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.)