4MA0 Foundation June 2015 Paper 2 Q15
15 Becky has a biased 6-sided spinner.
She spins the spinner 25 times.
She records the score for each spin.
The table shows information about her scores.
| Score | Frequency |
|---|---|
| 1 | 9 |
| 2 | 6 |
| 3 | 3 |
| 4 | 2 |
| 5 | 1 |
| 6 | 4 |
Becky spins the spinner one more time.
| Scheme | Marks |
|---|---|
| 1 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| 6 − 1 | M1 |
| 5 | A1 |
| (2) |
Notes
M1: For 6 and 1 identified
| Scheme | Marks |
|---|---|
| (25 + 1) ÷ 2 or 13 or 12.5 | M1 |
| 2 | A1 |
| (2) |
Notes
M1: or listing scores and clear attempt to find middle value
| Scheme | Marks |
|---|---|
| 1×9 + 2×6 + 3×3 + 4×2 + 5×1 + 6×4 oe (= 67) | M1 |
| “67” ÷ 25 or \(\dfrac{9 + 12 + 9 + 8 + 5 + 24}{25}\) oe (allow one error in a product) | M1 |
| 2.68 or \(2\dfrac{17}{25}\) | A1 |
| (3) |
Notes
M1: sight of at least 4 products and intention to add (can be their 25 if evidence of adding frequencies)
M1: dep for division of sum of products by 25 (can be their 25 if evidence of adding frequencies)
A1: accept 2.7 or 3 if preceded by \(\dfrac{67}{25}\)
(corrected from the printed mark scheme: the first line is printed as 1×9 + 2×6 + 3×3 + 5×1 + 4×6, which leaves out 4×2; the total 67 needs all six products)
| Scheme | Marks |
|---|---|
| M1 | |
| \(\dfrac{2}{25}\) oe | A1 |
| (2) | |
| Total 10 marks |
Notes
M1: \(\dfrac{x}{25}\); \(x \lt 25\) or \(\dfrac{2}{y}\); \(y \gt 2\)
A1: accept 0.08, 8%