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]