| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Standard +0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: z-score calculations, inverse normal for percentiles, sum of normals for part (iii), and symmetry properties for part (b). All parts are routine applications 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 transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.761\) or \(0.762\) (3 sf) | B1 [1] | BC Allow 0.76 |
| Answer | Marks | Guidance |
|---|---|---|
| \(62.0\) (3 sf) | B1 [1] | BC Allow 62 or 61.9; Allow \(m \geq 62.0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(\bar{X}\) e.g. "\(\bar{X}\)" or "mean" or \(\frac{18}{10}\) or \(\sqrt{\frac{18}{10}}\) | M1 | \(\mu = 550\) seen or implied |
| \(\bar{X} \sim N(55, \frac{18}{10})\) | M1 | \(\Sigma X \sim N(550, 180)\) Correct; May be implied |
| \(P(\bar{X} < \frac{530}{10})\) dep \(\sigma^2 = \frac{18}{10}\) | M1 | \(P(\Sigma X < 530)\) dep \(\sigma^2 = 180\); Stated or implied |
| \(= 0.0680\) (3 sf) | A1 [4] | Allow 0.068; Correct answer from limited (or no) working: M1M1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Y < 72) = 0.75\); \(\Phi^{-1}(0.75)\) or \(0.674\) | M1 | oe; May be implied e.g. on diagram; NB \(P(62 < Y < 72) = 0.5\) no marks yet |
| \(P(Y < 62) = 0.25\); \(\Phi^{-1}(0.25)\) or \(-0.674\) | M1 | \(\pm 0.674\) implies M1M1; Allow 0.67 |
| \(\frac{72-67}{\sigma}\) and \(\frac{62-67}{\sigma}\) | M1 | M1M1M1 may be implied by A1 |
| \(\frac{72-67}{\sigma} = 0.674\) and \(\frac{62-67}{\sigma} = -0.674\) | A1 | oe e.g. \(5 = 0.674\sigma\); A1 for correct equation, allow 0.67; SC correct answer with no working |
| \(\sigma = 7.41\) or \(7.42\) (3 sf) | A1 [5] | or SC B2 if correct to 2 sf |
| Trial and Improvement: \(\Phi^{-1}(0.75)\) or \(0.674\) or \(\Phi^{-1}(0.25)\) or \(-0.674\) | M2 | May be implied |
| e.g. \(\sigma = 8\): \(67 - 8 \times 0.674 = 61.6\); \(\sigma = 7\): \(67 - 7 \times 0.674 = 62.3\) | M1, A1 | At least one correct trial; Trials leading to values either side of 62 |
| \(\sigma = 7.41\): \(67 - 7.41 \times 0.674 = 62.0 \Rightarrow \sigma = 7.41\); or \(\sigma = 7.42\): \(67 - 7.42 \times 0.674 = 62.0 \Rightarrow \sigma = 7.42\) | A1 | Correct trial using \(\sigma = 7.41\) or \(7.42\) and conclusion \(\sigma = 7.41\) or \(7.42\) |
# Question 9(a)(i):
| $0.761$ or $0.762$ (3 sf) | B1 [1] | BC Allow 0.76 |
# Question 9(a)(ii):
| $62.0$ (3 sf) | B1 [1] | BC Allow 62 or 61.9; Allow $m \geq 62.0$ |
# Question 9(a)(iii):
| Use of $\bar{X}$ e.g. "$\bar{X}$" or "mean" or $\frac{18}{10}$ or $\sqrt{\frac{18}{10}}$ | M1 | $\mu = 550$ seen or implied |
| $\bar{X} \sim N(55, \frac{18}{10})$ | M1 | $\Sigma X \sim N(550, 180)$ Correct; May be implied |
| $P(\bar{X} < \frac{530}{10})$ dep $\sigma^2 = \frac{18}{10}$ | M1 | $P(\Sigma X < 530)$ dep $\sigma^2 = 180$; Stated or implied |
| $= 0.0680$ (3 sf) | A1 [4] | **Allow 0.068**; Correct answer from limited (or no) working: M1M1M1A1 |
# Question 9(b):
| $P(Y < 72) = 0.75$; $\Phi^{-1}(0.75)$ or $0.674$ | M1 | oe; May be implied e.g. on diagram; NB $P(62 < Y < 72) = 0.5$ no marks yet |
| $P(Y < 62) = 0.25$; $\Phi^{-1}(0.25)$ or $-0.674$ | M1 | $\pm 0.674$ implies M1M1; **Allow 0.67** |
| $\frac{72-67}{\sigma}$ and $\frac{62-67}{\sigma}$ | M1 | M1M1M1 may be implied by A1 |
| $\frac{72-67}{\sigma} = 0.674$ and $\frac{62-67}{\sigma} = -0.674$ | A1 | oe e.g. $5 = 0.674\sigma$; A1 for correct equation, allow 0.67; **SC correct answer with no working** |
| $\sigma = 7.41$ or $7.42$ (3 sf) | A1 [5] | or SC B2 if correct to 2 sf |
| **Trial and Improvement:** $\Phi^{-1}(0.75)$ or $0.674$ or $\Phi^{-1}(0.25)$ or $-0.674$ | M2 | May be implied |
| e.g. $\sigma = 8$: $67 - 8 \times 0.674 = 61.6$; $\sigma = 7$: $67 - 7 \times 0.674 = 62.3$ | M1, A1 | At least one correct trial; Trials leading to values either side of 62 |
| $\sigma = 7.41$: $67 - 7.41 \times 0.674 = 62.0 \Rightarrow \sigma = 7.41$; or $\sigma = 7.42$: $67 - 7.42 \times 0.674 = 62.0 \Rightarrow \sigma = 7.42$ | A1 | Correct trial using $\sigma = 7.41$ or $7.42$ and conclusion $\sigma = 7.41$ or $7.42$ |
---9
\begin{enumerate}[label=(\alph*)]
\item The masses, in grams, of plums of a certain kind have the distribution $\mathrm { N } ( 55,18 )$.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that a plum chosen at random has a mass between 50.0 and 60.0 grams.
\item The heaviest $5 \%$ of plums are classified as extra large.
Find the minimum mass of extra large plums.
\item The plums are packed in bags, each containing 10 randomly selected plums.
Find the probability that a bag chosen at random has a total mass of less than 530 g .
\end{enumerate}\item The masses, in grams, of apples of a certain kind have the distribution $\mathrm { N } \left( 67 , \sigma ^ { 2 } \right)$. It is given that half of the apples have masses between 62 g and 72 g .
Determine $\sigma$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2019 Q9 [11]}}