| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from given distribution |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing standard expectation and variance calculations with discrete distributions. Parts (a)-(d) are direct formula applications, (e)-(g) require careful enumeration of outcomes but follow standard methods, and (h) is a simple comparison of expectations. The multi-part structure adds length but not conceptual difficulty—all techniques are routine for S1 students. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 0 | 3 | 6 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 12 }\) | \(\frac { 2 } { 3 }\) | \(\frac { 1 } { 4 }\) |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \([E(X)] = [0 × \frac{1}{12} + 3 × \frac{2}{3} + 6 × \frac{1}{4}] = \frac{2}{3}\) or \(3.5\) | M1, A1 |
| (b) | \([E(X^2)] = [0^2 × \frac{1}{12} + 3^2 × \frac{2}{3} + 6^2 × \frac{1}{4}\) (\(= 15\)) | M1 |
| \([\text{Var}(X)] = "\)15"\( - \)(\frac{2}{3})^2\( | M1 | 2nd M1 for their \)E(X^2) -\( their \)E(X)^2$ |
| \(= \frac{11}{4}\) or \(2.75\) | A1 | |
| (c) | \(5p + 2(1 - p) = 3\) or \([\) allow \(p + q = 1\) and \(5p + 2q = 3\) for M1] | M1A1 |
| So \(p = \frac{1}{3}\) (\(*\)) | A1cso | |
| (d) | \(P(Y = 2) = \frac{2}{3}\) and \(P(Y = 5) = \frac{1}{3}\) | B1 |
| (e) | \(P(S = 30) = P(X = 6\) and \(Y = 5)\) | M1 |
| \(= \frac{1}{4} × \frac{1}{3} = \frac{1}{12}\) | A1cso | Dep on M1 scored for with no incorrect working seen e.g. \(30 = \frac{1}{4} × \frac{1}{3}\) is A0 |
| (f) | M1A1A1 | |
| \([s]\) | 4 | 6 |
| \([P(S = s)]\) | \(\frac{2}{36}\) | \(\frac{16}{36}\) |
| (g) | \(E(S) = \frac{1}{36}[4 × 2 + 6 × 16 + 12 × 6 + 15 × 8 + 25 × 1 + 30 × 3]\) | M1 |
| \(= 11\frac{2}{12}\) or \(\frac{137}{12}\) or \(11.416\) | A1 | For \(11\frac{2}{12}\) or \(\frac{137}{12}\) or any exact equivalent. (Correct ans. only 2/2, awrt 11.4 only M1A0) |
| (h) | \(E(X^2) = 15\) and \(E(S) = 11.416\ldots\) or | B1ft |
| \(E(X^2) > E(S)\) | dB1ft | 2nd d B1 dependent on a correct comparison of their values for choosing correct player. |
| \(\ldots\) so Charlotte has the higher total score |
(a) | $[E(X)] = [0 × \frac{1}{12} + 3 × \frac{2}{3} + 6 × \frac{1}{4}] = \frac{2}{3}$ or $3.5$ | M1, A1 | M1 for a fully correct expression (allow missing 0 term). Correct ans only is 2/2 |
(b) | $[E(X^2)] = [0^2 × \frac{1}{12} + 3^2 × \frac{2}{3} + 6^2 × \frac{1}{4}$ ($= 15$) | M1 | 1st M1 for a fully correct expression (allow missing 0 term) for $E(X^2)$. Allow Var(X) label |
| $[\text{Var}(X)] = "$15"$ - $(\frac{2}{3})^2$ | M1 | 2nd M1 for their $E(X^2) -$ their $E(X)^2$ |
| $= \frac{11}{4}$ or $2.75$ | A1 | |
(c) | $5p + 2(1 - p) = 3$ or $[$ allow $p + q = 1$ and $5p + 2q = 3$ for M1] | M1A1 | 1st M1 for attempting a linear eq'n in $p$(or $x$ etc). Must see $= 3$ and have 2 terms in $p$, 1 correct. 1st A1 for a fully correct equation for $p$ or for solving their eqns leading to correct eqn in $p$. 2nd A1 for $p = \frac{1}{3}$ with M1 scored and no incorrect working seen. |
| So $p = \frac{1}{3}$ ($*$) | A1cso | |
