| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Difficulty | Standard +0.8 This S3 question requires understanding of linear combinations of normal distributions and difference of independent normals. Part (a) needs recognizing that the difference of two normals has variance 2σ², a non-trivial conceptual step. Part (b) involves summing multiple independent normals with different parameters. Both parts require careful variance calculations and standardization, going beyond routine application of the normal distribution. |
| Spec | 2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| Mean | Standard Deviation | |
| \(P1\) | 252 | 17 |
| \(M1\) | 314 | 42 |
| \(S1\) | 284 | 29 |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a): let \(X =\) time to mark P1 paper | M1 A1 | |
| let \(A = X_1 - X_2\) \(\therefore A \sim N(0, 2 \times 17^2) = N(0, 578)\) | M1 A1 | |
| \(P(\text{-}5 < A < 5) = P(\frac{-5-0}{\sqrt{578}} < Z < \frac{5-0}{\sqrt{578}})\) | M1 A1 | |
| \(= P(-0.21 < Z < 0.21) = 0.5832 - (1 - 0.5832) = 0.166\) | M1 A1 | |
| Part (b): let \(M =\) time to mark M1 paper, let \(S =\) time to mark S1 paper | ||
| let \(T = M_1 + \ldots + M_{45} + S_1 + \ldots + S_{80}\) | M2 A2 | |
| \(\therefore T \sim N(45 \times 314 + 80 \times 284, 45 \times 42^2 + 80 \times 29^2) = N(36850, 146660)\) | M2 A2 | |
| \(P(\text{time} < 10 \text{ hours}) = P(T < 36000) = P(Z < \frac{36000-36850}{\sqrt{146660}})\) | M1 | |
| \(= P(Z < -2.22) = 1 - 0.9868 = 0.0132\) | M1 A1 | (13 marks total) |
**Part (a):** let $X =$ time to mark P1 paper | M1 A1 |
let $A = X_1 - X_2$ $\therefore A \sim N(0, 2 \times 17^2) = N(0, 578)$ | M1 A1 |
$P(\text{-}5 < A < 5) = P(\frac{-5-0}{\sqrt{578}} < Z < \frac{5-0}{\sqrt{578}})$ | M1 A1 |
$= P(-0.21 < Z < 0.21) = 0.5832 - (1 - 0.5832) = 0.166$ | M1 A1 |
**Part (b):** let $M =$ time to mark M1 paper, let $S =$ time to mark S1 paper | |
let $T = M_1 + \ldots + M_{45} + S_1 + \ldots + S_{80}$ | M2 A2 |
$\therefore T \sim N(45 \times 314 + 80 \times 284, 45 \times 42^2 + 80 \times 29^2) = N(36850, 146660)$ | M2 A2 |
$P(\text{time} < 10 \text{ hours}) = P(T < 36000) = P(Z < \frac{36000-36850}{\sqrt{146660}})$ | M1 |
$= P(Z < -2.22) = 1 - 0.9868 = 0.0132$ | M1 A1 | (13 marks total)
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**Total: 75 marks**An examiner believes that once she has marked the first 20 papers the time it takes her to mark one paper for a particular exam follows a Normal distribution. Having already marked more than 20 papers for each of the $P1$, $M1$ and $S1$ modules set one summer, the mean and standard deviation, in seconds, of the time it takes her to mark a paper for each module are as shown in the table below.
\begin{center}
\begin{tabular}{|c|c|c|}
\hline
& Mean & Standard Deviation \\
\hline
$P1$ & 252 & 17 \\
\hline
$M1$ & 314 & 42 \\
\hline
$S1$ & 284 & 29 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the difference in the time it takes her to mark two randomly chosen $P1$ papers is less than 5 seconds. [6]
\item Find the probability that it takes her less than 10 hours to mark 45 $M1$ and 80 $S1$ papers. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q7 [13]}}