| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Expected frequency with unknown parameter |
| Difficulty | Standard +0.3 This is a multi-part normal distribution question requiring finding an unknown mean from a given percentile, calculating probabilities, applying binomial distribution, and performing a one-tailed hypothesis test. While it involves several standard techniques and multiple steps, each component uses routine A-level methods without requiring novel insight or particularly complex reasoning. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05d Sample mean as random variable2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Diagram with 49 and 50.75 marked | B1cao | |
| \(P(L > 50.98) = 0.025\) | B1cao | |
| \(\frac{50.98 - \mu}{0.5} = 1.96\) | M1 | Standardising with \(\mu\) and 0.5, setting equal to a \(z\) value (\( |
| \(\therefore \mu = 50\) | A1cao | |
| \(P(49 < L < 50.75)\) | M1 | Attempting correct probability for strips that can be used |
| \(= 0.9104\ldots\) awrt 0.910 | A1ft | awrt 0.910, allow ft of their \(\mu\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(S\) = number of strips that cannot be used, \(S \sim B(10, 0.090)\) | M1 | Identifying suitable binomial distribution |
| \(= P(S \leqslant 3) = 0.991166\ldots\) awrt 0.991 | A1 | awrt 0.991 (from calculator) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 50.1 \quad H_1: \mu > 50.1\) | B1 | Hypotheses stated correctly |
| \(\bar{X} \sim N\!\left(50.1, \frac{0.6^2}{15}\right)\) and \(\bar{X} > 50.4\) | M1 | Selecting correct model (stated or implied) |
| \(P(\bar{X} > 50.4) = 0.0264\) | A1 | \(p =\) awrt 0.0264, allow \(z =\) awrt 1.94 |
| \(p = 0.0264 > 0.01\) or \(z = 1.936\ldots < 2.3263\), not significant | A1 | Correct calculation, comparison and correct statement |
| There is insufficient evidence that the mean length of strips is greater than 50.1 | A1 | Correct conclusion in context mentioning "mean length" and 50.1 |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Diagram with 49 and 50.75 marked | B1cao | |
| $P(L > 50.98) = 0.025$ | B1cao | |
| $\frac{50.98 - \mu}{0.5} = 1.96$ | M1 | Standardising with $\mu$ and 0.5, setting equal to a $z$ value ($|z|>1$) |
| $\therefore \mu = 50$ | A1cao | |
| $P(49 < L < 50.75)$ | M1 | Attempting correct probability for strips that can be used |
| $= 0.9104\ldots$ awrt **0.910** | A1ft | awrt 0.910, allow ft of their $\mu$ |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $S$ = number of strips that cannot be used, $S \sim B(10, 0.090)$ | M1 | Identifying suitable binomial distribution |
| $= P(S \leqslant 3) = 0.991166\ldots$ awrt 0.991 | A1 | awrt 0.991 (from calculator) |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 50.1 \quad H_1: \mu > 50.1$ | B1 | Hypotheses stated correctly |
| $\bar{X} \sim N\!\left(50.1, \frac{0.6^2}{15}\right)$ and $\bar{X} > 50.4$ | M1 | Selecting correct model (stated or implied) |
| $P(\bar{X} > 50.4) = 0.0264$ | A1 | $p =$ awrt 0.0264, allow $z =$ awrt 1.94 |
| $p = 0.0264 > 0.01$ or $z = 1.936\ldots < 2.3263$, not significant | A1 | Correct calculation, comparison and correct statement |
| There is insufficient evidence that the **mean length** of strips is **greater than 50.1** | A1 | Correct conclusion in context mentioning "mean length" and 50.1 |
---\begin{enumerate}
\item A machine cuts strips of metal to length $L \mathrm {~cm}$, where $L$ is normally distributed with standard deviation 0.5 cm .
\end{enumerate}
Strips with length either less than 49 cm or greater than 50.75 cm cannot be used.\\
Given that 2.5\% of the cut lengths exceed 50.98 cm ,\\
(a) find the probability that a randomly chosen strip of metal can be used.
Ten strips of metal are selected at random.\\
(b) Find the probability fewer than 4 of these strips cannot be used.
A second machine cuts strips of metal of length $X \mathrm {~cm}$, where $X$ is normally distributed with standard deviation 0.6 cm
A random sample of 15 strips cut by this second machine was found to have a mean length of 50.4 cm\\
(c) Stating your hypotheses clearly and using a $1 \%$ level of significance, test whether or not the mean length of all the strips, cut by the second machine, is greater than 50.1 cm
\hfill \mbox{\textit{Edexcel Paper 3 Q3 [12]}}