| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Known variance confidence intervals |
| Difficulty | Standard +0.8 Part (a) is a standard confidence interval calculation with known variance. Part (b) requires recognizing that the number of intervals not containing μ follows a binomial distribution with p=0.05, then calculating P(X>3) - this conceptual leap and multi-step binomial calculation elevates this above routine questions. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = 230.5\); \(230.5 \pm 1.96 \times \dfrac{1.2}{\sqrt{5}}\) | M1 | \(\bar{x} \pm z \times \dfrac{1.2}{\sqrt{5}}\) |
| \(z = 1.96\) | B1 | |
| \(= (229.44815...,\ 231.55185...) =\) awrt\((229.4,\ 231.6)\) | A1, A1 | At least one endpoint correct; both endpoints correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(X\) = number of confidence intervals that don't contain \(\mu\); \(X \sim \text{B}(20, 0.05)\) | M1 | Writing or using \(X \sim \text{B}(20, 0.05)\) or \(Y \sim \text{B}(20, 0.95)\) |
| \(\{P(X>3)\} = 1 - P(X \leq 3)\) or \(1 - 0.9841\) | A1 | \(1 - P(X \leq 3)\) or \(1 - 0.9841\) or \(P(Y \leq 16)\). Can be implied by final answer |
| \(= 0.0159\) | A1 | awrt 0.0159 |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = 230.5$; $230.5 \pm 1.96 \times \dfrac{1.2}{\sqrt{5}}$ | M1 | $\bar{x} \pm z \times \dfrac{1.2}{\sqrt{5}}$ |
| $z = 1.96$ | B1 | |
| $= (229.44815...,\ 231.55185...) =$ awrt$(229.4,\ 231.6)$ | A1, A1 | At least one endpoint correct; both endpoints correct |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $X$ = number of confidence intervals that **don't contain** $\mu$; $X \sim \text{B}(20, 0.05)$ | M1 | Writing or using $X \sim \text{B}(20, 0.05)$ or $Y \sim \text{B}(20, 0.95)$ |
| $\{P(X>3)\} = 1 - P(X \leq 3)$ or $1 - 0.9841$ | A1 | $1 - P(X \leq 3)$ or $1 - 0.9841$ or $P(Y \leq 16)$. Can be implied by final answer |
| $= 0.0159$ | A1 | awrt **0.0159** |
---6. A company produces a certain type of mug. The masses of these mugs are normally distributed with mean $\mu$ and standard deviation 1.2 grams. A random sample of 5 mugs is taken and the mass, in grams, of each mug is measured. The results are given below.
\section*{$\begin{array} { l l l l l } 229.1 & 229.6 & 230.9 & 231.2 & 231.7 \end{array}$}
\begin{enumerate}[label=(\alph*)]
\item Find a $95 \%$ confidence interval for $\mu$, giving your limits correct to 1 decimal place.
Sonia plans to take 20 random samples, each of 5 mugs. A 95\% confidence interval for $\mu$ is to be determined for each sample.
\item Find the probability that more than 3 of these intervals will not contain $\mu$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2017 Q6 [7]}}