AQA S2 2007 June — Question 4 7 marks

Exam BoardAQA
ModuleS2 (Statistics 2)
Year2007
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeMeasurement error modeling
DifficultyModerate -0.8 This is a straightforward application of the continuous uniform distribution to a familiar context (rounding error). Part (a) requires explaining why the pdf has the given form (routine reasoning about rounding), part (b) is a direct probability calculation using the rectangular distribution, and part (c) involves recalling and applying standard formulas for mean and variance of a uniform distribution. All parts are textbook-standard with no novel problem-solving required.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration

4 Students are each asked to measure the distance between two points to the nearest tenth of a metre.
  1. Given that the rounding error, \(X\) metres, in these measurements has a rectangular distribution, explain why its probability density function is $$f ( x ) = \left\{ \begin{array} { c c } 10 & - 0.05 < x \leqslant 0.05 \\ 0 & \text { otherwise } \end{array} \right.$$
  2. Calculate \(\mathrm { P } ( - 0.01 < X < 0.02 )\).
  3. Find the mean and the standard deviation of \(X\).

Question 4(a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(x) = \begin{cases} \frac{1}{b-a} & a \leq x \leq b \\ 0 & \text{otherwise} \end{cases}\) with \((-0.05, 0.05)\)B1 explain error \(\pm 0.05\)
\(\frac{1}{b-a} = \frac{1}{0.05-(-0.05)} = \frac{1}{0.1} = 10\)M1, A1
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(-0.01 < X < 0.02) = 0.03 \times 10 = 0.3\)M1, A1
Question 4(c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Mean \(= 0\)B1 CAO
Standard deviation \(= 0.0289\)B1 \(\frac{1}{20\sqrt{3}}\) OE
# Question 4(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = \begin{cases} \frac{1}{b-a} & a \leq x \leq b \\ 0 & \text{otherwise} \end{cases}$ with $(-0.05, 0.05)$ | B1 | explain error $\pm 0.05$ |
| $\frac{1}{b-a} = \frac{1}{0.05-(-0.05)} = \frac{1}{0.1} = 10$ | M1, A1 | |

# Question 4(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(-0.01 < X < 0.02) = 0.03 \times 10 = 0.3$ | M1, A1 | |

# Question 4(c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean $= 0$ | B1 | CAO |
| Standard deviation $= 0.0289$ | B1 | $\frac{1}{20\sqrt{3}}$ OE |

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4 Students are each asked to measure the distance between two points to the nearest tenth of a metre.
\begin{enumerate}[label=(\alph*)]
\item Given that the rounding error, $X$ metres, in these measurements has a rectangular distribution, explain why its probability density function is

$$f ( x ) = \left\{ \begin{array} { c c } 
10 & - 0.05 < x \leqslant 0.05 \\
0 & \text { otherwise }
\end{array} \right.$$
\item Calculate $\mathrm { P } ( - 0.01 < X < 0.02 )$.
\item Find the mean and the standard deviation of $X$.
\end{enumerate}

\hfill \mbox{\textit{AQA S2 2007 Q4 [7]}}