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Date	November 2018	Marks available	6	Reference code	18N.2.SL.TZ0.S_9
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	S_9	Adapted from	N/A

Question

A nationwide study on reaction time is conducted on participants in two age groups. The participants in Group X are less than 40 years old. Their reaction times are normally distributed with mean 0.489 seconds and standard deviation 0.07 seconds.

The participants in Group Y are 40 years or older. Their reaction times are normally distributed with mean 0.592 seconds and standard deviation σ seconds.

In the study, 38 % of the participants are in Group X.

A person is selected at random from Group X. Find the probability that their reaction time is greater than 0.65 seconds.

[2]

The probability that the reaction time of a person in Group Y is greater than 0.65 seconds is 0.396. Find the value of σ.

[4]

A randomly selected participant has a reaction time greater than 0.65 seconds. Find the probability that the participant is in Group X.

[6]

Ten of the participants with reaction times greater than 0.65 are selected at random. Find the probability that at least two of them are in Group X.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

0.010724

0.0107 A2 N2

[2 marks]

correct z-value (A1)

0.263714…

evidence of appropriate approach (M1)

eg $\frac{{0.65 - 0.592}}{\sigma }$ , $0.264 = \frac{{x - u}}{\sigma }$

correct substitution (A1)

eg $0.263714 = \frac{{0.65 - 0.592}}{\sigma }$ , $\sigma = \frac{{0.65 - 0.592}}{{0.264}}$

0.219934

σ = 0.220 A1 N3

[4 marks]

correct work for P(group X and $t$ > 0.65) or P(group Y and $t$ > 0.65) (may be seen anywhere) (A1)

eg ${\text{P}}\left( {{\text{group X}}} \right) \times {\text{P}}\left( {t > 0.65\left| {\text{X}} \right.} \right)$ , ${\text{P}}\left( {{\text{X}} \cap t > 0.65} \right) = 0.0107 \times 0.38\left( { = 0.004075} \right)$ ,

${\text{P}}\left( {{\text{Y}} \cap t > 0.65} \right) = 0.396 \times 0.62$

recognizing conditional probability (seen anywhere) (M1)

eg ${\text{P}}\left( {\left. {\text{X}} \right|t > 0.65} \right)$ , ${\text{P}}\left( {\left. A \right|B} \right) = \frac{{{\text{P}}\left( {A \cap B} \right)}}{{{\text{P}}\left( B \right)}}$

valid approach to find ${\text{P}}\left( {t > 0.65} \right)$ (M1)

eg , ${\text{P}}\left( {{\text{X and }}t > 0.65} \right) + {\text{P}}\left( {{\text{Y and }}t > 0.65} \right)$

correct work for ${\text{P}}\left( {t > 0.65} \right)$ (A1)

eg 0.0107 × 0.38 + 0.396 × 0.62, 0.249595

correct substitution into conditional probability formula A1

eg $\frac{{0.0107 \times 0.38}}{{0.0107 \times 0.38 + 0.396 \times 0.62}}$ , $\frac{{0.004075}}{{0.249595}}$

0.016327

${\text{P}}\left( {\left. {\text{X}} \right|t > 0.65} \right) = 0.0163270$ A1 N3

[6 marks]

recognizing binomial probability (M1)

eg $X \sim B\left( {n,\,\,p} \right)$ , $\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{q^{n - r}}$ , (0.016327)²(0.983672)⁸, $\left( {\begin{array}{*{20}{c}} {10} \\ 2 \end{array}} \right)$

valid approach (M1)

eg ${\text{P}}\left( {X \geqslant 2} \right) = 1 - {\text{P}}\left( {X \leqslant 1} \right)$ , $1 - {\text{P}}\left( {X < a} \right)$ , summing terms from 2 to 10 (accept binomcdf(10, 0.0163, 2, 10))