19N.1.SL.TZ0.S_7

pestleMathematicsAASLPaper 119N· sl-4-9-normal-distribution-and-calculationssource ↗

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

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

[2]

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

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

[4]

Markscheme / solution

METHOD 1

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

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

$b = 18$ A1 N2

METHOD 2

valid attempt to compare distributions (M1)

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

$b = 18$ A1 N2

[2 marks]

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

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

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

valid approach using symmetry (M1)

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

correct working (A1)

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

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

[4 marks]

Examiners’ report

[N/A]