Determine $P(0.25 \leqslant X \leqslant 0.75)$;

Determine the median of $X$.

Show that the probability density function $f$ of $X$ is given, for $0 \leqslant x \leqslant 1$, by

$$f(x) = (x + 1){{\text{e}}^{x - 1}}.$$

Hence determine the mean and the variance of $X$.

State the central limit theorem. 

A random sample of 100 observations is obtained from the distribution of $X$. If $\bar X$ denotes the sample mean, use the central limit theorem to find an approximate value of $P(\bar X > 0.65)$. Give your answer correct to two decimal places.

$P(0.25 \leqslant X \leqslant 0.75) = F(0.75) - F(0.25)$     (M1)

$= 0.466$     A1

Note:     Accept any answer that rounds correctly to 0.466.

[2 marks]

the median $m$ satisfies $F(m) = 0.5$     (M1)

$m = 0.685$     A1

Note:     Accept any answer that rounds correctly to 0.685.

[2 marks]

$f(x) = F’(x)$     (M1)

$= {{\text{e}}^{x - 1}} + x{{\text{e}}^{x - 1}}$     A1

$= (x + 1){{\text{e}}^{x - 1}}$     AG

[2 marks]

$\mu  = \int\limits_0^1 {x\left( {x + 1} \right){{\text{e}}^{x - 1}}{\text{d}}x}$    (M1)

$= 0.632\,\,\,\left( {1 - \frac{1}{{\text{e}}}} \right)$     A1

Note:     Accept any answer that rounds correctly to 0.632.

${\sigma ^2} = \int\limits_0^1 {x\left( {x + 1} \right){{\text{e}}^{x - 1}}{\text{d}}x}  - 0.632{ \ldots ^2}$     (M1)

$= 0.0719\,\,\,\left( {\frac{6}{{\text{e}}} - 2 - \frac{1}{{{{\text{e}}^2}}}} \right)$     A1

Note:     Accept any answer that rounds correctly to 0.072.

[4 marks]

the central limit theorem states that the mean of a large sample from any distribution (with a finite variance) is approximately normally distributed     A1

[1 mark]

$\bar X$ is approximately $N(0.632 \ldots ,{\text{ }}0.000719 \ldots )$     (M1)(A1)

$P(\bar X > 0.65) = 0.25$ (2 dps required)     A1

[3 marks]