Hint
Classes are disjoint and their boundaries are explicit. Divide the middle frequency by its class width.
Worked solution
(a)
\(\displaystyle \frac{f}{w}= \frac{19}{8}= \frac{19}{8}\)
Answer: \(\displaystyle \frac{19}{8}\).
Answer: \(\displaystyle \frac{19}{8}\)
(b)
\(\displaystyle \bar{t}\approx \sum \frac{fm}{\sum}f= \frac{220}{19}\)
Answer: \(\displaystyle \frac{220}{19}\).
Answer: \(\displaystyle \frac{220}{19}\)
(c)
\(\displaystyle \sum \frac{fm^{2}}{\sum}f- \bar{t}^{2}= \frac{14984}{361}\)
Answer: \(\displaystyle \frac{14984}{361}\).
Answer: \(\displaystyle \frac{14984}{361}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Divide the middle frequency by its class width.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Use frequency density rather than frequency for unequal-width bars.
(b) · 2 marks
- 1 mark: Weight each class midpoint by its frequency.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not average the midpoints without frequency weights.
(c) · 2 marks
- 1 mark: Use the weighted midpoint second moment.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Use frequency density rather than frequency for unequal-width bars.
Total: 6 marks.