19M.1.SL.TZ1.S_9

pestleMathematicsAISLPaper 119M· sl-4-6-combined-mutually-exclusive-conditional-independence-prob-diagramssource ↗

A random variable Z is normally distributed with mean 0 and standard deviation 1. It is known that P( $z$ < −1.6) = $a$ and P( $z$ > 2.4) = $b$ . This is shown in the following diagram.

A second random variable $X$ is normally distributed with mean $m$ and standard deviation $s$ .

It is known that P( $x$ < 1) = $a$ .

Find P(−1.6 < $z$ < 2.4). Write your answer in terms of $a$ and $b$ .

[2]

Given that $z$ > −1.6, find the probability that z < 2.4 . Write your answer in terms of $a$ and $b$ .

[4]

Write down the standardized value for $x = 1$ .

[1]

It is also known that P( $x$ > 2) = $b$ .

Find $s$ .

[6]

Markscheme / solution

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

recognizing area under curve = 1 (M1)

eg $a + x + b = 1$ , $100 - a - b$ , $1 - a + b$

${\text{P}}\left( { - 1.6 < z < 2.4} \right) = 1 - a - b\,\,\left( { = 1 - \left( {a + b} \right)} \right)$ A1 N2

[2 marks]

${\text{P}}\left( {z > - 1.6} \right) = 1 - a$ (seen anywhere) (A1)

recognizing conditional probability (M1)

eg ${\text{P}}\left( {A\left| B \right.} \right)$ , ${\text{P}}\left( {B\left| A \right.} \right)$

correct working (A1)

eg $\frac{{{\text{P}}\left( {z < 2.4 \cap z > - 1.6} \right)}}{{{\text{P}}\left( {z > - 1.6} \right)}}$ , $\frac{{{\text{P}}\left( { - 1.6 < z < 2.4} \right)}}{{{\text{P}}\left( {z > - 1.6} \right)}}$

${\text{P}}\left( {z < 2.4\left| z \right. > - 1.6} \right) = \frac{{1 - a - b}}{{1 - a}}$ A1 N4

Note: Do not award the final A1 if correct answer is seen followed by incorrect simplification.

[4 marks]

$z = - 1.6$ (may be seen in part (d)) A1 N1

Note: Depending on the candidate’s interpretation of the question, they may give $\frac{{1 - m}}{s}$ as the answer to part (c). Such answers should be awarded the first (M1) in part (d), even when part (d) is left blank. If the candidate goes on to show $z = - 1.6$ as part of their working in part (d), the A1 in part (c) may be awarded.

[1 mark]

attempt to standardize $x$ (do not accept $\frac{{x - \mu }}{\sigma }$ ) (M1)

eg $\frac{{1 - m}}{s}$ (may be seen in part (c)), $\frac{{m - 2}}{s}$ , $\frac{{x - m}}{\sigma }$

correct equation with each $z$ -value (A1)(A1)

eg $- 1.6 = \frac{{1 - m}}{s}$ , $2.4 = \frac{{2 - m}}{s}$ , $m + 2.4s = 2$

valid approach (to set up equation in one variable) M1

eg $2.4 = \frac{{2 - \left( {1.6s + 1} \right)}}{s}$ , $\frac{{1 - m}}{{ - 1.6}} = \frac{{2 - m}}{{2.4}}$

correct working (A1)

eg $1.6s + 1 = 2 - 2.4s$ , $4s = 1$ , $m = \frac{7}{5}$

$s = \frac{1}{4}$ A1 N2

[6 marks]

Examiners’ report

[N/A]