| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.3 This is a standard S2 normal distribution question requiring routine application of z-tables and inverse normal calculations. Part (i) involves straightforward probability lookups, part (ii) requires solving simultaneous equations from two z-score relationships (a common textbook exercise), and part (iii) asks for a symmetric confidence interval. All techniques are core S2 content 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 transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 120) = P\!\left(Z < \frac{120-115.3}{21.9}\right)\) | M1 | For standardizing |
| \(= P(Z < 0.2146)\) | A1 | For \(z = 0.2146\) |
| \(= \Phi(0.2146) = 0.5849\) | A1 CAO | Min 3 sf, to include use of difference column |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\!\left(\frac{100-115.3}{21.9} < Z < \frac{110-115.3}{21.9}\right)\) | M1 | For standardizing both 100 & 110 |
| \(= P(-0.6986 < Z < -0.2420)\) | M1 | For correct structure in calculation |
| \(= \Phi(0.6986) - \Phi(0.2420) = 0.7577 - 0.5956 = 0.1621\) | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| From tables \(\Phi^{-1}(0.1) = -1.282\) | B1 | For \(\pm 1.282\) seen |
| \(\frac{k - 115.3}{21.9} = -1.282\) | M1 | For equation in \(k\) and negative z-value |
| \(k = 115.3 - 1.282 \times 21.9 = 87.22\) | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\Phi^{-1}(0.70) = 0.5244,\quad \Phi^{-1}(0.15) = -1.036\) | B1 | For 0.5244 or \(\pm 1.036\) seen |
| \(180 = \mu + 0.5244\,\sigma\) | M1 | For at least one equation in \(\mu\) and \(\sigma\) and \(\Phi^{-1}\) value |
| \(140 = \mu - 1.036\,\sigma\) | ||
| \(40 = 1.5604\,\sigma\) | M1 dep | For attempt to solve two equations |
| \(\sigma = 25.63,\quad \mu = 166.55\) | A1 CAO | For both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\Phi^{-1}(0.975) = 1.96\) | B1 | For \(\pm 1.96\) seen |
| \(a = 166.55 - 1.96 \times 25.63 = 116.3\) | M1 | For either equation |
| \(b = 166.55 + 1.96 \times 25.63 = 216.8\) | A1, A1 | Allow other correct intervals |
# Question 3:
## Part (i)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 120) = P\!\left(Z < \frac{120-115.3}{21.9}\right)$ | M1 | For standardizing |
| $= P(Z < 0.2146)$ | A1 | For $z = 0.2146$ |
| $= \Phi(0.2146) = 0.5849$ | A1 CAO | Min 3 sf, to include use of difference column |
## Part (i)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\!\left(\frac{100-115.3}{21.9} < Z < \frac{110-115.3}{21.9}\right)$ | M1 | For standardizing both 100 & 110 |
| $= P(-0.6986 < Z < -0.2420)$ | M1 | For correct structure in calculation |
| $= \Phi(0.6986) - \Phi(0.2420) = 0.7577 - 0.5956 = 0.1621$ | A1 CAO | |
## Part (i)(C)
| Answer | Marks | Guidance |
|--------|-------|----------|
| From tables $\Phi^{-1}(0.1) = -1.282$ | B1 | For $\pm 1.282$ seen |
| $\frac{k - 115.3}{21.9} = -1.282$ | M1 | For equation in $k$ and negative z-value |
| $k = 115.3 - 1.282 \times 21.9 = 87.22$ | A1 CAO | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Phi^{-1}(0.70) = 0.5244,\quad \Phi^{-1}(0.15) = -1.036$ | B1 | For 0.5244 or $\pm 1.036$ seen |
| $180 = \mu + 0.5244\,\sigma$ | M1 | For at least one equation in $\mu$ and $\sigma$ and $\Phi^{-1}$ value |
| $140 = \mu - 1.036\,\sigma$ | | |
| $40 = 1.5604\,\sigma$ | M1 dep | For attempt to solve two equations |
| $\sigma = 25.63,\quad \mu = 166.55$ | A1 CAO | For both |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Phi^{-1}(0.975) = 1.96$ | B1 | For $\pm 1.96$ seen |
| $a = 166.55 - 1.96 \times 25.63 = 116.3$ | M1 | For either equation |
| $b = 166.55 + 1.96 \times 25.63 = 216.8$ | A1, A1 | Allow other correct intervals |
---3 The number of minutes, $X$, for which a particular model of laptop computer will run on battery power is Normally distributed with mean 115.3 and standard deviation 21.9.
\begin{enumerate}[label=(\roman*)]
\item (A) Find $\mathrm { P } ( X < 120 )$.\\
(B) Find $\mathrm { P } ( 100 < X < 110 )$.\\
(C) Find the value of $k$ for which $\mathrm { P } ( X > k ) = 0.9$.
The number of minutes, $Y$, for which a different model of laptop computer will run on battery power is known to be Normally distributed with mean $\mu$ and standard deviation $\sigma$.
\item Given that $\mathrm { P } ( Y < 180 ) = 0.7$ and $\mathrm { P } ( Y < 140 ) = 0.15$, find the values of $\mu$ and $\sigma$.
\item Find values of $a$ and $b$ for which $\mathrm { P } ( a < Y < b ) = 0.95$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2009 Q3 [17]}}