(i)     Determine an expression for \(F(x)\), valid for \(1 \leqslant x \leqslant 2\), where F denotes the cumulative distribution function of X.

(ii)     Hence, or otherwise, determine the median of X.

(i)     State the central limit theorem.

(ii)     A random sample of 150 observations is taken from the distribution of X and \(\bar X\) denotes the sample mean. Use the central limit theorem to find, approximately, the probability that \(\bar X\) is greater than 1.6.

(i)     \(F(x) = \int_1^x {\frac{{3{u^2} + 2u}}{{10}}{\text{d}}u} \)     (M1)

\( = \left[ {\frac{{{u^3} + {u^2}}}{{10}}} \right]_1^x\)     A1

Note: Do not penalise missing or wrong limits at this stage.

Accept the use of x in the integrand.

\( = \frac{{{x^3} + {x^2} - 2}}{{10}}\)     A1

(ii)     the median m satisfies the equation \(F(m) = \frac{1}{2}\) so     (M1)

\({m^3} + {m^2} - 7 = 0\)     (A1)

Note: Do not FT from an incorrect \(F(x)\).

\(m = 1.63\)     A1

Note: Accept any answer that rounds to 1.6.

[6 marks]

(i)     the mean of a large sample from any distribution is approximately 

normal     A1

Note: This is the minimum acceptable explanation.

(ii)     we require the mean \(\mu \) and variance \({\sigma ^2}\) of X

\(\mu  = \int_1^2 {\left( {\frac{{3{x^3} + 2{x^2}}}{{10}}} \right){\text{d}}x} \)     (M1)

\( = \frac{{191}}{{120}}{\text{ }}(1.591666 \ldots )\)     A1

\({\sigma ^2} = \int_1^2 {\left( {\frac{{3{x^4} + 2{x^3}}}{{10}}} \right){\text{d}}x - {\mu ^2}} \)     (M1)

\( = 0.07659722 \ldots \)     A1

the central limit theorem states that

\(\bar X \approx N\left( {\mu ,\frac{{{\sigma ^2}}}{n}} \right),\) i.e. \(N(1.591666 \ldots ,{\text{ }}0.0005106481 \ldots )\)     M1A1

\({\text{P}}(\bar X > 1.6) = 0.356\)     A1

Note: Accept any answer that rounds to 0.36.

[8 marks]

Solutions to (a)(i) were disappointing in general, suggesting that many candidates are unfamiliar with the concept of the cumulative distribution function. Many candidates knew that it was something to do with the integral of the probability density function but some thought it was \(\int\limits_1^2 {f(x){\text{d}}x} \) which they then evaluated as \(1\) while others thought it was just \(\int {f(x){\text{d}}x}  = \frac{{\left( {{x^2} + {x^3}} \right)}}{{10}}\) which is not, in general, a valid method. However, most candidates solved (a)(ii) correctly, usually by integrating the probability density function from \(1\) to \(m\).

In (b)(i), the statement of the central limit theorem was often quite dreadful. The term ‘sample mean’ was often not mentioned and a common misconception appears to be that the actual distribution rather than the sample mean tends to normality as the sample size increases. Solutions to (b)(ii) often failed to go beyond finding the mean and variance of \(X\) . In calculating the variance, some candidates rounded the mean from \(1.5916666..\) to \(1.59\) which resulted in an incorrect value for the variance. It is important to note that calculating a variance usually involves a small difference of two large numbers so that full accuracy must be maintained.