OCR S2 — Question 5 9 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeQuadratic relationship μ = kσ²
DifficultyChallenging +1.2 This question requires standardizing a normal distribution with a non-standard relationship between μ and σ (σ² = μ²/4), then recognizing the standardization simplifies to remove μ. Part (i) requires algebraic manipulation and insight that the result is parameter-free. Parts (ii-iii) are more routine (binomial probability and independence concept). The unusual variance parameterization and the insight needed in part (i) elevate this above average difficulty, but it remains accessible with solid S2 knowledge.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

5 In an investment model the increase, \(Y \%\), in the value of an investment in one year is modelled as a continuous random variable with the distribution \(\mathrm { N } \left( \mu , \frac { 1 } { 4 } \mu ^ { 2 } \right)\). The value of \(\mu\) depends on the type of investment chosen.
  1. Find \(\mathrm { P } ( Y < 0 )\), showing that it is independent of the value of \(\mu\).
  2. Given that \(\mu = 6\), find the probability that \(Y < 9\) in each of three randomly chosen years.
  3. Explain why the calculation in part (ii) might not be valid if applied to three consecutive years.

(i) Answer: \(\frac{0 - \mu}{\mu/2} = -2\), independent of \(\mu\)
AnswerMarks Guidance
\(1 - \Phi(2) = 1 - 0.9772 = 0.0228\)Marks: M1 A1 A1 A1 Guidance: Standardise, allow –, allow \(\mu^2/4\). \(z = 2\) or –2. \(z\)-value independent of \(\mu\) and any relevant statement. Answer, a.r.t. 0.023
(ii) Answer: \(\Phi\left(\frac{9-6}{3}\right) = \Phi(1.0) = 0.8413\)
[not 0.1587]
Cube previous answer
AnswerMarks Guidance
\(= 0.59546\)Marks: M1 A1 M1 A1 Guidance: Standardise and use \(\Phi\) [no \(\sqrt{n}\)]. 0.8413 [not 0.1587]. Cube previous answer. Answer, in range [0.595, 0.596]
(iii) Answer: Annual increases not independentMarks: B1 Guidance: Independence mentioned, in context. Allow "one year affects the next" but not "years not random"
**(i)** **Answer:** $\frac{0 - \mu}{\mu/2} = -2$, independent of $\mu$
$1 - \Phi(2) = 1 - 0.9772 = 0.0228$ | **Marks:** M1 A1 A1 A1 | **Guidance:** Standardise, allow –, allow $\mu^2/4$. $z = 2$ or –2. $z$-value independent of $\mu$ and any relevant statement. Answer, a.r.t. 0.023

**(ii)** **Answer:** $\Phi\left(\frac{9-6}{3}\right) = \Phi(1.0) = 0.8413$
[not 0.1587]
Cube previous answer
$= 0.59546$ | **Marks:** M1 A1 M1 A1 | **Guidance:** Standardise and use $\Phi$ [no $\sqrt{n}$]. 0.8413 [not 0.1587]. Cube previous answer. Answer, in range [0.595, 0.596]

**(iii)** **Answer:** Annual increases not independent | **Marks:** B1 | **Guidance:** Independence mentioned, in context. Allow "one year affects the next" but not "years not random"
5 In an investment model the increase, $Y \%$, in the value of an investment in one year is modelled as a continuous random variable with the distribution $\mathrm { N } \left( \mu , \frac { 1 } { 4 } \mu ^ { 2 } \right)$. The value of $\mu$ depends on the type of investment chosen.\\
(i) Find $\mathrm { P } ( Y < 0 )$, showing that it is independent of the value of $\mu$.\\
(ii) Given that $\mu = 6$, find the probability that $Y < 9$ in each of three randomly chosen years.\\
(iii) Explain why the calculation in part (ii) might not be valid if applied to three consecutive years.

\hfill \mbox{\textit{OCR S2  Q5 [9]}}