Question 1 - A-Level Maths

CAIE S1 2023 November — Question 1 6 marks

Exam Board	CAIE
Module	S1 (Statistics 1)
Year	2023
Session	November
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Discrete Probability Distributions
Type	Simple algebraic expression for P(X=x)
Difficulty	Moderate -0.8 This is a straightforward probability distribution question requiring only basic techniques: finding k using ΣP(X=x)=1, constructing a table, then applying standard E(X) and Var(X) formulas with simple arithmetic. No problem-solving insight needed, just routine application of S1 definitions.
Spec	2.04a Discrete probability distributions 5.02b Expectation and variance: discrete random variables

1 Becky sometimes works in an office and sometimes works at home. The random variable $X$ denotes the number of days that she works at home in any given week. It is given that $$\mathrm { P } ( X = x ) = k x ( x + 1 )$$ where $k$ is a constant and $x = 1,2,3$ or 4 only.

Draw up the probability distribution table for $X$, giving the probabilities as numerical fractions.
Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.

Show mark scheme Show mark scheme source

Question 1(a):

Answer	Marks	Guidance
$2k + 6k + 12k + 20k = 1$, $\left[k = \frac{1}{40}\right]$	M1	Using sum of probabilities $= 1$ to form an equation in $k$. Accept $1\times2\times k + 2\times3\times k + 3\times4\times k + 4\times5\times k = 1$
$X$	$1$	$2$
$P(X)$	$\frac{2}{40}$	$\frac{6}{40}$
$0.05$	$0.15$	$0.3$
M1	Two correctly linked, accurate probabilities. May be in terms of $k$. May not be in a table.
A1	Table with correct $X$ values and correct probabilities.

Total: 3 marks

Question 1(b):

Answer	Marks	Guidance
$[\text{E}(X) =]\ \frac{1\times2 + 2\times6 + 3\times12 + 4\times20}{40}$, $\frac{2+12+36+80}{40}$	M1	$[\text{E}(X) = 1\times2k + 2\times6k + 3\times12k + 4\times20k = 130k]$. Accept unsimplified expression. May be calculated in variance. FT their table with 3 or more probabilities summing to $1\ (0 < p < 1)$. If there are outcomes in the table without probabilities, condone and treat as $p = 0$.
$\left[\text{Var}(X) = \frac{1^2\times2 + 2^2\times6 + 3^2\times12 + 4^2\times20}{40} - \left(\textit{their}\ \text{E}(X)\right)^2\right]$	M1	$[\text{Var}(X) = 1^2\times2k + 2^2\times6k + 3^2\times12k + 4^2\times20k - (130k)^2]$. Appropriate variance formula using their $(\text{E}(X))^2$ value. FT their table with 3 or more probabilities $(0 < p < 1)$ which need not sum to 1. Note: if table is correct, $\frac{454}{40}$ $\left(\text{or}\ \frac{227}{20}\ \text{or any calculation}\right) - (\textit{their}\ \text{E}(X))^2$ implies M1.
$\text{E}(X) = \frac{13}{4},\ 3\frac{1}{4},\ 3.05\quad \text{Var}(X) = \frac{63}{80},\ 0.7875$	A1	Answers for $\text{E}(X)$ and $\text{Var}(X)$ must be identified. $\text{E}(X)$ may be identified by correct use in variance. Condone E, V, $\mu$, $\sigma$ etc. If A0 earned, SC B1 for identified correct final solutions.

Total: 3 marks

## Question 1(a):

$2k + 6k + 12k + 20k = 1$, $\left[k = \frac{1}{40}\right]$ | **M1** | Using sum of probabilities $= 1$ to form an equation in $k$. Accept $1\times2\times k + 2\times3\times k + 3\times4\times k + 4\times5\times k = 1$

| $X$ | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|
| $P(X)$ | $\frac{2}{40}$ | $\frac{6}{40}$ | $\frac{12}{40}$ | $\frac{20}{40}$ |
| | $0.05$ | $0.15$ | $0.3$ | $0.5$ |

| **M1** | Two correctly linked, accurate probabilities. May be in terms of $k$. May not be in a table.

| **A1** | Table with correct $X$ values and correct probabilities.

**Total: 3 marks**

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## Question 1(b):

$[\text{E}(X) =]\ \frac{1\times2 + 2\times6 + 3\times12 + 4\times20}{40}$, $\frac{2+12+36+80}{40}$ | **M1** | $[\text{E}(X) = 1\times2k + 2\times6k + 3\times12k + 4\times20k = 130k]$. Accept unsimplified expression. May be calculated in variance. FT *their* table with 3 or more probabilities summing to $1\ (0 < p < 1)$. If there are outcomes in the table without probabilities, condone and treat as $p = 0$.

