IBDP Maths AA: Topic: SL 4.2: Presentation of data: IB style Questions SL Paper 2

Question

A random sample of nine adults were selected to see whether sleeping well affected their reaction times to a visual stimulus. Each adult’s reaction time was measured twice. The first measurement for reaction time was taken on a morning after the adult had slept well. The second measurement was taken on a morning after the same adult had not slept well.
The box and whisker diagrams for the reaction times, measured in seconds, are shown below.

Consider the box and whisker diagram representing the reaction times after sleeping well.
(a) State the median reaction time after sleeping well.
(b) Verify that the measurement of 0.46 seconds is not an outlier.
(c) State why it appears that the mean reaction time is greater than the median reaction time.
Now consider the two box and whisker diagrams.
(d) Comment on whether these box and whisker diagrams provide any evidence that might suggest that not sleeping well causes an increase in reaction time.

Answer/Explanation

Ans:

(b) IQR = 0.35 – 0.27 (=0.08)(s)
substituting their IQR into correct expression for upper fence
0.35 + 1.5 × 0.08 (=0.47) (s)
0.46 < 0.47
so 0.46 (s) is not an outlier
(c) EITHER
the median is closer to the lower quartile (positively skewed)
OR
The distribution is positively skewed
OR
the range of reaction times below the median is smaller than the range of reaction times above the median
Note: These are sample answers from a range of acceptable correct answers. Award R1 for any correct statement that explains this. Do not award R1 if there is also an incorrect statement, even if another statement in the answer is correct. Accept a correctly and clearly labelled diagram.

(d) EITHER
the distribution for ‘not sleeping well’ is centred at a higher reaction time
OR
The median reaction time after not sleeping well is equal to the upper quartile reaction time after sleeping well
OR
75% of reaction times are <0.35 seconds after sleeping well, compared with 50% after not sleeping well
OR
the sample size of 9 is too small to draw any conclusions
Note: These are sample answers from a range of acceptable correct answers. Accept any relevant correct statement that relates to the median and/or quartiles shown in the box plots. Do not accept a comparison of means. Do not award R1 if there is also an incorrect statement, even if another statement in the answer is correct. Award R0 to “correlation does not imply causation”.

Question

The following diagram is a box and whisker plot for a set of data.


The interquartile range is 20 and the range is 40.

Write down the median value.[1]

a.

Find the value of

(i)     \(a\) ;

(ii)    \(b\) .[4]

b.
Answer/Explanation

Markscheme

18     A1     N1

[1 mark]

a.

(i) 10     A2      N2

(ii) 44     A2      N2

[4 marks]

b.

Question

On one day 180 flights arrived at a particular airport. The distance travelled and the arrival status for each incoming flight was recorded. The flight was then classified as on time, slightly delayed, or heavily delayed.

The results are shown in the following table.

A χ2 test is carried out at the 10 % significance level to determine whether the arrival status of incoming flights is independent of the distance travelled.

The critical value for this test is 7.779.

A flight is chosen at random from the 180 recorded flights.

State the alternative hypothesis.[1]

a.

Calculate the expected frequency of flights travelling at most 500 km and arriving slightly delayed.[2]

b.

Write down the number of degrees of freedom.[1]

c.

Write down the χ2 statistic.[2]

d.i.

Write down the associated p-value.[1]

d.ii.

State, with a reason, whether you would reject the null hypothesis.[2]

e.

Write down the probability that this flight arrived on time.[2]

f.

Given that this flight was not heavily delayed, find the probability that it travelled between 500 km and 5000 km.[2]

g.

Two flights are chosen at random from those which were slightly delayed.

Find the probability that each of these flights travelled at least 5000 km.[3]

h.
Answer/Explanation

Markscheme

The arrival status is dependent on the distance travelled by the incoming flight     (A1)

Note: Accept “associated” or “not independent”.[1 mark]

a.

\(\frac{{60 \times 45}}{{180}}\)  OR  \(\frac{{60}}{{180}} \times \frac{{45}}{{180}} \times 180\)     (M1)

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

= 15     (A1) (G2)[2 marks]

b.

4     (A1)

Note: Award (A0) if “2 + 2 = 4” is seen.[1 mark]

c.

9.55 (9.54671…)    (G2)

Note: Award (G1) for an answer of 9.54.[2 marks]

d.i.

0.0488 (0.0487961…)     (G1)[1 mark]

d.ii.

Reject the Null Hypothesis     (A1)(ft)

Note: Follow through from their hypothesis in part (a).

9.55 (9.54671…) > 7.779     (R1)(ft)

OR

0.0488 (0.0487961…) < 0.1     (R1)(ft)

Note: Do not award (A1)(ft)(R0)(ft). Follow through from part (d). Award (R1)(ft) for a correct comparison, (A1)(ft) for a consistent conclusion with the answers to parts (a) and (d). Award (R1)(ft) for χ2calc > χ2crit , provided the calculated value is explicitly seen in part (d)(i).[2 marks]

e.

\(\frac{{52}}{{180}}\,\,\left( {0.289,\,\,\frac{{13}}{{45}},\,\,28.9\,{\text{% }}} \right)\)     (A1)(A1) (G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.[2 marks]

f.

\(\frac{{35}}{{97}}\,\,\left( {0.361,\,\,36.1\,{\text{% }}} \right)\)     (A1)(A1) (G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.[2 marks]

g.

\(\frac{{14}}{{45}} \times \frac{{13}}{{44}}\)     (A1)(M1)

Note: Award (A1) for two correct fractions and (M1) for multiplying their two fractions.

\( = \frac{{182}}{{1980}}\,\,\left( {0.0919,\,\,\frac{{91}}{{990}},\,0.091919 \ldots ,\,9.19\,{\text{% }}} \right)\)     (A1) (G2)[3 marks]

h.
Scroll to Top