(a)     A random variable, X , has probability density function defined by

\[f(x) = \left\{ {\begin{array}{*{20}{l}}
  {100,}&{{\text{for }} - 0.005 \leqslant x < 0.005} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

Determine E(X) and Var(X) .

(b)     When a real number is rounded to two decimal places, an error is made.

Show that this error can be modelled by the random variable X .

(c)     A list contains 20 real numbers, each of which has been given to two decimal places. The numbers are then added together.

(i)     Write down bounds for the resulting error in this sum.

(ii)     Using the central limit theorem, estimate to two decimal places the probability that the absolute value of the error exceeds 0.01.

(iii)     State clearly any assumptions you have made in your calculation.

(a)     f(x)is even (symmetrical about the origin)     (M1)

\({\text{E}}(X) = 0\)     A1

\({\text{Var}}(X) = {\text{E}}({X^2}) = \int_{ - 0.005}^{0.005} {100{x^2}{\text{d}}x} \)     (M1)(A1)

\( = 8.33 \times {10^{ - 6}}\left( {{\text{accept }}0.83 \times {{10}^{ - 5}}{\text{ or }}\frac{1}{{120\,000}}} \right)\)     A1

[5 marks]

(b)     rounding errors to 2 decimal places are uniformly distributed     R1

and lie within the interval \( - 0.005 \leqslant x < 0.005.\)     R1

this defines X     AG

[2 marks]

(c)     (i)     using the symbol y to denote the error in the sum of 20 real numbers each rounded to 2 decimal places

\( - 0.1 \leqslant y( = 20 \times x) < 0.1\)     A1

(ii)     \(Y \approx {\text{N}}(20 \times 0,{\text{ }}20 \times 8.3 \times {10^{ - 6}}) = {\text{N}}(0,{\text{ }}0.00016)\)     (M1)(A1)

\({\text{P}}\left( {\left| Y \right| > 0.01} \right) = 2\left( {1 - {\text{P}}(Y < 0.01)} \right)\)     (M1)(A1)

\( = 2\left( {1 - {\text{P}}\left( {Z < \frac{{0.01}}{{0.0129}}} \right)} \right)\)

\( = 0.44\) to 2 decimal places     A1     N4

(iii)     it is assumed that the errors in rounding the 20 numbers are independent     R1

and, by the central limit theorem, the sum of the errors can be modelled approximately by a normal distribution     R1

[8 marks]

Total [15 marks]

This was the only question on the paper with a conceptually ‘hard’ final part. Part(a) was generally well done, either by integration or by use of the standard formulae for a uniform distribution. Many candidates were not able to provide convincing reasoning in parts (b) and (c)(iii). Part(c)(ii), the application of the Central Limit Theorem was only very rarely tackled competently.