Foundation November 2023 Paper 3 Q22
22 Seija works at a weather station.
The table gives information about the temperature, \(T\) °C, at midday for each of 50 cities in the UK on Tuesday.
| Temperature (\(T\) °C) | Frequency |
|---|---|
| \(10 \lt T \leqslant 15\) | 2 |
| \(15 \lt T \leqslant 20\) | 8 |
| \(20 \lt T \leqslant 25\) | 13 |
| \(25 \lt T \leqslant 30\) | 21 |
| \(30 \lt T \leqslant 35\) | 6 |
(a) Calculate an estimate for the mean temperature. (3)
Seija says,
“The median temperature is 22.5 °C because 22.5 is the middle number in the middle group.”
(b) Is Seija correct?
Give a reason for your answer. (1)
Give a reason for your answer. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 24.6 | M1 | for finding 5 products within intervals (including end points) with not more than one error, may be seen near table. eg \(2 \times 12.5\ (= 25)\), \(8 \times 17.5\ (= 140)\), \(13 \times 22.5\ (= 292.5)\), \(21 \times 27.5\ (= 577.5)\), \(6 \times 32.5\ (= 195)\) or for 1230 |
| M1 | for \(\Sigma fx \div \Sigma f\) eg \((\text{``}25\text{''} + \text{``}140\text{''} + \text{``}292.5\text{''} + \text{``}577.5\text{''} + \text{``}195\text{''}) \div \text{``}50\text{''}\) or \(\text{``}1230\text{''} \div \text{``}50\text{''}\) | |
| A1 | for 24.6 oe |
Additional guidance
| Min \(fx\) | Max \(fx\) |
|---|---|
| 20 | 30 |
| 120 | 160 |
| 260 | 325 |
| 525 | 630 |
| 180 | 210 |
\(\Sigma fx\) must come from 5 products,
\(fx\) within intervals (including end points)
| Answer | Mark | Mark scheme |
|---|---|---|
| No, with reason | C1 | for No and reason Acceptable No, the median is in the interval \(25 \lt T \leqslant 30\) No, the median is in the group containing the 25(.5)th temperature No, she did not take into account frequency No, the frequencies are not the same for each group. Not acceptable No, the median is 27.5 No, the median is higher than 22.5 \(25 \lt T \leqslant 30\) Yes, … |
Additional guidance
Any incorrect statement as part of a correct response can be ignored unless it contradicts the statement.