| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Percentage/proportion exceeding threshold |
| Difficulty | Standard +0.3 This is a multi-part normal distribution question requiring standard techniques: inverse normal calculation, binomial approximation to normal, and a straightforward one-sample z-test. While it involves several steps and different statistical concepts, each part uses routine A-level procedures without requiring novel insight or complex problem-solving. The hypothesis test in part (c) is particularly standard with given values making calculations straightforward. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{24.63 - 25}{\sigma} = -1.0364\) | M1 | Standardising 24.63, 25 and \(\sigma\); setting equal to \(z\) where \(1 < |
| \([\sigma =]\ 0.357\) (must come from compatible signs) | A1 | awrt 0.36; do not award if signs not compatible |
| \(P(D > k) = 0.4\) or \(P(D < k) = 0.6\) | B1 | Either correct probability statement |
| \(\frac{k-25}{0.357} = 0.2533\) | M1 | Correct expression with \(z =\) awrt 0.253 |
| \(k =\) awrt 25.09 | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([Y \sim B(200, 0.45) \rightarrow]\ W \sim N(90, 49.5)\) | B1 | Setting up normal approximation; look out for \(\sigma = \frac{3\sqrt{22}}{2}\) or \(\sigma =\) awrt 7.04 |
| \(P(Y < 100) \approx P(W < 99.5) \left[= P\left(Z < \frac{99.5-90}{\sqrt{49.5}}\right)\right]\) | M1 | Continuity correction: \(P(W < 100.5)\), \(P(W < 99.5)\) or \(P(W < 98.5)\) condone \(\leq\) |
| \(= 0.9115...\) awrt 0.912 | A1 | [Note: 0.911299... from binomial scores 0/3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu = 25 \quad H_1: \mu < 25\) | B1 | Both hypotheses in terms of \(\mu\) |
| \([\bar{D} \sim] N\!\left(25, \frac{0.16^2}{20}\right)\) | M1 | Selecting suitable model: Normal, mean 25, sd \(= \frac{0.16}{\sqrt{20}}\) or var \(= \frac{4}{3125}\) |
| \(P(\bar{D} < 24.94) [= P(Z < -1.677...)] = 0.046766...\) | A1 | \(p\) value awrt 0.047 or test statistic awrt \(-1.68\) or CV awrt 24.941 |
| \(p = 0.047 < 0.05\) or \(z = -1.677... < -1.6449\) or \(24.94 < 24.94115...\) or reject \(H_0\)/in critical region/significant | M1 | Correct comparison (including compatible signs) or correct non-contextual conclusion |
| There is sufficient evidence to support Hannah's belief | A1 | Correct conclusion in context mentioning Hannah's belief or mean amount less than 25ml (dep on M1A1M1) |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{24.63 - 25}{\sigma} = -1.0364$ | M1 | Standardising 24.63, 25 and $\sigma$; setting equal to $z$ where $1 < |z| < 2$ |
| $[\sigma =]\ 0.357$ (must come from compatible signs) | A1 | awrt 0.36; do not award if signs not compatible |
| $P(D > k) = 0.4$ or $P(D < k) = 0.6$ | B1 | Either correct probability statement |
| $\frac{k-25}{0.357} = 0.2533$ | M1 | Correct expression with $z =$ awrt 0.253 |
| $k =$ awrt **25.09** | A1 | — |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[Y \sim B(200, 0.45) \rightarrow]\ W \sim N(90, 49.5)$ | B1 | Setting up normal approximation; look out for $\sigma = \frac{3\sqrt{22}}{2}$ or $\sigma =$ awrt 7.04 |
| $P(Y < 100) \approx P(W < 99.5) \left[= P\left(Z < \frac{99.5-90}{\sqrt{49.5}}\right)\right]$ | M1 | Continuity correction: $P(W < 100.5)$, $P(W < 99.5)$ or $P(W < 98.5)$ condone $\leq$ |
| $= 0.9115...$ awrt **0.912** | A1 | [Note: 0.911299... from binomial scores 0/3] |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 25 \quad H_1: \mu < 25$ | B1 | Both hypotheses in terms of $\mu$ |
| $[\bar{D} \sim] N\!\left(25, \frac{0.16^2}{20}\right)$ | M1 | Selecting suitable model: Normal, mean 25, sd $= \frac{0.16}{\sqrt{20}}$ or var $= \frac{4}{3125}$ |
| $P(\bar{D} < 24.94) [= P(Z < -1.677...)] = 0.046766...$ | A1 | $p$ value awrt 0.047 **or** test statistic awrt $-1.68$ **or** CV awrt 24.941 |
| $p = 0.047 < 0.05$ or $z = -1.677... < -1.6449$ or $24.94 < 24.94115...$ or reject $H_0$/in critical region/significant | M1 | Correct comparison (including compatible signs) **or** correct non-contextual conclusion |
| There is sufficient evidence to support Hannah's belief | A1 | Correct conclusion in context mentioning Hannah's belief or mean amount less than 25ml (dep on M1A1M1) |\begin{enumerate}
\item A machine puts liquid into bottles of perfume. The amount of liquid put into each bottle, $D \mathrm { ml }$, follows a normal distribution with mean 25 ml
\end{enumerate}
Given that 15\% of bottles contain less than 24.63 ml\\
(a) find, to 2 decimal places, the value of $k$ such that $\mathrm { P } ( 24.63 < D < k ) = 0.45$
A random sample of 200 bottles is taken.\\
(b) Using a normal approximation, find the probability that fewer than half of these bottles contain between 24.63 ml and $k \mathrm { ml }$
The machine is adjusted so that the standard deviation of the liquid put in the bottles is now 0.16 ml
Following the adjustments, Hannah believes that the mean amount of liquid put in each bottle is less than 25 ml
She takes a random sample of 20 bottles and finds the mean amount of liquid to be 24.94 ml\\
(c) Test Hannah's belief at the $5 \%$ level of significance.
You should state your hypotheses clearly.
\hfill \mbox{\textit{Edexcel Paper 3 2019 Q5 [13]}}