June 2023 Paper 2 Q11

OCR ACurrent spec9 marksNormal Distribution

11 The random variable \(Y\) has the distribution \(\mathrm{N}(\mu, \sigma^2)\).

(a) Find \(\mathrm{P}(Y \gt \mu - \sigma)\). [1]

(b) Given that \(\mathrm{P}(Y \gt 45) = 0.2\) and \(\mathrm{P}(Y \lt 25) = 0.3\), determine the values of \(\mu\) and \(\sigma\). [6]

The random variables \(U\) and \(V\) have the distributions N(10, 4) and N(12, 9) respectively.

(c) It is given that \(\mathrm{P}(U \lt b) = \mathrm{P}(V \gt c)\), where \(b \gt 10\) and \(c \lt 12\).
Determine \(b\) in terms of \(c\). [2]

Scheme	Marks	AO
0.841	B1	1.2
	[1]

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{45 - \mu}{\sigma} = \Phi^{-1}(0.8)\qquad \dfrac{25 - \mu}{\sigma} = -\Phi^{-1}(0.7)\)	M1	3.1a
\(\dfrac{45 - \mu}{\sigma} = 0.84162\)	A1	1.1
\(\dfrac{25 - \mu}{\sigma} = -0.52440\)	A1	1.1
\(\dfrac{45 - \mu}{25 - \mu} = -\dfrac{0.84162}{0.52440}\quad (= -1.60492)\)	M1dep	2.1
\(\mu \in [32.6, 32.8]\) and \(\sigma \in [14.5, 14.7]\)	A1	1.1
\(\mu = 32.7\) (3 sf) and \(\sigma = 14.6\) (3 sf)	A1	1.1
	[6]

Notes

M1: One of these attempted. Or \(\mathrm{P}\left(Z \lt \frac{45 - \mu}{\sigma}\right) = 0.8\) seen or this standardised form clearly shown on a diagram

A1: Not a required answer so allow truncated e.g. 0.841, 0.84…

A1: Not a required answer so allow truncated e.g. −0.52…

M1dep: dep previous M1. Attempt to solve their equations simultaneously, any method (may be implied by correct answers)

A1: For obtaining both values in the given intervals.

A1: For both \(\mu\) and \(\sigma\) correct to 3sf BC cao
(\(\mu = 32.6778\), \(\sigma = 14.6411\))
SC B1 for either \(\mu = 32.7\) (3sf) or \(\sigma = 14.6\) (3sf) if neither A mark gained, but must be correct to 3sf. (max [5/6])

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{b - 10}{2} = -\dfrac{c - 12}{3}\)	M1	3.1a
\(b = 18 - \dfrac{2}{3}c\)	A1	1.1
	[2]

Notes

M1: oe but signs must be correct (e.g. accept \(\frac{b - 10}{2} = \frac{12 - c}{3}\))

A1: oe but must have \(b\) as the subject