| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Probability distributions with parameters |
| Difficulty | Standard +0.3 This is a standard Further Statistics 1 question requiring systematic application of probability axioms and expectation formulas. Part (a) is trivial, (b) is routine calculation, (c) involves solving simultaneous equations from probability sum = 1 and the given E(X³) condition, and (d) requires algebraic manipulation of an inequality. While it has multiple parts and requires careful algebra, it follows predictable patterns with no novel insight needed—slightly easier than average A-level difficulty. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.03c Calculate mean/variance: by integration |
| \(x\) | - 3 | - 1 | 1 | 2 | 4 |
| \(\mathrm { P } ( X = x )\) | \(q\) | \(\frac { 7 } { 30 }\) | \(\frac { 7 } { 30 }\) | \(q\) | \(r\) |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \(q + \frac{r}{30}\) | B1 |
| (b) | \(\text{E}(X^2) = (-3)^2 \times q + (-1)^2 \times \frac{r}{30} + 1^2 \times \frac{r}{30} + 2^2 \times q + 4^2 \times r = \frac{7}{15} + 13q + 16r\) (*) | M1, A1*cso |
| (c) | \(\text{E}(X) = -3q + -\frac{r}{30} + \frac{r}{30} + 2q + 4r = \{= 4r - q\}\); \(\text{E}(X^2 + 6X) = \frac{7}{15} + 7q + 40r\); \(\text{E}(X^3) = (-3)^3 \times q + (-1)^3 \times \frac{r}{30} + 1^3 \times \frac{r}{30} + 2^3 \times q + 4^3 \times r = 64r - 19q\). Sum of probabilities = 1 gives: \(2q + r = \frac{16}{30}\) (o.e.); Solve: \(24r - 26q = \frac{7}{15}\) and \(r + 2q = \frac{8}{15}\) e.g. \(37r = \frac{111}{15}\). So \(r = \frac{1}{5}\) and \(q = \frac{1}{6}\) | dM1, M1, A1, M1, dM1, A1 |
| (d) | \(X^3 > X^2 + 6X \Rightarrow X(X-3)(X+2) > 0\); Use of sketch or table to see: \(-2 < X < 0\) or \(X > 3\). So \(P(X^3 > X^2 + 6X) = P(X = -1 \text{ or } 4) = \frac{7}{30} + "r" = \frac{13}{30}\) | M1, A1, M1, A1ft |
| (14 marks) |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $q + \frac{r}{30}$ | B1 | |
| (b) | $\text{E}(X^2) = (-3)^2 \times q + (-1)^2 \times \frac{r}{30} + 1^2 \times \frac{r}{30} + 2^2 \times q + 4^2 \times r = \frac{7}{15} + 13q + 16r$ (*) | M1, A1*cso | M1 for at least 3 correct terms of the expression for E(X²). A1*cso evidence of M1 scored with no incorrect working seen leading to correct answer (*). Allow $-3^2 \times q + -1^2 \times \frac{30}{30}$ etc if followed by 9q + ... but not if simply followed by given answer |
| (c) | $\text{E}(X) = -3q + -\frac{r}{30} + \frac{r}{30} + 2q + 4r = \{= 4r - q\}$; $\text{E}(X^2 + 6X) = \frac{7}{15} + 7q + 40r$; $\text{E}(X^3) = (-3)^3 \times q + (-1)^3 \times \frac{r}{30} + 1^3 \times \frac{r}{30} + 2^3 \times q + 4^3 \times r = 64r - 19q$. Sum of probabilities = 1 gives: $2q + r = \frac{16}{30}$ (o.e.); Solve: $24r - 26q = \frac{7}{15}$ and $r + 2q = \frac{8}{15}$ e.g. $37r = \frac{111}{15}$. So $r = \frac{1}{5}$ and $q = \frac{1}{6}$ | dM1, M1, A1, M1, dM1, A1 | 1st M1 for realising the need to find E(X) – a correct attempt with at least 3 correct terms seen. 1st A1 for the correct expression (needn't be simplified at this stage). 2nd M1 for a correct attempt at E(X³) with at least 3 correct terms seen. Treat no $-\frac{r}{30}$ terms as one correct term. 3rd M1 for using sum of probabilities = 1 to form an equation in q and r (needn't be simplified). Must be correct or clearly state that Σprobs = 1 being attempted with only one slip. 4th dM1 for solving their 2 linear equations in q and r (dep on 3rd M1 and 1st or 2nd M1). Must see correct method to reduce to a linear equation in one variable. 3rd A1 for $r = \frac{1}{5}$ and $q = \frac{1}{6}$ or any exact equivalents (dep on 2 correct equations seen) |
