| Date | May 2016 | Marks available | 3 | Reference code | 16M.2.hl.TZ0.2 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Show that | Question number | 2 | Adapted from | N/A |
Question
The random variable \(X\) has probability density function given by
\[f(x) = \left\{ {\begin{array}{*{20}{l}}
{x{{\text{e}}^{ - x}},}&{{\text{for }}x \geqslant 0,} \\
{0,}&{{\text{otherwise}}}
\end{array}} \right..\]
A sample of size 50 is taken from the distribution of \(X\).
Use l’Hôpital’s rule to show that \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{{\text{e}}^x}}} = 0\).
(i) Find \({\text{E}}({X^2})\).
(ii) Show that \({\text{Var}}(X) = 2\).
State the central limit theorem.
Find the probability that the sample mean is less than 2.3.
Markscheme
attempt to apply l’Hôpital’s rule M1
\(\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{{{\text{e}}^x}}}\) A1
then \(\mathop {\lim }\limits_{x \to \infty } \frac{{6x}}{{{{\text{e}}^x}}}\)
then \(\mathop {\lim }\limits_{x \to \infty } \frac{6}{{{{\text{e}}^x}}}\) A1
\( = 0\) AG
[3 marks]
(i) \({\text{E}}({X^2}) = \mathop {\lim }\limits_{R \to \infty } \int\limits_0^R {{x^3}{{\text{e}}^{ - x}}{\text{d}}x} \) M1
attempt at integration by parts M1
the integral \( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + \mathop \smallint \limits_0^R 3{x^2}{{\text{e}}^{ - x}}{\text{d}}x\) A1A1
\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + \int\limits_0^R {6x{{\text{e}}^{ - x}}{\text{d}}x} \) M1
\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + [ - 6x{{\text{e}}^{ - x}}]_0^R + \int\limits_0^R {6{{\text{e}}^{ - x}}{\text{d}}x} \) A1
\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + [ - 6x{{\text{e}}^{ - x}}]_0^R + [ - 6{{\text{e}}^{ - x}}]_0^R\) A1
\( = 6\) when \(R \to \infty \) R1
(ii) \({\text{E}}(X) = 2\) A1
\({\text{Var}}(X) = {\text{E}}({X^2}) - {\left( {{\text{E}}(X)} \right)^2} = 6 - {2^2}\) M1
\( = 2\) AG
[10 marks]
if a random sample of size \(n\) is taken from any distribution \(X\), with \({\text{E}}(X) = \mu \) and \({\text{Var}}(X) = {\sigma ^2}\), then, for large n, A1
the sample mean \(\bar X\) has approximate distribution \({\text{N}}\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}} \right)\) A1
[2 marks]
\(\bar X \sim {\text{N}}\left( {2,{\text{ }}\frac{2}{{50}} = {{(0.2)}^2}} \right)\) (A1)
\({\text{P}}(\bar X < 2.3) = \left( {{\text{P}}(Z < 1.5)} \right) = 0.933\) A1
[2 marks]
Examiners report
In part (b) the infinite upper limit was rarely treated rigorously.
In answering part (c) many failed to say that the Central Limit Theorem is valid for large samples and for any initial distribution. The parameters of the distribution were often not stated.
