| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Known variance confidence interval |
| Difficulty | Challenging +1.2 This is a multi-part S3 question requiring standard confidence interval calculations and linear combinations of normal distributions. Part (a) is routine verification, parts (b-c)(i) are textbook applications, but part (c)(ii) requires translating 'non-overlapping intervals' into probability conditions involving the difference of sample means, which demands some problem-solving insight beyond pure recall. |
| Spec | 5.04b Linear combinations: of normal distributions5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x}\pm 1.6449\times\frac{5}{\sqrt{10}}\) | M1 B1 | M1 for \(\bar{x}\pm z\text{-value}\times\frac{5}{\sqrt{10}}\); B1 for \(z=1.6449\) or better |
| \(\bar{x}\pm 2.60 \Rightarrow (\bar{x}-2.60,\ \bar{x}+2.60)\) | A1* | Answer given; no incorrect working (condone use of 1.645) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{y}\pm 1.96\times\frac{3}{\sqrt{20}}\) | M1 B1 | M1 for \(\bar{y}\pm z\text{-value}\times\frac{3}{\sqrt{20}}\); B1 for \(z=1.96\) or better |
| \(\bar{y}\pm 1.31 \Rightarrow (\bar{y}-1.31,\ \bar{y}+1.31)\) | A1 | Allow 1.315 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{X}-\bar{Y}\sim N\!\left(\mu-\mu,\ \frac{5^2}{10}+\frac{3^2}{20}\right)\Rightarrow \bar{X}-\bar{Y}\sim N(0,\ 2.95)\) | M1 A1 | M1 for correct method to find variance; A1 for \(N(0,2.95)\); allow \(N\!\left(0,\frac{5^2}{10}+\frac{3^2}{20}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Do not overlap when \(\bar{x}-2.60>\bar{y}+1.31\) or \(\bar{x}+2.60<\bar{y}-1.31\) | M1 | oe ft part (b) |
| \(\bar{x}-\bar{y}>3.91\) or \(\bar{x}-\bar{y}<-3.91\) | A1ft | ft part (b) |
| \(2\times P(\bar{X}-\bar{Y}>3.91)=2\times P\!\left(Z>\frac{3.91}{\sqrt{2.95}}\right)=2\times P(Z>2.276\ldots)\) | M1 M1 | First M1 for multiplying by 2; second M1 for standardising ft their 3.91 and their SD |
| \(2\times 0.0113=0.0226\) (calculator: \(2\times 0.0114\ldots=0.0228\)) | A1 | awrt \(0.0226\) – awrt \(0.0228\) |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x}\pm 1.6449\times\frac{5}{\sqrt{10}}$ | M1 B1 | M1 for $\bar{x}\pm z\text{-value}\times\frac{5}{\sqrt{10}}$; B1 for $z=1.6449$ or better |
| $\bar{x}\pm 2.60 \Rightarrow (\bar{x}-2.60,\ \bar{x}+2.60)$ | A1* | Answer given; no incorrect working (condone use of 1.645) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{y}\pm 1.96\times\frac{3}{\sqrt{20}}$ | M1 B1 | M1 for $\bar{y}\pm z\text{-value}\times\frac{3}{\sqrt{20}}$; B1 for $z=1.96$ or better |
| $\bar{y}\pm 1.31 \Rightarrow (\bar{y}-1.31,\ \bar{y}+1.31)$ | A1 | Allow 1.315 |
### Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{X}-\bar{Y}\sim N\!\left(\mu-\mu,\ \frac{5^2}{10}+\frac{3^2}{20}\right)\Rightarrow \bar{X}-\bar{Y}\sim N(0,\ 2.95)$ | M1 A1 | M1 for correct method to find variance; A1 for $N(0,2.95)$; allow $N\!\left(0,\frac{5^2}{10}+\frac{3^2}{20}\right)$ |
### Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Do not overlap when $\bar{x}-2.60>\bar{y}+1.31$ or $\bar{x}+2.60<\bar{y}-1.31$ | M1 | oe ft part (b) |
| $\bar{x}-\bar{y}>3.91$ or $\bar{x}-\bar{y}<-3.91$ | A1ft | ft part (b) |
| $2\times P(\bar{X}-\bar{Y}>3.91)=2\times P\!\left(Z>\frac{3.91}{\sqrt{2.95}}\right)=2\times P(Z>2.276\ldots)$ | M1 M1 | First M1 for multiplying by 2; second M1 for standardising ft their 3.91 and their SD |
| $2\times 0.0113=0.0226$ (calculator: $2\times 0.0114\ldots=0.0228$) | A1 | awrt $0.0226$ – awrt $0.0228$ |
---\begin{enumerate}
\item The continuous random variable $X$ is normally distributed with
\end{enumerate}
$$X \sim \mathrm {~N} \left( \mu , 5 ^ { 2 } \right)$$
A random sample of 10 observations of $X$ is taken and $\bar { X }$ denotes the sample mean.\\
(a) Show that a $90 \%$ confidence interval for $\mu$, in terms of $\bar { x }$, is given by
$$( \bar { x } - 2.60 , \bar { x } + 2.60 )$$
The continuous random variable $Y$ is normally distributed with
$$Y \sim \mathrm {~N} \left( \mu , 3 ^ { 2 } \right)$$
A random sample of 20 observations of $Y$ are taken and $\bar { Y }$ denotes the sample mean.\\
(b) Find a 95\% confidence interval for $\mu$, in terms of $\bar { y }$\\
(c) Given that $X$ and $Y$ are independent,\\
(i) find the distribution of $\bar { X } - \bar { Y }$\\
(ii) calculate the probability that the two confidence intervals from part (a) and part (b) do not overlap.
\hfill \mbox{\textit{Edexcel S3 2023 Q5 [13]}}