17M.2.SL.TZ1.T_5

pestleMathematicsAISLPaper 217M· sl-4-3-mean-median-mode-mean-of-grouped-data-standard-deviation-quartiles-iqrsource ↗

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]

Find the interquartile range.

[2]

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

[2]

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]

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 / solution

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

$6 - 4$ (M1)

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

$= 2$ (A1) (G2)

[2 marks]

$\frac{{11 + 10 + 5}}{{50}}$ (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]

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

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

$\frac{{\frac{{10}}{{50}} \times \frac{9}{{49}}}}{{\frac{{26}}{{50}}}}$

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]

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

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]

f.i.

[N/A]

f.ii.