| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Probability multiple CIs contain mean |
| Difficulty | Standard +0.3 This is a straightforward S3 confidence interval question requiring standard procedures: (a) checking if a value lies in a given interval, (b) calculating a CI using the formula with critical values, and (c) applying binomial probability. All steps are routine applications of learned techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu = 328\), \(H_1: \mu \neq 328\) | B1 | Both hypotheses correct in terms of \(\mu\) |
| Significance level is 5% | dB1 | Dependent on two-tail test indicated; allow \((\alpha=)\ 0.05\) |
| 328 is within the interval, no evidence to support that \(\mu\) is not 328 | B1 | Allow "do not reject \(H_0\)"; condone poor null hypothesis notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1.96 \times \dfrac{\sigma}{\sqrt{150}} = \dfrac{329.76 - 327.84}{2} (= 0.96)\) | M1 B1 | M1 forms equation; B1 use of 1.96 (may be seen in (a) or implied by \(\sigma\)) |
| \(\sigma = 5.9987\) | — | — |
| \(328 - 2.3263 \times \dfrac{5.9987}{\sqrt{200}}\) or \(328 + 2.3263 \times \dfrac{5.9987}{\sqrt{200}}\) | M1 B1 | M1 correct method for one end; B1 awrt 2.3263; allow \(2.3 < |
| \(328 \pm 2.3263 \times \dfrac{5.9987}{\sqrt{200}}\) | dM1 | Dependent on both previous M marks; \(n=200\); fully correct method |
| \((327.0132...,\ 328.986...)\) awrt \((327, 329)\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim B(5, 0.98)\) or \(Y \sim B(5, 0.02)\) | M1 | Use of Binomial e.g. \(B(5, 0.98)\) or \(B(5, 0.02)\); may be implied by awrt 0.996 |
| \(P(X \geq 4)\) or \(P(Y \leq 1) = {}^5C_4(0.98)^4(0.02) + (0.98)^5\) | A1 | awrt 0.0922 + awrt 0.904 |
| \(= 0.9961...\) awrt \(0.996\) | A1 | — |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 328$, $H_1: \mu \neq 328$ | B1 | Both hypotheses correct in terms of $\mu$ |
| Significance level is 5% | dB1 | Dependent on two-tail test indicated; allow $(\alpha=)\ 0.05$ |
| 328 is within the interval, no evidence to support that $\mu$ is not 328 | B1 | Allow "do not reject $H_0$"; condone poor null hypothesis notation |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1.96 \times \dfrac{\sigma}{\sqrt{150}} = \dfrac{329.76 - 327.84}{2} (= 0.96)$ | M1 B1 | M1 forms equation; B1 use of 1.96 (may be seen in (a) or implied by $\sigma$) |
| $\sigma = 5.9987$ | — | — |
| $328 - 2.3263 \times \dfrac{5.9987}{\sqrt{200}}$ or $328 + 2.3263 \times \dfrac{5.9987}{\sqrt{200}}$ | M1 B1 | M1 correct method for one end; B1 awrt 2.3263; allow $2.3 < |z| < 3.1$ |
| $328 \pm 2.3263 \times \dfrac{5.9987}{\sqrt{200}}$ | dM1 | Dependent on both previous M marks; $n=200$; fully correct method |
| $(327.0132...,\ 328.986...)$ awrt $(327, 329)$ | A1 | — |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(5, 0.98)$ or $Y \sim B(5, 0.02)$ | M1 | Use of Binomial e.g. $B(5, 0.98)$ or $B(5, 0.02)$; may be implied by awrt 0.996 |
| $P(X \geq 4)$ or $P(Y \leq 1) = {}^5C_4(0.98)^4(0.02) + (0.98)^5$ | A1 | awrt 0.0922 + awrt 0.904 |
| $= 0.9961...$ awrt $0.996$ | A1 | — |
---\begin{enumerate}
\item The volume of water in a bottle has a normal distribution with unknown mean, $\mu$ millilitres, and known standard deviation, $\sigma$ millilitres.
\end{enumerate}
A random sample of 150 of the bottles of water gave a 95\% confidence interval for $\mu$ of\\
(327.84, 329.76)\\
(a) Using the confidence interval given, test whether or not $\mu = 328$
State your hypotheses clearly and write down the significance level you have used.
A second random sample, of 200 of these bottles of water, had a mean volume of 328 millilitres.\\
(b) Calculate a 98\% confidence interval for $\mu$ based on this second sample.
You must show all steps in your working.\\
(Solutions relying entirely on calculator technology are not acceptable.)
Using five different random samples of 200 of these bottles of water, five $98 \%$ confidence intervals for $\mu$ are to be found.\\
(c) Calculate the probability that more than 3 of these intervals will contain $\mu$
\hfill \mbox{\textit{Edexcel S3 2024 Q3 [12]}}