IB Mathematics SL 4.2 Understanding of box and whisker diagrams AA SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
For the original 28 scores, the median is the mean of the 14th and 15th scores when the results are arranged in ascending order.
Aiden’s score is below every original score, while Brett’s score is above every original score. Adding these two scores places one new value at each end of the ordered data.
For the 30 scores, the middle values are the 15th and 16th scores. These are the same two original scores that were previously in the 14th and 15th positions.
Therefore, their average remains 10.5.
✅ Answer: The new scores are added on opposite sides of the median, so the two values used to calculate the median remain unchanged.
(b)
The sum of the original 28 scores is:
\(\text{Original sum}=28\times10.5=294\)
The sum of all 30 scores is:
\(\text{New sum}=30\times10.6=318\)
Therefore, the sum of Aiden’s and Brett’s scores is:
\(A+B=318-294\)
\(A+B=24\)
From the box-and-whisker diagram, the original minimum score is 6 and the original maximum score is 17.
Since Aiden scores less than 6, his score becomes the new minimum. Since Brett scores more than 17, his score becomes the new maximum.
The new range is 14, so:
\(B-A=14\)
We now have the simultaneous equations:
\(A+B=24\)
\(B-A=14\)
Add the equations.
\(2B=38\)
\(B=19\)
Substitute \(B=19\) into \(A+B=24\).
\(A+19=24\)
\(A=5\)
These values satisfy the conditions \(A<6\), \(B>17\), and \(19-5=14\).
✅ Answer: Aiden scores \(5\) and Brett scores \(19\).