Hint
Use the stated units and precision. Subtract adjacent cumulative frequencies.
Worked solution
(a)
Subtract adjacent cumulative frequencies.
The difference between cumulative counts is \(\displaystyle 15-5=10\).
Answer: \(10\)
(b)
Locate half the total frequency and interpolate within its class.
The half-total position is 10, halfway through the middle class, giving 10.
Answer: \(10\)
(c)
Locate the two quartile count positions and subtract their boundary values.
\(\displaystyle Q_{1}=5\) and \(\displaystyle Q_{3}=15\) at the quarter positions, giving IQR=10.
Answer: \(10\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 2 marks
- 1 mark: Subtract adjacent cumulative frequencies.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
(b) · 2 marks
- 1 mark: Locate half the total frequency and interpolate within its class.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Check the answer against the original conditions.
(c) · 2 marks
- 1 mark: Locate the two quartile count positions and subtract their boundary values.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
Total: 6 marks.