AQA S3 2012 June — Question 6 17 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2012
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeMultiple stage process probability
DifficultyStandard +0.8 This question requires systematic application of variance formulas for correlated variables (including the non-trivial Var(U+V) with negative correlation and Var(W-2U)), understanding that continuous normal distributions have zero probability at exact points, and computing probabilities for linear combinations of normals. While methodical rather than requiring deep insight, the correlation handling, multiple parts, and conceptual understanding of continuous distributions place it moderately above average difficulty.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions
6 Alyssa lives in the country but works in a city centre.
Her journey to work each morning involves a car journey, a walk and wait, a train journey, and a walk. Her car journey time, \(U\) minutes, from home to the village car park has a mean of 13 and a standard deviation of 3 . Her time, \(V\) minutes, to walk from the village car park to the village railway station and wait for a train to depart has a mean of 15 and a standard deviation of 6 . Her train journey time, \(W\) minutes, from the village railway station to the city centre railway station has a mean of 24 and a standard deviation of 4 . Her time, \(X\) minutes, to walk from the city centre railway station to her office has a mean of 9 and a standard deviation of 2 . The values of the product moment correlation coefficient for the above 4 variables are $$\rho _ { U V } = - 0.6 \quad \text { and } \quad \rho _ { U W } = \rho _ { U X } = \rho _ { V W } = \rho _ { V X } = \rho _ { W X } = 0$$
  1. Determine values for the mean and the variance of:
    1. \(M = U + V\);
    2. \(D = W - 2 U\);
    3. \(T = M + W + X\), given that \(\rho _ { M W } = \rho _ { M X } = 0\).
  2. Assuming that the variables \(M , D\) and \(T\) are normally distributed, determine the probability that, on a particular morning:
    1. Alyssa's journey time from leaving home to leaving the village railway station is exactly 30 minutes;
    2. Alyssa's train journey time is more than twice her car journey time;
    3. Alyssa's total journey time is between 50 minutes and 70 minutes.
Question 6:
Part (a)(i): \(M = U + V\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(M) = 13 + 15 = 28\)B1 Correct mean
\(\text{Var}(M) = \text{Var}(U) + \text{Var}(V) + 2\rho_{UV}\sigma_U\sigma_V\)M1 Correct variance formula with covariance term
\(= 9 + 36 + 2(-0.6)(3)(6)\)A1 Correct substitution
\(= 45 - 21.6 = 23.4\)A1 Correct variance
Part (a)(ii): \(D = W - 2U\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(D) = 24 - 2(13) = -2\)B1 Correct mean
\(\text{Var}(D) = \text{Var}(W) + 4\text{Var}(U) - 4\rho_{WU}\sigma_W\sigma_U\)M1 Correct formula
\(= 16 + 4(9) - 0 = 52\)A1 Correct variance (since \(\rho_{WU}=0\))
Part (a)(iii): \(T = M + W + X\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(T) = 28 + 24 + 9 = 61\)B1 Correct mean
\(\text{Var}(T) = 23.4 + 16 + 4 = 43.4\)B1 Correct variance (using \(\rho_{MW}=\rho_{MX}=0\))
Part (b)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(M = 30) = 0\)B1 Since \(M\) is continuous/normally distributed
Part (b)(ii): \(P(W > 2U)\), i.e. \(P(D > 0)\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(D \sim N(-2, 52)\)
\(P(D > 0) = P\left(Z > \frac{0-(-2)}{\sqrt{52}}\right)\)M1 Standardising
\(= P\left(Z > \frac{2}{\sqrt{52}}\right) = P(Z > 0.2774)\)A1 Correct z-value
\(= 1 - \Phi(0.2774) = 1 - 0.6093 = 0.3907\)A1 Correct probability (awrt 0.391)
Part (b)(iii): \(P(50 < T < 70)\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(T \sim N(61, 43.4)\)
\(P(50 < T < 70) = P\left(\frac{50-61}{\sqrt{43.4}} < Z < \frac{70-61}{\sqrt{43.4}}\right)\)M1 Standardising both values
\(= P(-1.6698 < Z < 1.3663)\)A1 Correct z-values
\(= \Phi(1.3663) - (1 - \Phi(1.6698))\)M1 Correct probability method
\(= 0.9141 - 0.0475 = 0.8666\)A1 Correct answer (awrt 0.867) 
# Question 6:

