Edexcel S1 2014 June — Question 5 10 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2014
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypePiecewise or conditional probability function
DifficultyModerate -0.8 This is a routine S1 probability distribution question requiring standard techniques: summing probabilities to find k, calculating cumulative distribution F(x), and computing E(X), E(X²), and Var(aX+b). All parts follow textbook procedures with straightforward arithmetic—no problem-solving insight needed, just methodical application of formulas.
Spec2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance

5. The discrete random variable \(X\) has the probability function $$\mathrm { P } ( X = x ) = \begin{cases} k x & x = 2,4,6 \\ k ( x - 2 ) & x = 8 \\ 0 & \text { otherwise } \end{cases}$$ where \(k\) is a constant.
  1. Show that \(k = \frac { 1 } { 18 }\)
  2. Find the exact value of \(\mathrm { F } ( 5 )\).
  3. Find the exact value of \(\mathrm { E } ( X )\).
  4. Find the exact value of \(\mathrm { E } \left( X ^ { 2 } \right)\).
  5. Calculate \(\operatorname { Var } ( 3 - 4 X )\) giving your answer to 3 significant figures.

Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
(a) \(2k + 4k + 6k + k(8-2) = 1\) giving \(k = \frac{1}{18}\) (*)M1 A1cso (2) M1 for \(2k+4k+6k+k(8-2)=1\). A1 for \(k=\frac{1}{18}\). NB cso so no incorrect working seen
(b) \([2k+4k] = \frac{6}{18} = \frac{1}{3}\)B1 (1) \(\frac{1}{3}\) or any exact numerical equivalent
(c) \(E(X) = \left(2\times\frac{1}{9}\right)+\left(4\times\frac{2}{9}\right)+\left(6\times\frac{1}{3}\right)+\left(8\times\frac{1}{3}\right) = 5\frac{7}{9}\) (or exact equiv. e.g. \(\frac{52}{9}\))M1 A1 (2) M1 for an expression for \(E(X)\) with at least 3 correct terms (products), allow use of \(k\) e.g. \(104k\)
(d) \(E(X^2) = \left(4\times\frac{1}{9}\right)+\left(16\times\frac{2}{9}\right)+\left(36\times\frac{1}{3}\right)+\left(64\times\frac{1}{3}\right) = 37\frac{1}{3}\) (or exact equiv. e.g. \(\frac{112}{3}\))M1 A1 (2) M1 for an expression for \(E(X^2)\) with at least 3 correct terms. A1 for any exact equivalent only. E.g. 37.3 is A0 but \(37.\dot{3}\) is OK
(e) \(\text{Var}(X) = 37\frac{1}{3} - \left(5\frac{7}{9}\right)^2\) \([= 3.95\ldots\) or \(\frac{320}{81}]\) \(\text{Var}(3-4X) = 16 \times 3.95\ldots =\) awrt 63.2 (allow \(\frac{5120}{81}\))M1 M1 A1 (3) 1st M1 for \(E(X^2)-[E(X)]^2\) ft their (c) and (d). 2nd M1 for statement "\(4^2 \times \text{Var}(X)\)" seen or \(4^2\times\) their \(\text{Var}(X)\) provided their \(\text{Var}(X)>0\). Do not allow \(16\times E(X^2)\). NB condone \(-4^2\times\text{Var}(X)\) if answer later becomes positive. A1 for exact fraction \(\frac{5120}{81}\) o.e. or decimal approximation awrt 63.2
# Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $2k + 4k + 6k + k(8-2) = 1$ giving $k = \frac{1}{18}$ (*) | M1 A1cso **(2)** | M1 for $2k+4k+6k+k(8-2)=1$. A1 for $k=\frac{1}{18}$. NB cso so no incorrect working seen |
| **(b)** $[2k+4k] = \frac{6}{18} = \frac{1}{3}$ | B1 **(1)** | $\frac{1}{3}$ or any exact numerical equivalent |
| **(c)** $E(X) = \left(2\times\frac{1}{9}\right)+\left(4\times\frac{2}{9}\right)+\left(6\times\frac{1}{3}\right)+\left(8\times\frac{1}{3}\right) = 5\frac{7}{9}$ (or exact equiv. e.g. $\frac{52}{9}$) | M1 A1 **(2)** | M1 for an expression for $E(X)$ with at least 3 correct terms (products), allow use of $k$ e.g. $104k$ |
| **(d)** $E(X^2) = \left(4\times\frac{1}{9}\right)+\left(16\times\frac{2}{9}\right)+\left(36\times\frac{1}{3}\right)+\left(64\times\frac{1}{3}\right) = 37\frac{1}{3}$ (or exact equiv. e.g. $\frac{112}{3}$) | M1 A1 **(2)** | M1 for an expression for $E(X^2)$ with at least 3 correct terms. A1 for any exact equivalent only. E.g. 37.3 is A0 but $37.\dot{3}$ is OK |
| **(e)** $\text{Var}(X) = 37\frac{1}{3} - \left(5\frac{7}{9}\right)^2$ $[= 3.95\ldots$ or $\frac{320}{81}]$ $\text{Var}(3-4X) = 16 \times 3.95\ldots =$ awrt **63.2** (allow $\frac{5120}{81}$) | M1 M1 A1 **(3)** | 1st M1 for $E(X^2)-[E(X)]^2$ ft their (c) and (d). 2nd M1 for statement "$4^2 \times \text{Var}(X)$" seen or $4^2\times$ their $\text{Var}(X)$ provided their $\text{Var}(X)>0$. Do not allow $16\times E(X^2)$. NB condone $-4^2\times\text{Var}(X)$ if answer later becomes positive. A1 for exact fraction $\frac{5120}{81}$ o.e. or decimal approximation awrt 63.2 |

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5. The discrete random variable $X$ has the probability function

$$\mathrm { P } ( X = x ) = \begin{cases} k x & x = 2,4,6 \\ k ( x - 2 ) & x = 8 \\ 0 & \text { otherwise } \end{cases}$$

where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 18 }$
\item Find the exact value of $\mathrm { F } ( 5 )$.
\item Find the exact value of $\mathrm { E } ( X )$.
\item Find the exact value of $\mathrm { E } \left( X ^ { 2 } \right)$.
\item Calculate $\operatorname { Var } ( 3 - 4 X )$ giving your answer to 3 significant figures.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2014 Q5 [10]}}