Date | November 2016 | Marks available | 2 | Reference code | 16N.2.sl.TZ0.8 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Ten students were surveyed about the number of hours, x, they spent browsing the Internet during week 1 of the school year. The results of the survey are given below.
10∑i=1xi=252, σ=5 and median=27.
During week 4, the survey was extended to all 200 students in the school. The results are shown in the cumulative frequency graph:
Find the mean number of hours spent browsing the Internet.
During week 2, the students worked on a major project and they each spent an additional five hours browsing the Internet. For week 2, write down
(i) the mean;
(ii) the standard deviation.
During week 3 each student spent 5% less time browsing the Internet than during week 1. For week 3, find
(i) the median;
(ii) the variance.
(i) Find the number of students who spent between 25 and 30 hours browsing the Internet.
(ii) Given that 10% of the students spent more than k hours browsing the Internet, find the maximum value of k.
Markscheme
attempt to substitute into formula for mean (M1)
egΣx10, 252n, 25210
mean =25.2 (hours) A1 N2
[2 marks]
(i) mean =30.2 (hours) A1 N1
(ii) σ=5 (hours) A1 N1
[2 marks]
(i) valid approach (M1)
eg95%, 5% of 27
correct working (A1)
eg0.95×27, 27−(5% of 27)
median =25.65 (exact), 25.7 (hours) A1 N2
(ii) METHOD 1
variance =(standard deviation)2 (seen anywhere) (A1)
valid attempt to find new standard deviation (M1)
egσnew=0.95×5, 4.75
variance =22.5625 (exact), 22.6 A1 N2
METHOD 2
variance =(standard deviation)2 (seen anywhere) (A1)
valid attempt to find new variance (M1)
eg0.952 , 0.9025×σ2
new variance =22.5625 (exact), 22.6 A1 N2
[6 marks]
(i) both correct frequencies (A1)
eg80, 150
subtracting their frequencies in either order (M1)
eg150−80, 80−150
70 (students) A1 N2
(ii) evidence of a valid approach (M1)
eg10% of 200, 90%
correct working (A1)
eg0.90×200, 200−20, 180 students
k=35 A1 N3
[6 marks]