AQA S3 2013 June — Question 5 10 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2013
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeMultiple stage process probability
DifficultyStandard +0.3 This is a straightforward application of standard results for linear combinations of normal variables. Part (a) requires routine use of variance formulas with correlation (Var(X+Y) = Var(X) + Var(Y) + 2ρσ_Xσ_Y), and part (b) involves standard normal probability calculations. The correlation coefficient is given, eliminating any conceptual challenge. This is slightly easier than average because it's a direct textbook-style application with no problem-solving insight required, though the correlation aspect adds minor complexity.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions
5 The schedule for an organisation's afternoon meeting is as follows.
Session A (Speaker 1) 2.00 pm to 3.15 pm
Session B (Discussion) 3.15 pm to 3.45 pm
Session C (Speaker 2) \(\quad 3.45 \mathrm { pm }\) to 5.00 pm
Records show that:
the duration, \(X\), of Session A has mean 68 minutes and standard deviation 10 minutes;
the duration, \(Y\), of Session B has mean 25 minutes and standard deviation 5 minutes;
the duration, \(Z\), of Session C has mean 73 minutes and standard deviation 15 minutes;
and that: $$\rho _ { X Z } = 0 \quad \rho _ { X Y } = - 0.8 \quad \rho _ { Y Z } = 0$$
  1. Determine the means and the variances of:
    1. \(L = X + Z\);
    2. \(M = X + Y\).
  2. Assuming that \(L\) and \(M\) are each normally distributed, determine the probability that:
    1. the total time for the two speaker sessions is less than \(2 \frac { 1 } { 2 }\) hours;
    2. Session C is late in starting.
Question 5:
Part (a)(i): \(L = X + Z\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(L) = 68 + 73 = 141\) (minutes)B1
\(\text{Var}(L) = 10^2 + 15^2 = 325\) (since \(\rho_{XZ} = 0\))B1 Must state or use \(\rho_{XZ}=0\)
Part (a)(ii): \(M = X + Y\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(M) = 68 + 25 = 93\) (minutes)B1
\(\text{Var}(M) = \sigma_X^2 + \sigma_Y^2 + 2\rho_{XY}\sigma_X\sigma_Y\)M1 Correct formula with covariance term
\(= 100 + 25 + 2(-0.8)(10)(5)\)A1ft
\(= 125 - 80 = 45\)A1
Part (b)(i): Total time for two speaker sessions less than \(2\frac{1}{2}\) hours
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(2\frac{1}{2}\) hours \(= 150\) minutes; use \(L \sim N(141, 325)\)M1
\(P(L < 150) = P\!\left(Z < \dfrac{150-141}{\sqrt{325}}\right) = P(Z < 0.4994...)\)A1ft
\(= 0.6913\) (awrt \(0.691\))A1
Part (b)(ii): Session C is late in starting
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Session C is late if \(M > 90\) minutes (scheduled duration of A+B is \(3.45-2.00 = 105\) min, but Sessions A and B together must exceed 105 min)M1 Need to identify correct threshold: \(M > 105\)
\(M \sim N(93, 45)\)
\(P(M > 105) = P\!\left(Z > \dfrac{105-93}{\sqrt{45}}\right) = P(Z > 1.789)\)M1 Standardising correctly
\(= 1 - \Phi(1.789) = 1 - 0.9633\)A1ft
\(= 0.0367\) (awrt \(0.0367\))A1 
## Question 5:

### Part (a)(i): $L = X + Z$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(L) = 68 + 73 = 141$ (minutes) | B1 | |
| $\text{Var}(L) = 10^2 + 15^2 = 325$ (since $\rho_{XZ} = 0$) | B1 | Must state or use $\rho_{XZ}=0$ |

### Part (a)(ii): $M = X + Y$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(M) = 68 + 25 = 93$ (minutes) | B1 | |
| $\text{Var}(M) = \sigma_X^2 + \sigma_Y^2 + 2\rho_{XY}\sigma_X\sigma_Y$ | M1 | Correct formula with covariance term |
| $= 100 + 25 + 2(-0.8)(10)(5)$ | A1ft | |
| $= 125 - 80 = 45$ | A1 | |

### Part (b)(i): Total time for two speaker sessions less than $2\frac{1}{2}$ hours

| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\frac{1}{2}$ hours $= 150$ minutes; use $L \sim N(141, 325)$ | M1 | |
| $P(L < 150) = P\!\left(Z < \dfrac{150-141}{\sqrt{325}}\right) = P(Z < 0.4994...)$ | A1ft | |
| $= 0.6913$ (awrt $0.691$) | A1 | |

### Part (b)(ii): Session C is late in starting

| Answer/Working | Marks | Guidance |
|---|---|---|
| Session C is late if $M > 90$ minutes (scheduled duration of A+B is $3.45-2.00 = 105$ min, but Sessions A and B together must exceed 105 min) | M1 | Need to identify correct threshold: $M > 105$ |
| $M \sim N(93, 45)$ | |
| $P(M > 105) = P\!\left(Z > \dfrac{105-93}{\sqrt{45}}\right) = P(Z > 1.789)$ | M1 | Standardising correctly |
| $= 1 - \Phi(1.789) = 1 - 0.9633$ | A1ft | |
| $= 0.0367$ (awrt $0.0367$) | A1 | |

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5 The schedule for an organisation's afternoon meeting is as follows.\\
Session A (Speaker 1) 2.00 pm to 3.15 pm\\
Session B (Discussion) 3.15 pm to 3.45 pm\\
Session C (Speaker 2) $\quad 3.45 \mathrm { pm }$ to 5.00 pm\\
Records show that:\\
the duration, $X$, of Session A has mean 68 minutes and standard deviation 10 minutes;\\
the duration, $Y$, of Session B has mean 25 minutes and standard deviation 5 minutes;\\
the duration, $Z$, of Session C has mean 73 minutes and standard deviation 15 minutes;\\
and that:

$$\rho _ { X Z } = 0 \quad \rho _ { X Y } = - 0.8 \quad \rho _ { Y Z } = 0$$
\begin{enumerate}[label=(\alph*)]
\item Determine the means and the variances of:
\begin{enumerate}[label=(\roman*)]
\item $L = X + Z$;
\item $M = X + Y$.
\end{enumerate}\item Assuming that $L$ and $M$ are each normally distributed, determine the probability that:
\begin{enumerate}[label=(\roman*)]
\item the total time for the two speaker sessions is less than $2 \frac { 1 } { 2 }$ hours;
\item Session C is late in starting.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA S3 2013 Q5 [10]}}
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