trees | 4 | 6 | 4 | 5 | 6 | 2 |
b. calculate and correct to one decimal place the: i. mean; ii. standard deviation.
Explanation
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)