S1 January 2011 Q2
2. Keith records the amount of rainfall, in mm, at his school, each day for a week. The results are given below.
2.8 5.6 2.3 9.4 0.0 0.5 1.8
Jenny then records the amount of rainfall, \(x\) mm, at the school each day for the following 21 days. The results for the 21 days are summarised below.
\[\sum x = 84.6\]Keith realises that he has transposed two of his figures. The number 9.4 should have been 4.9 and the number 0.5 should have been 5.0
Keith corrects these figures.
| Scheme | Marks |
|---|---|
| 2.8 + 5.6 + 2.3 + 9.4 + 0.5 + 1.8 + 84.6 = 107 | M1 |
| mean = 107 / 28 (= 3.821… ) (awrt 3.8) | A1 |
| (2) |
Notes
M1 for a clear attempt to add the two sums. Accept a full expression or
2.8 + 5.6 + …+ 84.6 = \(x\) where \(100 \lt x \lt 110\)
i.e. seeing at least two correct terms of Keith’s and the 84.6 with a slip.
A1 for awrt 3.8 (Condone 1 dp/2sf here since data is given to 1 dp or 2 sf)
Accept \(\dfrac{107}{28}\) or \(3\dfrac{23}{28}\) or any exact equivalent
Correct answer implies M1A1
| Scheme | Marks |
|---|---|
| It will have no effect since one is 4.5 under what it should be and the other is 4.5 above what it should be. | B1 dB1 |
| (2) | |
| (4 marks) |
Notes
1st B1 for clearly stating that it will have no effect. (“roughly the same” is B0 B0)
2nd dB1 for a supporting reason that mentions the fact that the increase and decrease are the same and gives some numerical value(s) to support this.
e.g. Sum of Keith’s observations is still 22.4 ( or mean is still 3.2)
or Sum is still 107
or \(9.4 - 4.9 = 5 - 0.5\) (o.e.)
This second B1 is dependent on their saying there is no effect so B0B1 is not possible.