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Date	November 2019	Marks available	2	Reference code	19N.1.SL.TZ0.S_7
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Find	Question number	S_7	Adapted from	N/A

Question

Let $X$ $X$ and $Y$ $Y$ be normally distributed with $X \sim N (14, a^{2})$ $X \sim {\text{N}}\left( {14{\text{, }}{a^2}} \right)$ and $Y \sim N (22, a^{2})$ $Y \sim {\text{N}}\left( {22{\text{, }}{a^2}} \right)$ , $a > 0$ $a > 0$ .

Find $b$ $b$ so that $P (X > b) = P (Y < b)$ ${\text{P}}\left( {X > b} \right) = {\text{P}}\left( {Y < b} \right)$ .

[2]

It is given that $P (X > 20) = 0.112$ ${\text{P}}\left( {X > 20} \right) = 0.112$ .

Find $P (16 < Y < 28)$ ${\text{P}}\left( {16 < Y < 28} \right)$ .

[4]

Markscheme

METHOD 1

recognizing that $b$ $b$ is midway between the means of $14$ $14$ and $22$ $22$ . (M1)

eg , $b = \frac{14 + 22}{2}$ $b = \frac{{14 + 22}}{2}$

$b = 18$ $b = 18$ A1 N2

METHOD 2

valid attempt to compare distributions (M1)

eg $\frac{b - 14}{a} = \frac{b - 22}{a}, b - 14 = 22 - b$ $\frac{{b - 14}}{a} = \frac{{b - 22}}{a}{\text{, }}b - 14 = 22 - b$

$b = 18$ $b = 18$ A1 N2

[2 marks]

valid attempt to compare distributions (seen anywhere) (M1)

eg $Y$ $Y$ is a horizontal translation of $X$ $X$ of $8$ $8$ units to the right,

$P (16 < Y < 28) = P (8 < X < 20), P (Y > 22 + 6) = P (X > 14 + 6)$ ${\text{P}}\left( {16 < Y < 28} \right) = {\text{P}}\left( {8 < X < 20} \right){\text{, P}}\left( {Y > 22 + 6} \right) = {\text{P}}\left( {X > 14 + 6} \right)$

valid approach using symmetry (M1)

eg $1 - 2 P (X > 20), 1 - 2 P (Y < 16), 2 \times P (14 < x < 20), P (X < 8) = P (X > 20)$ $1 - 2{\text{P}}\left( {X > 20} \right){\text{, }}1 - 2{\text{P}}\left( {Y < 16} \right){\text{, }}2 \times {\text{P}}\left( {14 < X < 20} \right){\text{, P}}\left( {X < 8} \right) = {\text{P}}\left( {X > 20} \right)$

correct working (A1)

eg $1 - 2 (0.112), 2 \times (0.5 - 0.112), 2 \times 0.388, 0.888 - 0.112$ $1 - 2\left( {0.112} \right){\text{, }}2 \times \left( {0.5 - 0.112} \right){\text{, }}2 \times 0.388{\text{, }}0.888 - 0.112$

$P (16 < Y < 28) = 0.776$ ${\text{P}}\left( {16 < Y < 28} \right) = 0.776$ A1 N3

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.9—Normal distribution and calculations

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Topic 4—Statistics and probability

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