User interface language: English | Español

Date May 2017 Marks available 3 Reference code 17M.2.sl.TZ1.5
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 5 Adapted from N/A

Question

The table below shows the distribution of test grades for 50 IB students at Greendale School.

M17/5/MATSD/SP2/ENG/TZ1/05

A student is chosen at random from these 50 students.

A second student is chosen at random from these 50 students.

The number of minutes that the 50 students spent preparing for the test was normally distributed with a mean of 105 minutes and a standard deviation of 20 minutes.

Calculate the mean test grade of the students;

[2]
a.i.

Calculate the standard deviation.

[1]
a.ii.

Find the median test grade of the students.

[1]
b.

Find the interquartile range.

[2]
c.

Find the probability that this student scored a grade 5 or higher.

[2]
d.

Given that the first student chosen at random scored a grade 5 or higher, find the probability that both students scored a grade 6.

[3]
e.

Calculate the probability that a student chosen at random spent at least 90 minutes preparing for the test.

[2]
f.i.

Calculate the expected number of students that spent at least 90 minutes preparing for the test.

[2]
f.ii.

Markscheme

\(\frac{{1(1) + 3(2) + 7(3) + 13(4) + 11(5) + 10(6) + 5(7)}}{{50}} = \frac{{230}}{{50}}\)     (M1)

 

Note:     Award (M1) for correct substitution into mean formula.

 

\( = 4.6\)     (A1)     (G2)

[2 marks]

a.i.

\(1.46{\text{ }}(1.45602 \ldots )\)     (G1)

[1 mark]

a.ii.

5     (A1)

[1 mark]

b.

\(6 - 4\)     (M1)

 

Note:     Award (M1) for 6 and 4 seen.

 

\( = 2\)     (A1)     (G2)

[2 marks]

c.

\(\frac{{11 + 10 + 5}}{{50}}\)     (M1)

 

Note:     Award (M1) for \(11 + 10 + 5\) seen.

 

\( = \frac{{26}}{{50}}{\text{ }}\left( {\frac{{13}}{{25}},{\text{ }}0.52,{\text{ }}52\% } \right)\)     (A1)     (G2)

[2 marks]

d.

\(\frac{{10}}{{{\text{their }}26}} \times \frac{9}{{49}}\)     (M1)(M1)

 

Note:     Award (M1) for \(\frac{{10}}{{{\text{their }}26}}\) seen, (M1) for multiplying their first probability by \(\frac{9}{{49}}\).

 

OR

\(\frac{{\frac{{10}}{{50}} \times \frac{9}{{49}}}}{{\frac{{26}}{{50}}}}\)

 

Note:     Award (M1) for \({\frac{{10}}{{50}} \times \frac{9}{{49}}}\) seen, (M1) for dividing their first probability by \(\frac{{{\text{their }}26}}{{50}}\).

 

\( = \frac{{45}}{{637}}{\text{ (}}0.0706,{\text{ }}0.0706436 \ldots ,{\text{ }}7.06436 \ldots \% )\)     (A1)(ft)     (G3)

 

Note:     Follow through from part (d).

 

[3 marks]

e.

\({\text{P}}(X \geqslant 90)\)     (M1)

OR

M17/5/MATSD/SP2/ENG/TZ1/05.f.i/M     (M1)

 

Note:     Award (M1) for a diagram showing the correct shaded region \(( > 0.5)\).

 

\(0.773{\text{ }}(0.773372 \ldots ){\text{ }}0.773{\text{ }}(0.773372 \ldots ,{\text{ }}77.3372 \ldots \% )\)     (A1)     (G2)

[2 marks]

f.i.

\(0.773372 \ldots  \times 50\)     (M1)

\( = 38.7{\text{ }}(38.6686 \ldots )\)     (A1)(ft)     (G2)

 

Note:     Follow through from part (f)(i).

 

[2 marks]

f.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.i.
[N/A]
f.ii.

Syllabus sections

Topic 4 - Statistical applications » 4.1
Show 62 related questions

View options