| (d) | $X^3 > X^2 + 6X \Rightarrow X(X-3)(X+2) > 0$; Use of sketch or table to see: $-2 < X < 0$ or $X > 3$. So $P(X^3 > X^2 + 6X) = P(X = -1 \text{ or } 4) = \frac{7}{30} + "r" = \frac{13}{30}$ | M1, A1, M1, A1ft | 1st M1 for 1st stage towards solving the inequality (factorising the cubic). 1st A1 for solving the inequality. 2nd M1 for identifying the values of X required i.e. –1 and 4. 2nd A1ft for $\frac{13}{30}$ or exact equivalent e.g. 0.43 (Allow ft of "their $r$" + $\frac{7}{30}$) |
| | | **(14 marks)** | |
**Notes for Question 4:**
**(b)** M1 for at least 3 correct terms of the expression for E(X²). A1*cso evidence of M1 scored with no incorrect working seen leading to correct answer (*). Allow $-3^2 \times q + -1^2 \times \frac{r}{30}$ etc if followed by 9q + ... but not if simply followed by given answer.
**(c)** 1st M1 for realising the need to find E(X) – a correct attempt with at least 3 correct terms seen. 1st A1 for the correct expression (needn't be simplified at this stage). 2nd M1 for a correct attempt at E(X³) with at least 3 correct terms seen. Treat no $-\frac{r}{30}$ terms as one correct term. 3rd M1 for using sum of probabilities = 1 to form an equation in q and r (needn't be simplified). Must be correct or clearly state that Σprobs = 1 being attempted with only one slip. 4th dM1 for solving their 2 linear equations in q and r (dep on 3rd M1 and 1st or 2nd M1). Must see correct method to reduce to a linear equation in one variable. 3rd A1 for $r = \frac{1}{5}$ and $q = \frac{1}{6}$ or any exact equivalents (dep on 2 correct equations seen).
**(d)** 1st M1 for 1st stage towards solving the inequality (factorising the cubic). 1st A1 for solving the inequality. 2nd M1 for identifying the values of X required i.e. –1 and 4. 2nd A1ft for $\frac{13}{30}$ or exact equivalent e.g. 0.43 (Allow ft of "their $r$" + $\frac{7}{30}$).
**ALT** Table 1st M1 for at least 4 correct values for $X^3$ and $X^2 + 6X$ (must be labelled). 1st A1 for all 10 correct values. [NB Can score M1A0M1A1ft in (d)].\begin{enumerate}
\item The discrete random variable $X$ has probability distribution
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 3 & - 1 & 1 & 2 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $q$ & $\frac { 7 } { 30 }$ & $\frac { 7 } { 30 }$ & $q$ & $r$ \\
\hline
\end{tabular}
\end{center}
where $q$ and $r$ are probabilities.\\
(a) Write down, in terms of $q , \mathrm { P } ( X \leqslant 0 )$\\
(b) Show that $\mathrm { E } \left( X ^ { 2 } \right) = \frac { 7 } { 15 } + 13 q + 16 r$
Given that $\mathrm { E } \left( X ^ { 3 } \right) = \mathrm { E } \left( X ^ { 2 } \right) + \mathrm { E } ( 6 X )$\\
(c) find the value of $q$ and the value of $r$\\
(d) Hence find $\mathrm { P } \left( X ^ { 3 } > X ^ { 2 } + 6 X \right)$
\hfill \mbox{\textit{Edexcel FS1 AS 2019 Q4 [14]}}