User interface language: English | Español

Date November 2016 Marks available 2 Reference code 16N.2.sl.TZ0.8
Level SL only Paper 2 Time zone TZ0
Command term Write down 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.

\[\sum\limits_{i = 1}^{10} {{x_i} = 252,{\text{ }}\sigma  = 5{\text{ and median}} = 27.} \]

During week 4, the survey was extended to all 200 students in the school. The results are shown in the cumulative frequency graph:

N16/5/MATME/SP2/ENG/TZ0/08.d

Find the mean number of hours spent browsing the Internet.

[2]
a.

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.

[2]
b.

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.

[6]
c.

(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 \(k\).

[6]
d.

Markscheme

attempt to substitute into formula for mean     (M1)

eg\(\,\,\,\,\,\)\(\frac{{\Sigma x}}{{10}},{\text{ }}\frac{{252}}{n},{\text{ }}\frac{{252}}{{10}}\)

mean \( = 25.2{\text{ (hours)}}\)     A1     N2

[2 marks]

a.

(i)     mean \( = 30.2{\text{ (hours)}}\)     A1 N1

(ii)     \(\sigma  = 5{\text{ (hours)}}\)     A1     N1

[2 marks]

b.

(i)     valid approach     (M1)

eg\(\,\,\,\,\,\)95%, 5% of 27

correct working     (A1)

eg\(\,\,\,\,\,\)\(0.95 \times 27,{\text{ }}27 - (5\% {\text{ of }}27)\)

median \( = 25.65{\text{ (exact), }}25.7{\text{ (hours)}}\)     A1     N2

(ii)     METHOD 1

variance \( = {({\text{standard deviation}})^2}\) (seen anywhere)     (A1)

valid attempt to find new standard deviation     (M1)

eg\(\,\,\,\,\,\)\({\sigma _{new}} = 0.95 \times 5,{\text{ }}4.75\)

variance \( = 22.5625{\text{ }}({\text{exact}}),{\text{ }}22.6\)     A1     N2

METHOD 2

variance \( = {({\text{standard deviation}})^2}\) (seen anywhere)     (A1)

valid attempt to find new variance     (M1)

eg\(\,\,\,\,\,\)\({0.95^2}{\text{ }},{\text{ }}0.9025 \times {\sigma ^2}\)

new variance \( = 22.5625{\text{ }}({\text{exact}}),{\text{ }}22.6\)     A1     N2

[6 marks]

c.

(i)     both correct frequencies     (A1)

eg\(\,\,\,\,\,\)80, 150

subtracting their frequencies in either order     (M1)

eg\(\,\,\,\,\,\)\(150 - 80,{\text{ }}80 - 150\)

70 (students)     A1     N2

(ii)     evidence of a valid approach     (M1)

eg\(\,\,\,\,\,\)10% of 200, 90%

correct working     (A1)

eg\(\,\,\,\,\,\)\(0.90 \times 200,{\text{ }}200 - 20\), 180 students

\(k = 35\)     A1     N3

[6 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 5 - Statistics and probability » 5.2 » Statistical measures and their interpretations.

View options