| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Explain why not valid PDF |
| Difficulty | Standard +0.3 This is a standard S2 question testing routine PDF properties: sketching, checking validity, finding mode/median/constant k. Part (a) requires expanding and integrating to show ∫f(x)dx ≠ 1. Parts (b)-(e) are textbook exercises in differentiation and integration with no novel problem-solving. Slightly easier than average due to straightforward algebraic manipulation. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | [Graph showing decreasing quadratic] | B1 (shape) B1 (domain) B1 (3) |
| (b) | \([g(y) = k(12y - y^3)]\) \([g'(y)] = k(12 - 3y^2)\) \(12 - 3y^2 = 0\) \(y = 2\) | M1 M1 A1 (3) |
| (c) | \(\int k(12y - y^3)dy = k\left[6y^2 - \frac{y^4}{4}\right]_1\) \(k\left(6 \times 3^2 - \frac{3^4}{4} - 6 + \frac{1}{4}\right) = 1\) \(k = \frac{1}{28}\) | M1 M1 A1 (3) |
| (d) | \(\int_1^{28}(12y - y^3)dy = \frac{1}{28}\left(6m^2 - \frac{m^4}{4} - 6 \times 1^2 - \frac{1^4}{4}\right) = 0.5\) \(m^4 - 24m^2 + 79 = 0 \to m = 1.98437\ldots\) awrt \(\mathbf{1.98}\) | M1 M1 A1 (3) |
| (e) | Median \(\approx\) Mode, therefore there is no skew. | M1 A1 ft (2) Total 14 |
(a) | [Graph showing decreasing quadratic] | B1 (shape) B1 (domain) B1 (3) | 1st B1 for correct shape (decreasing quadratic) for this mark ignore graph $x < 1$ and $x > 4$ must cross x axis / 2nd B1 for correct domain (graph starting at $x = 1$ and finishing at $x = 4$). If $x = 3$ is labelled, then it must be correctly placed at the $x$-intercept / 3rd B1 for Not a density function / Albert is incorrect and correct supporting reason. Allow any reference to pdf cannot be negative.
(b) | $[g(y) = k(12y - y^3)]$ $[g'(y)] = k(12 - 3y^2)$ $12 - 3y^2 = 0$ $y = 2$ | M1 M1 A1 (3) | 1st M1 for expanding and attempting to differentiate (or using product rule to differentiate) / 2nd M1 for equating to 0 and attempt to solve / A1 for 2 only
(c) | $\int k(12y - y^3)dy = k\left[6y^2 - \frac{y^4}{4}\right]_1$ $k\left(6 \times 3^2 - \frac{3^4}{4} - 6 + \frac{1}{4}\right) = 1$ $k = \frac{1}{28}$ | M1 M1 A1 (3) | 1st M1 for attempt to integrate ($y^n \to y^{n+1}$) / 2nd M1 for equating to 1 and use of correct limits
(d) | $\int_1^{28}(12y - y^3)dy = \frac{1}{28}\left(6m^2 - \frac{m^4}{4} - 6 \times 1^2 - \frac{1^4}{4}\right) = 0.5$ $m^4 - 24m^2 + 79 = 0 \to m = 1.98437\ldots$ awrt $\mathbf{1.98}$ | M1 M1 A1 (3) | 1st M1 for use of $\int g(y)dy = 0.5$ / 2nd M1 for arranging to a 3TQ = 0 and attempt to solve / A1 awrt 1.98 only
(e) | Median $\approx$ Mode, therefore there is no skew. | M1 A1 ft (2) Total 14 | M1 for a comparison of 1 $\leq$ 'their (b)' $\leq$ 3 with 1 $\leq$ 'their (d)' $\leq$ 3. Note: mean = 1.9857... / A1 ft for no skew (allow [slight] negative skew). Correct statement following from their comparison. Note: mean = 1.9857... comparisons based on the mean must use correct mean awrt 1.99
---3. The function $\mathrm { f } ( x )$ is defined as
$$f ( x ) = \begin{cases} \frac { 1 } { 9 } ( x + 5 ) ( 3 - x ) & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
Albert believes that $\mathrm { f } ( x )$ is a valid probability density function.
\begin{enumerate}[label=(\alph*)]
\item Sketch $\mathrm { f } ( x )$ and comment on Albert's belief.
The continuous random variable $Y$ has probability density function given by
$$g ( y ) = \begin{cases} k y \left( 12 - y ^ { 2 } \right) & 1 \leqslant y \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a positive constant.
\item Use calculus to find the mode of $Y$
\item Use algebraic integration to find the value of $k$
\item Find the median of $Y$ giving your answer to 3 significant figures.
\item Describe the skewness of the distribution of $Y$ giving a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q3 [14]}}