| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparison involving sums or multiples |
| Difficulty | Standard +0.8 This question requires understanding of linear combinations of independent normal random variables, including finding the distribution of a difference (part a) and a weighted sum (part b). While the concepts are standard for S3, students must correctly identify parameters, apply variance rules for independent variables, and work with non-trivial calculations (17m = 1700cm). The multi-step reasoning and potential for algebraic errors place this above average difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(M \sim N(177, 25)\), \(F \sim N(163, 16)\) | ||
| \(E(M-F) = 177 - 163 = 14\) | B1 | |
| \(\text{Var}(M-F) = 25 + 16 = 41\), so \(M-F \sim N(14, 41)\) | M1A1 | |
| \(P(M-F>0) = P\!\left(Z > \frac{-14}{\sqrt{41}}\right)\) or \(P\!\left(Z < \frac{14}{\sqrt{41}}\right)\) | M1 | M1 for identifying correct probability and standardising with their mean and sd |
| \(= P(Z < 2.186...)\) | ||
| \(= 0.9854\) (or 0.9856 by calculator) awrt 0.985 or 0.986 | A1 | Accept metres: 2.14 award M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(W = M_1 + M_2 + \ldots + M_6 + F_1 + F_2 + \ldots + F_4\) | ||
| \(E(W) = 6\times177 + 4\times163 = 1714\) | B1 | |
| \(\text{Var}(W) = 6\times25 + 4\times16 = 214\) | M1A1 | Condone reversed sds for method |
| \(P(W < 1700) = P\!\left(Z < \frac{1700-1714}{\sqrt{214}}\right)\) or \(P\!\left(Z > \frac{1714-1700}{\sqrt{214}}\right)\) | M1 | Require explicit sd or accept \(\sqrt{1156}\) for M1A0; implied by correct answer |
| \(= P(Z < -0.957...)\) awrt \(Z < -0.96\) or \(Z > 0.96\) | A1 | |
| \(= 1 - 0.8315 = 0.1685\) awrt 0.169 (0.1693 by calculator) | A1 |
# Question 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $M \sim N(177, 25)$, $F \sim N(163, 16)$ | | |
| $E(M-F) = 177 - 163 = 14$ | B1 | |
| $\text{Var}(M-F) = 25 + 16 = 41$, so $M-F \sim N(14, 41)$ | M1A1 | |
| $P(M-F>0) = P\!\left(Z > \frac{-14}{\sqrt{41}}\right)$ or $P\!\left(Z < \frac{14}{\sqrt{41}}\right)$ | M1 | M1 for identifying correct probability and standardising with their mean and sd |
| $= P(Z < 2.186...)$ | | |
| $= 0.9854$ (or 0.9856 by calculator) awrt 0.985 or 0.986 | A1 | Accept metres: 2.14 award M1A0 |
---
# Question 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $W = M_1 + M_2 + \ldots + M_6 + F_1 + F_2 + \ldots + F_4$ | | |
| $E(W) = 6\times177 + 4\times163 = 1714$ | B1 | |
| $\text{Var}(W) = 6\times25 + 4\times16 = 214$ | M1A1 | Condone reversed sds for method |
| $P(W < 1700) = P\!\left(Z < \frac{1700-1714}{\sqrt{214}}\right)$ or $P\!\left(Z > \frac{1714-1700}{\sqrt{214}}\right)$ | M1 | Require explicit sd or accept $\sqrt{1156}$ for M1A0; implied by correct answer |
| $= P(Z < -0.957...)$ awrt $Z < -0.96$ or $Z > 0.96$ | A1 | |
| $= 1 - 0.8315 = 0.1685$ awrt 0.169 (0.1693 by calculator) | A1 | |
The image provided appears to be only the back/contact information page of an Edexcel mark scheme document. It contains only publisher information (Edexcel Publications contact details, Order Code UA033146 Summer 2012, Pearson Education Limited registration details) and logos for Ofqual, Welsh Assembly Government, and CEA.
There is **no mark scheme content** visible on this page — no questions, answers, mark allocations, or guidance notes are present.
Could you please share the actual mark scheme pages (the ones containing the questions and marking criteria)? I'd be happy to extract and format that content for you once those pages are provided.7. The heights, in cm, of the male employees in a large company follow a normal distribution with mean 177 and standard deviation 5 The heights, in cm, of the female employees follow a normal distribution with mean 163 and standard deviation 4
A male employee and a female employee are chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the male employee is taller than the female employee.
Six male employees and four female employees are chosen at random.
\item Find the probability that their total height is less than 17 m .
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2012 Q7 [11]}}