| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | October |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (upper tail) |
| Difficulty | Standard +0.3 This is a standard hypothesis testing question with routine normal distribution calculations. Part (a) is a basic z-score lookup, part (b) is a textbook one-sample z-test with known variance, part (c) involves straightforward conditional probability, and part (d) requires finding a median of a truncated normal distribution. While multi-part, each component uses standard techniques without requiring novel insight or complex problem-solving, making it slightly easier than average. |
| 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 |
|---|---|---|
| (a) | {Let \(X =\) time spent, \(\text{P}(X > 15) =\ \)} \(0.105649...\) awrt \(\mathbf{0.106}\) | B1 |
| (b) | \(H_0 : \mu = 10\); \(H_1 : \mu > 10\). \(\bar{X} \sim \text{N}\left(10, \left(\frac{4}{\sqrt{20}}\right)^2\right); \text{P}(\bar{X} > 11.5) = 0.046766...[\) Condone 0.9532...] [This is significant (\(< 5\%\)) so ] there is evidence to support the complaint. | M1;A1; A1 |
| (c)(i) | \([\text{P}(T < 2) =\ ] 0.1956...\) awrt \(\mathbf{0.196}\) | B1 |
| (c)(ii) | Require \(\frac{\text{P}(0 < T < 2)}{\text{P}(T > 0)} = \frac{0.119119...}{0.923436...} = 0.1289955...\) awrt \(\mathbf{0.129}\) | M1; A1;A1 |
| (c)(iii) | The current model suggests non-negligible probability of \(T\) values \(< 0\) which is impossible. | B1 |
| (d) | Require \(t\) such that \(\text{P}(T > t \vert T > 2) = 0.5\) or \(\text{P}(T < t \vert T > 2) = 0.5\). e.g. \(\frac{\text{P}(T > t)}{\text{P}(T > 2)} = 0.5\); so \(\text{P}(T > t) = 0.5 \times [1 - (c)(i)]\) or \(\text{P}(T > t) = 0.5 \times 0.8043...\). [i.e. \(\text{P}(T > t) = 0.40...\) implies] \(\frac{t-5}{3.5} = 0.2533\) or \(\text{P}(T < t) = "0.5978..."\). \(t = 5.886...\) or from calculator 5.867... so awrt \(\mathbf{5.9}\) | M1; M1; A1ft; M1; A1 |
| (a) | {Let $X =$ time spent, $\text{P}(X > 15) =\ $} $0.105649...$ awrt $\mathbf{0.106}$ | B1 | B1 for awrt 0.106 (from calculator) [Allow 10.6%]. |
|---|---|---|---|
| (b) | $H_0 : \mu = 10$; $H_1 : \mu > 10$. $\bar{X} \sim \text{N}\left(10, \left(\frac{4}{\sqrt{20}}\right)^2\right); \text{P}(\bar{X} > 11.5) = 0.046766...[$ Condone 0.9532...] [This is significant ($< 5\%$) so ] there is evidence to support the complaint. | M1;A1; A1 | B1 for both hypotheses correct in terms of $\mu$. M1 for selection of a correct model (sight or use of correct normal- may not have label $\bar{X}$). 1st A1 for use of this model to get probability allow 0.046–0.047 [Condone awrt 0.953]. 2nd A1 (dep on 1st A1 or at least $\text{P}(\bar{X} > 11.5) < 0.05$ (o.e.)) for a correct conclusion in context -must mention **complaint/claim** or **time/mins is > 10**. |
| (c)(i) | $[\text{P}(T < 2) =\ ] 0.1956...$ awrt $\mathbf{0.196}$ | B1 | B1 for awrt 0.196 (from calculator) [Allow 19.6%]. |
| (c)(ii) | Require $\frac{\text{P}(0 < T < 2)}{\text{P}(T > 0)} = \frac{0.119119...}{0.923436...} = 0.1289955...$ awrt $\mathbf{0.129}$ | M1; A1;A1 | M1 for a correct probability ratio expression (may be implied by 1st A1 scored). 1st A1 for a correct ratio of probabilities (both correct or truncated to 2 dp). 2nd A1 for awrt 0.129. |
| (c)(iii) | The current model suggests **non-negligible probability** of $T$ values $< 0$ which is impossible. | B1 | B1 for a suitable explanation of why model is not suitable based on negative $T$ values. Must say that a **significant proportion** of values $< 0$ (o.e.) $e.g., \text{P}(T > 0)$ should be closer to 1 or Difference between $\text{P}(T < 2 \vert T > 0)$ and $\text{P}(T < 2)$ is too big (o.e.). |
| (d) | Require $t$ such that $\text{P}(T > t \vert T > 2) = 0.5$ or $\text{P}(T < t \vert T > 2) = 0.5$. e.g. $\frac{\text{P}(T > t)}{\text{P}(T > 2)} = 0.5$; so $\text{P}(T > t) = 0.5 \times [1 - (c)(i)]$ or $\text{P}(T > t) = 0.5 \times 0.8043...$. [i.e. $\text{P}(T > t) = 0.40...$ implies] $\frac{t-5}{3.5} = 0.2533$ or $\text{P}(T < t) = "0.5978..."$. $t = 5.886...$ or from calculator 5.867... so awrt $\mathbf{5.9}$ | M1; M1; A1ft; M1; A1 | 1st M1 for a correct conditional probability statement to start the problem or 0.5 × $\text{P}(T > 2)$. 2nd M1 for correct ratio of probability expressions [Must have $\text{P}(T > t)$ or $\text{P}(T < t)$]. 1st A1ft for a correct equation for $\text{P}(T > t)$ (o.e.) if their answer to part (c)(i) [May be in a diagram]. 3rd M1 for attempt to find $t$ (standardising and sight of 0.2533) or prepare to use calc (ft). Arriving at $\text{P}(T < \text{median}) = 1 - 0.5 \times$ "their 0.8043" will score 1st 4 marks. 2nd A1 for awrt 5.9 and at least one M mark scores 5/5 [Answer only send to review]. Sight of awrt 5.9 and at least one M mark scores 5/5. |\begin{enumerate}
\item A health centre claims that the time a doctor spends with a patient can be modelled by a normal distribution with a mean of 10 minutes and a standard deviation of 4 minutes.\\
(a) Using this model, find the probability that the time spent with a randomly selected patient is more than 15 minutes.
\end{enumerate}
Some patients complain that the mean time the doctor spends with a patient is more than 10 minutes.
The receptionist takes a random sample of 20 patients and finds that the mean time the doctor spends with a patient is 11.5 minutes.\\
(b) Stating your hypotheses clearly and using a $5 \%$ significance level, test whether or not there is evidence to support the patients' complaint.
The health centre also claims that the time a dentist spends with a patient during a routine appointment, $T$ minutes, can be modelled by the normal distribution where $T \sim \mathrm {~N} \left( 5,3.5 ^ { 2 } \right)$\\
(c) Using this model,\\
(i) find the probability that a routine appointment with the dentist takes less than 2 minutes\\
(ii) find $\mathrm { P } ( T < 2 \mid T > 0 )$\\
(iii) hence explain why this normal distribution may not be a good model for $T$.
The dentist believes that she cannot complete a routine appointment in less than 2 minutes.\\
She suggests that the health centre should use a refined model only including values of $T > 2$\\
(d) Find the median time for a routine appointment using this new model, giving your answer correct to one decimal place.
\hfill \mbox{\textit{Edexcel Paper 3 2020 Q5 [15]}}