| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Standard +0.3 This is a standard S1 question requiring systematic application of probability axioms (probabilities sum to 1) and expectation formulas to solve simultaneous equations for two unknowns. Part (c) uses the standard variance transformation rule Var(aX+b) = a²Var(X). While multi-step, each component is routine textbook material with no novel insight required, making it slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | - 2 | - 1 | 0 | 1 | 3 |
| \(\mathrm { P } ( X = x )\) | 0.15 | \(a\) | \(b\) | \(a\) | 0.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{E}(X) =]\ (-2 \times 0.15) + (-1 \times a) + 0 + (1 \times a) + (3 \times 0.4)\) or \(-0.3 - a + a + 1.2\) | M1 | At least 3 non-zero correct products seen |
| \(= \mathbf{0.9}\) | A1 | For 0.9 or any exact equivalent |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{E}(X^2) =] \{(-2)^2 \times 0.15\} + \{(-1)^2 \times a\} + \{1^2 \times a\} + \{3^2 \times 0.4\}\) | M1 | At least 3 non-zero correct products for \(\text{E}(X^2)\) |
| or \(0.6 + 2a + 3.6\) | ||
| So \(4.2 + 2a = 4.54\) | dM1 | Dependent on 1st M1; using \(\text{E}(X^2)\) and 4.54 to form linear equation in \(a\) |
| \(a = \mathbf{0.17}\) | A1 | For \(a = 0.17\) or exact equivalent |
| Use of sum of probabilities \(= 1\): e.g. \(0.15 +\) "0.34" \(+ 0.4 + b = 1\) | M1 | For use of sum of probabilities \(= 1\) to form linear equation for \(b\) |
| \(b = \mathbf{0.11}\) | A1 | For \(b = 0.11\) or exact equivalent |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{Var}(X) =]\ 4.54 - (\text{their } 0.9)^2\ [= 3.73]\) | M1 | For correct expression for \(\text{Var}(X)\) (ft their 0.9) |
| \(\text{Var}(Y) = (-2)^2 \text{Var}(X)\) | M1 | For \((-2)^2 \times (\text{their Var}(X))\); condone \(-2^2\text{Var}(X)\) if it later becomes \(4\text{Var}(X)\) |
| \(= \mathbf{14.92}\) (accept 14.9) | A1 | For 14.92 (accept 14.9) |
| (3) [10 marks] |
# Question 2:
## Part (a):
$[\text{E}(X) =]\ (-2 \times 0.15) + (-1 \times a) + 0 + (1 \times a) + (3 \times 0.4)$ or $-0.3 - a + a + 1.2$ | M1 | At least 3 non-zero correct products seen
$= \mathbf{0.9}$ | A1 | For 0.9 or any exact equivalent
| | (2)
## Part (b):
$[\text{E}(X^2) =] \{(-2)^2 \times 0.15\} + \{(-1)^2 \times a\} + \{1^2 \times a\} + \{3^2 \times 0.4\}$ | M1 | At least 3 non-zero correct products for $\text{E}(X^2)$
or $0.6 + 2a + 3.6$ | |
So $4.2 + 2a = 4.54$ | dM1 | Dependent on 1st M1; using $\text{E}(X^2)$ and 4.54 to form linear equation in $a$
$a = \mathbf{0.17}$ | A1 | For $a = 0.17$ or exact equivalent
Use of sum of probabilities $= 1$: e.g. $0.15 +$ "0.34" $+ 0.4 + b = 1$ | M1 | For use of sum of probabilities $= 1$ to form linear equation for $b$
$b = \mathbf{0.11}$ | A1 | For $b = 0.11$ or exact equivalent
| | (5)
## Part (c):
$[\text{Var}(X) =]\ 4.54 - (\text{their } 0.9)^2\ [= 3.73]$ | M1 | For correct expression for $\text{Var}(X)$ (ft their 0.9)
$\text{Var}(Y) = (-2)^2 \text{Var}(X)$ | M1 | For $(-2)^2 \times (\text{their Var}(X))$; condone $-2^2\text{Var}(X)$ if it later becomes $4\text{Var}(X)$
$= \mathbf{14.92}$ (accept 14.9) | A1 | For 14.92 (accept 14.9)
| | (3) **[10 marks]**
---2. The discrete random variable $X$ has the following probability distribution.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 2 & - 1 & 0 & 1 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & 0.15 & $a$ & $b$ & $a$ & 0.4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( X )$.
Given that $\mathrm { E } \left( X ^ { 2 } \right) = 4.54$
\item find the value of $a$ and the value of $b$.
The random variable $Y = 3 - 2 X$
\item Find $\operatorname { Var } ( Y )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2019 Q2 [10]}}