(d) | $P(Y = 2) = \frac{2}{3}$ and $P(Y = 5) = \frac{1}{3}$ | B1 | For correct values for $P(Y = 2)$ and $P(Y = 5)$. Needn't be in formal table but labelled. |
(e) | $P(S = 30) = P(X = 6$ and $Y = 5)$ | M1 | For $6 × 5 = 30$ or $P(30) = P(6,5)$ or $P(30) = P(6) × P(5)$ or $S = (XY) = 6 × 5$ and $X =$ 6 and $Y = 5$ |
| $= \frac{1}{4} × \frac{1}{3} = \frac{1}{12}$ | A1cso | Dep on M1 scored for with no incorrect working seen e.g. $30 = \frac{1}{4} × \frac{1}{3}$ is A0 |
(f) | | M1A1A1 | 1st M1 for an attempt at prob. distribution with at least 3 correct ($s$ and $P(S = s)$)Exc' $s = 30$. 1st A1 for 6 correct $s$ values. 2nd A1 for a fully correct prob. distribution including $s = 30$ |
| $[s]$ | 4 | 6 | 12 | 15 | 25 | (30) | |
| $[P(S = s)]$ | $\frac{2}{36}$ | $\frac{16}{36}$ | $\frac{6}{36}$ | $\frac{8}{36}$ | $\frac{1}{36}$ | $(\frac{1}{12})$ | |
(g) | $E(S) = \frac{1}{36}[4 × 2 + 6 × 16 + 12 × 6 + 15 × 8 + 25 × 1 + 30 × 3]$ | M1 | For attempting $E(S)$ using their values. Must see $\ldots 3$ products (correct ft) decimals to 3sf |
| $= 11\frac{2}{12}$ or $\frac{137}{12}$ or $11.416$ | A1 | For $11\frac{2}{12}$ or $\frac{137}{12}$ or any exact equivalent. (Correct ans. only 2/2, awrt 11.4 only M1A0) |
(h) | $E(X^2) = 15$ and $E(S) = 11.416\ldots$ or | B1ft | 1st B1 for correct comparison of their $E(S)$ and $E(X^2)$ labelled in (b) or (h) [expressions or values] |
| $E(X^2) > E(S)$ | dB1ft | 2nd d B1 dependent on a correct comparison of their values for choosing correct player. |
| $\ldots$ so Charlotte has the higher total score | | |6. The score, $X$, for a biased spinner is given by the probability distribution
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & 0 & 3 & 6 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 12 }$ & $\frac { 2 } { 3 }$ & $\frac { 1 } { 4 }$ \\
\hline
\end{tabular}
\end{center}
Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
A biased coin has one face labelled 2 and the other face labelled 5 The score, $Y$, when the coin is spun has
$$\mathrm { P } ( Y = 5 ) = p \quad \text { and } \quad \mathrm { E } ( Y ) = 3$$
\item Form a linear equation in $p$ and show that $p = \frac { 1 } { 3 }$
\item Write down the probability distribution of $Y$.
Sam plays a game with the spinner and the coin.\\
Each is spun once and Sam calculates his score, $S$, as follows
$$\begin{aligned}
& \text { if } X = 0 \text { then } S = Y ^ { 2 } \\
& \text { if } X \neq 0 \text { then } S = X Y
\end{aligned}$$
\item Show that $\mathrm { P } ( S = 30 ) = \frac { 1 } { 12 }$
\item Find the probability distribution of $S$.
\item Find $\mathrm { E } ( S )$.
Charlotte also plays the game with the spinner and the coin.\\
Each is spun once and Charlotte ignores the score on the coin and just uses $X ^ { 2 }$ as her score. Sam and Charlotte each play the game a large number of times.
\item State, giving a reason, which of Sam and Charlotte should achieve the higher total score.
\begin{center}
\begin{tabular}{|l|l|}
\hline
\hline
END & \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2017 Q6 [18]}}