## Part (a)(i): $M = U + V$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(M) = 13 + 15 = 28$ | B1 | Correct mean |
| $\text{Var}(M) = \text{Var}(U) + \text{Var}(V) + 2\rho_{UV}\sigma_U\sigma_V$ | M1 | Correct variance formula with covariance term |
| $= 9 + 36 + 2(-0.6)(3)(6)$ | A1 | Correct substitution |
| $= 45 - 21.6 = 23.4$ | A1 | Correct variance |

## Part (a)(ii): $D = W - 2U$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(D) = 24 - 2(13) = -2$ | B1 | Correct mean |
| $\text{Var}(D) = \text{Var}(W) + 4\text{Var}(U) - 4\rho_{WU}\sigma_W\sigma_U$ | M1 | Correct formula |
| $= 16 + 4(9) - 0 = 52$ | A1 | Correct variance (since $\rho_{WU}=0$) |

## Part (a)(iii): $T = M + W + X$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(T) = 28 + 24 + 9 = 61$ | B1 | Correct mean |
| $\text{Var}(T) = 23.4 + 16 + 4 = 43.4$ | B1 | Correct variance (using $\rho_{MW}=\rho_{MX}=0$) |

## Part (b)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(M = 30) = 0$ | B1 | Since $M$ is continuous/normally distributed |

## Part (b)(ii): $P(W > 2U)$, i.e. $P(D > 0)$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $D \sim N(-2, 52)$ | | |
| $P(D > 0) = P\left(Z > \frac{0-(-2)}{\sqrt{52}}\right)$ | M1 | Standardising |
| $= P\left(Z > \frac{2}{\sqrt{52}}\right) = P(Z > 0.2774)$ | A1 | Correct z-value |
| $= 1 - \Phi(0.2774) = 1 - 0.6093 = 0.3907$ | A1 | Correct probability (awrt 0.391) |

## Part (b)(iii): $P(50 < T < 70)$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $T \sim N(61, 43.4)$ | | |
| $P(50 < T < 70) = P\left(\frac{50-61}{\sqrt{43.4}} < Z < \frac{70-61}{\sqrt{43.4}}\right)$ | M1 | Standardising both values |
| $= P(-1.6698 < Z < 1.3663)$ | A1 | Correct z-values |
| $= \Phi(1.3663) - (1 - \Phi(1.6698))$ | M1 | Correct probability method |
| $= 0.9141 - 0.0475 = 0.8666$ | A1 | Correct answer (awrt 0.867) |
6 Alyssa lives in the country but works in a city centre.\\
Her journey to work each morning involves a car journey, a walk and wait, a train journey, and a walk.

Her car journey time, $U$ minutes, from home to the village car park has a mean of 13 and a standard deviation of 3 .

Her time, $V$ minutes, to walk from the village car park to the village railway station and wait for a train to depart has a mean of 15 and a standard deviation of 6 .

Her train journey time, $W$ minutes, from the village railway station to the city centre railway station has a mean of 24 and a standard deviation of 4 .

Her time, $X$ minutes, to walk from the city centre railway station to her office has a mean of 9 and a standard deviation of 2 .

The values of the product moment correlation coefficient for the above 4 variables are

$$\rho _ { U V } = - 0.6 \quad \text { and } \quad \rho _ { U W } = \rho _ { U X } = \rho _ { V W } = \rho _ { V X } = \rho _ { W X } = 0$$
\begin{enumerate}[label=(\alph*)]
\item Determine values for the mean and the variance of:
\begin{enumerate}[label=(\roman*)]
\item $M = U + V$;
\item $D = W - 2 U$;
\item $T = M + W + X$, given that $\rho _ { M W } = \rho _ { M X } = 0$.
\end{enumerate}\item Assuming that the variables $M , D$ and $T$ are normally distributed, determine the probability that, on a particular morning:
\begin{enumerate}[label=(\roman*)]
\item Alyssa's journey time from leaving home to leaving the village railway station is exactly 30 minutes;
\item Alyssa's train journey time is more than twice her car journey time;
\item Alyssa's total journey time is between 50 minutes and 70 minutes.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA S3 2012 Q6 [17]}}
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