| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparison involving sums or multiples |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal random variables with clear structure. Parts (a)-(c) involve standard normal distribution calculations and combining independent normals using the formula for sums. Part (d) tests understanding of independence assumptions. While it requires careful arithmetic with means and variances, it follows a well-practiced template with no novel problem-solving or geometric insight needed. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | P(Length ≥ 180) = 0.1587 |
| [1] | 1.1 | |
| 6 | (b) | Total length ⁓ N(5 × 179.2, 5 × 0.64) |
| Answer | Marks |
|---|---|
| P(Total < 895) = 0.2881 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | For method for either Can be implied by correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (c) | Mean = 5 × 179.2 + 6 × 9.8 |
Total length ⁓ N(954.8, 3.74)
| Answer | Marks |
|---|---|
| P(Total length < 950) = 0.0065 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | For method for mean Can be implied by correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (d) | The widths of individual fence posts must be |
| independent | E1 | |
| [1] | 2.4 |
Question 6:
6 | (a) | P(Length ≥ 180) = 0.1587 | B1
[1] | 1.1
6 | (b) | Total length ⁓ N(5 × 179.2, 5 × 0.64)
N(896, 3.2)
P(Total < 895) = 0.2881 | M1
A1
B1
[3] | 3.1b
1.1
1.1 | For method for either Can be implied by correct answer
For both correct
BC
6 | (c) | Mean = 5 × 179.2 + 6 × 9.8
Variance = 5 × 0.64 + 6 × 0.09
Total length ⁓ N(954.8, 3.74)
P(Total length < 950) = 0.0065 | M1
M1
A1
A1
[4] | 3.1b
1.1
1.1
1.1 | For method for mean Can be implied by correct answer
For method for variance Can be implied by correct answer
For both correct
BC
6 | (d) | The widths of individual fence posts must be
independent | E1
[1] | 2.46 The length $L$ of a particular type of fence panel is Normally distributed with mean 179.2 cm and standard deviation 0.8 cm . You should assume that the lengths of individual fence panels are independent of each other.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the length of a randomly chosen fence panel is at least 180 cm .
\item Find the probability that the total length of 5 randomly chosen fence panels is less than 895 cm .
The width $W$ of a fence post is Normally distributed with mean 9.8 cm and standard deviation 0.3 cm . A straight fence is constructed using 6 posts and 5 panels with no gaps between them.
Fig. 6 shows a view from above of the first two posts, the first panel and the start of the second panel. You should assume that the lengths of fence panels and widths of fence posts are independent.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4caa7409-cb32-41da-ad64-012a45753296-6_213_1522_934_244}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
\item Determine the probability that the total length of the fence, including the posts, is less than 9.5 m .
\item State another assumption that is necessary for the calculation of the probability in part (c) to be valid.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2022 Q6 [9]}}