| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Rounded or discrete from continuous |
| Difficulty | Standard +0.3 This is a standard S2 normal distribution question covering routine techniques: basic probability calculations, inverse normal for percentiles, continuity correction for rounding, and solving simultaneous equations with z-scores. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 50) = P\left(Z < \frac{50-45.3}{11.5}\right) = P(Z < 0.4087) = \Phi(0.4087) = 0.6585\) | M1 for standardising, M1 for correct structure, A1 CAO inc use of diff tables | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(45.3 < X < 50) = 0.6585 - 0.5 = 0.1585\) | M1, A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Phi^{-1}(0.9) = 1.282\); \(\frac{k-45.3}{11.5} = 1.282\); \(k = 45.3 + 1.282 \times 11.5 = 60.0\) | B1 for 1.282 seen, M1 for equation in \(k\), A1 CAO | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{score}=111) = P(110.5 < Y < 111.5) = P\left(\frac{110.5-100}{15} < Z < \frac{111.5-100}{15}\right) = P(0.7 < Z < 0.7667) = \Phi(0.7667) - \Phi(0.7) = 0.7784 - 0.7580 = 0.0204\) | B1 for both continuity corrections, M1 for standardising, M1 for correct structure, A1 CAO | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Phi^{-1}(0.3) = -0.5244\), \(\Phi^{-1}(0.8) = 0.8416\); \(22 = \mu + 0.8416\sigma\); \(15 = \mu - 0.5244\sigma\); \(7 = 1.3660\sigma\); \(\sigma = 5.124\), \(\mu = 17.69\) | B1 for 0.5244 or 0.8416, M1 for at least one equation, A1 for both correct, M1 for attempt to solve, A1 CAO for both | [5] |
# Question 3:
## Part (i)(A)
| $P(X < 50) = P\left(Z < \frac{50-45.3}{11.5}\right) = P(Z < 0.4087) = \Phi(0.4087) = 0.6585$ | M1 for standardising, M1 for correct structure, A1 CAO inc use of diff tables | [3] |
## Part (i)(B)
| $P(45.3 < X < 50) = 0.6585 - 0.5 = 0.1585$ | M1, A1 | [2] |
## Part (ii)
| $\Phi^{-1}(0.9) = 1.282$; $\frac{k-45.3}{11.5} = 1.282$; $k = 45.3 + 1.282 \times 11.5 = 60.0$ | B1 for 1.282 seen, M1 for equation in $k$, A1 CAO | [3] |
## Part (iii)
| $P(\text{score}=111) = P(110.5 < Y < 111.5) = P\left(\frac{110.5-100}{15} < Z < \frac{111.5-100}{15}\right) = P(0.7 < Z < 0.7667) = \Phi(0.7667) - \Phi(0.7) = 0.7784 - 0.7580 = 0.0204$ | B1 for both continuity corrections, M1 for standardising, M1 for correct structure, A1 CAO | [4] |
## Part (iv)
| $\Phi^{-1}(0.3) = -0.5244$, $\Phi^{-1}(0.8) = 0.8416$; $22 = \mu + 0.8416\sigma$; $15 = \mu - 0.5244\sigma$; $7 = 1.3660\sigma$; $\sigma = 5.124$, $\mu = 17.69$ | B1 for 0.5244 or 0.8416, M1 for at least one equation, A1 for both correct, M1 for attempt to solve, A1 CAO for both | [5] |
---3 In an English language test for 12-year-old children, the raw scores, $X$, are Normally distributed with mean 45.3 and standard deviation 11.5.
\begin{enumerate}[label=(\roman*)]
\item Find\\
(A) $\mathrm { P } ( X < 50 )$,\\
(B) $\mathrm { P } ( 45.3 < X < 50 )$.
\item Find the least raw score which would be obtained by the highest scoring $10 \%$ of children.
\item The raw score is then scaled so that the scaled score is Normally distributed with mean 100 and standard deviation 15. This scaled score is then rounded to the nearest integer. Find the probability that a randomly selected child gets a rounded score of exactly 111 .
\item In a Mathematics test for 12-year-old children, the raw scores, $Y$, are Normally distributed with mean $\mu$ and standard deviation $\sigma$. Given that $\mathrm { P } ( Y < 15 ) = 0.3$ and $\mathrm { P } ( Y < 22 ) = 0.8$, find the values of $\mu$ and $\sigma$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2010 Q3 [17]}}