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 ...
Question 1 Report
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.
Answer Details
(a) Frequency table
Tallying the 50 weights gives:
| Weight, \(x\) (kg) | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 5 | 12 | 14 | 10 | 6 | 2 |
Total frequency \(=1+5+12+14+10+6+2=50\).
(b)(i) Mean
| \(x\) | \(f\) | \(fx\) | \(fx^{2}\) |
|---|---|---|---|
| 62 | 1 | 62 | 3844 |
| 63 | 5 | 315 | 19845 |
| 64 | 12 | 768 | 49152 |
| 65 | 14 | 910 | 59150 |
| 66 | 10 | 660 | 43560 |
| 67 | 6 | 402 | 26934 |
| 68 | 2 | 136 | 9248 |
| Total | 50 | 3253 | 211733 |
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3253}{50}=65.06\text{ kg}\]
(b)(ii) Standard deviation
\[\text{S.D.}=\sqrt{\frac{\sum fx^{2}}{\sum f}-\left(\frac{\sum fx}{\sum f}\right)^{2}}=\sqrt{\frac{211733}{50}-(65.06)^{2}}\]
\[=\sqrt{4234.66-4232.8036}=\sqrt{1.8564}=1.36\text{ kg (2 d.p.)}\]
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