The weight (in kg) of 50 contestants at a competition is as follows: 65 66
The weight (in kg) of 50 contestants at a competition is as follows:
65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64.
(a) Construct a frequenct table for the discrete data.
(b) Calculate, correct to 2 decimal places, the;
(i) mean ; (ii) standard deviation of the data.
Explanation
(a)
| Weight (kg) | Frequency |
| 62 | 1 |
| 63 | 5 |
| 64 | 12 |
| 65 | 14 |
| 66 | 10 |
| 67 | 6 |
| 68 | 2 |
(b) Using an assumed mean of 65kg
| \(x\) | \(f\) | \(fx\) | \(d = x - A\) | \(fd\) | \(fd^{2}\) |
| 62 | 1 | 62 | -3 | -3 | 9 |
| 63 | 5 | 315 | -2 | -10 | 20 |
| 64 | 12 | 768 | -1 | -12 | 12 |
| 65 | 14 | 910 | 0 | 0 | 0 |
| 66 | 10 | 660 | 1 | 10 | 10 |
| 67 | 6 | 402 | 2 | 12 | 24 |
| 68 | 2 | 136 | 3 | 6 | 18 |
| \(\sum\) = | 50 | 3253 | 3 | 93 |
(i) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)
= \(\frac{3253}{50}\)
= 65.06 kg
(ii) \(SD = \sqrt{\frac{\sum fd^{2}}{\sum f} - (\frac{\sum fd^{2}}{\sum f})^{2}}\)
\(SD = \sqrt{\frac{93}{50} - (\frac{3}{50})^{2}}\)
= \(\sqrt{\frac{93}{50} - \frac{9}{2500}}\)
= \(\sqrt{\frac{4650 - 9}{2500}}\)
= \(\sqrt{\frac{4641}{2500}}\)
= \(1.3625 \approxeq 1.36\) (2 d.p.)