$\left[\text{Var}(X) = \frac{1^2\times2 + 2^2\times6 + 3^2\times12 + 4^2\times20}{40} - \left(\textit{their}\ \text{E}(X)\right)^2\right]$ | **M1** | $[\text{Var}(X) = 1^2\times2k + 2^2\times6k + 3^2\times12k + 4^2\times20k - (130k)^2]$. Appropriate variance formula using *their* $(\text{E}(X))^2$ value. FT *their* table with 3 or more probabilities $(0 < p < 1)$ which need not sum to 1. Note: if table is correct, $\frac{454}{40}$ $\left(\text{or}\ \frac{227}{20}\ \text{or any calculation}\right) - (\textit{their}\ \text{E}(X))^2$ implies M1.

$\text{E}(X) = \frac{13}{4},\ 3\frac{1}{4},\ 3.05\quad \text{Var}(X) = \frac{63}{80},\ 0.7875$ | **A1** | Answers for $\text{E}(X)$ and $\text{Var}(X)$ must be identified. $\text{E}(X)$ may be identified by correct use in variance. Condone E, V, $\mu$, $\sigma$ etc. If A0 earned, **SC B1** for identified correct final solutions.

**Total: 3 marks**

Show LaTeX source

1 Becky sometimes works in an office and sometimes works at home. The random variable $X$ denotes the number of days that she works at home in any given week. It is given that

$$\mathrm { P } ( X = x ) = k x ( x + 1 )$$

where $k$ is a constant and $x = 1,2,3$ or 4 only.
\begin{enumerate}[label=(\alph*)]
\item Draw up the probability distribution table for $X$, giving the probabilities as numerical fractions.
\item Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2023 Q1 [6]}}

This paper (6 questions)

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Q1 6 Q2 6 Q3 7 Q4 10 Q5 11 Q6 10

Answer	Marks	Guidance
\(2k + 6k + 12k + 20k = 1\), \(\left[k = \frac{1}{40}\right]\)	M1	Using sum of probabilities \(= 1\) to form an equation in \(k\). Accept \(1\times2\times k + 2\times3\times k + 3\times4\times k + 4\times5\times k = 1\)
\(X\)	\(1\)	\(2\)
\(P(X)\)	\(\frac{2}{40}\)	\(\frac{6}{40}\)
\(0.05\)	\(0.15\)	\(0.3\)
M1	Two correctly linked, accurate probabilities. May be in terms of \(k\). May not be in a table.
A1	Table with correct \(X\) values and correct probabilities.

Answer	Marks	Guidance
\([\text{E}(X) =]\ \frac{1\times2 + 2\times6 + 3\times12 + 4\times20}{40}\), \(\frac{2+12+36+80}{40}\)	M1	\([\text{E}(X) = 1\times2k + 2\times6k + 3\times12k + 4\times20k = 130k]\). Accept unsimplified expression. May be calculated in variance. FT their table with 3 or more probabilities summing to \(1\ (0 < p < 1)\). If there are outcomes in the table without probabilities, condone and treat as \(p = 0\).
\(\left[\text{Var}(X) = \frac{1^2\times2 + 2^2\times6 + 3^2\times12 + 4^2\times20}{40} - \left(\textit{their}\ \text{E}(X)\right)^2\right]\)	M1	\([\text{Var}(X) = 1^2\times2k + 2^2\times6k + 3^2\times12k + 4^2\times20k - (130k)^2]\). Appropriate variance formula using their \((\text{E}(X))^2\) value. FT their table with 3 or more probabilities \((0 < p < 1)\) which need not sum to 1. Note: if table is correct, \(\frac{454}{40}\) \(\left(\text{or}\ \frac{227}{20}\ \text{or any calculation}\right) - (\textit{their}\ \text{E}(X))^2\) implies M1.
\(\text{E}(X) = \frac{13}{4},\ 3\frac{1}{4},\ 3.05\quad \text{Var}(X) = \frac{63}{80},\ 0.7875\)	A1	Answers for \(\text{E}(X)\) and \(\text{Var}(X)\) must be identified. \(\text{E}(X)\) may be identified by correct use in variance. Condone E, V, \(\mu\), \(\sigma\) etc. If A0 earned, SC B1 for identified correct final solutions.