| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from table |
| Difficulty | Moderate -0.3 This is a straightforward S1 question requiring standard variance calculations from a given probability distribution. Part (a) is routine expectation calculation, (b) uses the standard Var(X) = E(X²) - [E(X)]² formula, (c) applies linear transformation rules (E(aX) = aE(X), Var(aX) = a²Var(X)), and (d) is basic conditional probability. All techniques are direct applications of formulas with no problem-solving insight required, making it slightly easier than average but not trivial due to the multiple parts and arithmetic involved. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 10 }\) | \(\frac { 1 } { 5 }\) | \(\frac { 3 } { 10 }\) | \(\frac { 2 } { 5 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E\!\left(\frac{1}{X}\right) = 1\times\frac{1}{10}+\frac{1}{2}\times\frac{1}{5}+\frac{1}{3}\times\frac{3}{10}+\frac{1}{4}\times\frac{2}{5} = \frac{2}{5}\) | B1* | Value given, so must see sight of a correct expression with no incorrect working. Allow equivalent expressions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E\!\left(\!\left(\frac{1}{X}\right)^{\!2}\right) = 1^2\times\frac{1}{10}+\left(\frac{1}{2}\right)^{\!2}\times\frac{1}{5}+\left(\frac{1}{3}\right)^{\!2}\times\frac{3}{10}+\left(\frac{1}{4}\right)^{\!2}\times\frac{2}{5} = \frac{5}{24}\) | M1 | For attempt at expression for \(E\!\left(\!\left(\frac{1}{X}\right)^{\!2}\right)\) with at least 3 correct terms |
| \(\text{Var}\!\left(\frac{1}{X}\right) = \frac{5}{24} - \left(\frac{2}{5}\right)^{\!2} = \frac{29}{600}\) | M1 A1 | For correct expression for \(\text{Var}\!\left(\frac{1}{X}\right)\) ft stated \(E\!\left(\!\left(\frac{1}{X}\right)^{\!2}\right)\); cao allow awrt 0.0483 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(Y) = 12\) | B1 | For \(E(Y) = 12\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(Y) = 30^2 \text{Var}\!\left(\frac{1}{X}\right) = \frac{87}{2}\) or if \(y\): 30, 15, 10, 7.5 then \(\text{Var}(Y) = \frac{375}{2} - 12^2 = \frac{87}{2}\) | M1 A1 | For correct use of \(30^2\text{Var}\!\left(\frac{1}{X}\right)\) ft their \(\text{Var}\!\left(\frac{1}{X}\right)\) or \(\frac{375}{2} - 12^2\); for \(\text{Var}(Y) = \frac{87}{2}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([Y < 20 \Rightarrow] \frac{30}{X} < 20 \Rightarrow X > 1.5\) or \(y\): 30, 15, 10, 7.5 | M1 | For a correct inequality for \(Y < 20\) or all 4 values of \(Y\) found |
| \(P(Y < 20) = P(X > 1.5) = \frac{9}{10}\) | A1 | For \(P(Y<20) = \frac{9}{10}\); may be seen as denominator (e.g. \(0.2+0.3+0.4\)) in ratio |
| \([P(X<3\mid Y<20)] = \frac{P(X=2)}{P(X>1.5)} = \frac{\frac{1}{5}}{\frac{9}{10}} = \frac{2}{9}\) | dM1 | Dependent on 1st M1; for \(\frac{P(X=2)}{P(X>1.5)}\) or \(\frac{P(Y=15)}{P(Y<20)}\); allow \(\frac{P(1.5 |
| \(\frac{2}{9}\) | A1 A1 | For correct numerator; for \(\frac{2}{9}\) oe (allow decimal 3sf or better e.g. 0.222) |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E\!\left(\frac{1}{X}\right) = 1\times\frac{1}{10}+\frac{1}{2}\times\frac{1}{5}+\frac{1}{3}\times\frac{3}{10}+\frac{1}{4}\times\frac{2}{5} = \frac{2}{5}$ | B1* | Value given, so must see sight of a correct expression with no incorrect working. Allow equivalent expressions |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E\!\left(\!\left(\frac{1}{X}\right)^{\!2}\right) = 1^2\times\frac{1}{10}+\left(\frac{1}{2}\right)^{\!2}\times\frac{1}{5}+\left(\frac{1}{3}\right)^{\!2}\times\frac{3}{10}+\left(\frac{1}{4}\right)^{\!2}\times\frac{2}{5} = \frac{5}{24}$ | M1 | For attempt at expression for $E\!\left(\!\left(\frac{1}{X}\right)^{\!2}\right)$ with at least 3 correct terms |
| $\text{Var}\!\left(\frac{1}{X}\right) = \frac{5}{24} - \left(\frac{2}{5}\right)^{\!2} = \frac{29}{600}$ | M1 A1 | For correct expression for $\text{Var}\!\left(\frac{1}{X}\right)$ ft stated $E\!\left(\!\left(\frac{1}{X}\right)^{\!2}\right)$; cao allow awrt 0.0483 |
## Part (c)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(Y) = 12$ | B1 | For $E(Y) = 12$ |
## Part (c)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(Y) = 30^2 \text{Var}\!\left(\frac{1}{X}\right) = \frac{87}{2}$ or if $y$: 30, 15, 10, 7.5 then $\text{Var}(Y) = \frac{375}{2} - 12^2 = \frac{87}{2}$ | M1 A1 | For correct use of $30^2\text{Var}\!\left(\frac{1}{X}\right)$ ft their $\text{Var}\!\left(\frac{1}{X}\right)$ or $\frac{375}{2} - 12^2$; for $\text{Var}(Y) = \frac{87}{2}$ oe |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[Y < 20 \Rightarrow] \frac{30}{X} < 20 \Rightarrow X > 1.5$ or $y$: 30, 15, 10, 7.5 | M1 | For a correct inequality for $Y < 20$ or all 4 values of $Y$ found |
| $P(Y < 20) = P(X > 1.5) = \frac{9}{10}$ | A1 | For $P(Y<20) = \frac{9}{10}$; may be seen as denominator (e.g. $0.2+0.3+0.4$) in ratio |
| $[P(X<3\mid Y<20)] = \frac{P(X=2)}{P(X>1.5)} = \frac{\frac{1}{5}}{\frac{9}{10}} = \frac{2}{9}$ | dM1 | Dependent on 1st M1; for $\frac{P(X=2)}{P(X>1.5)}$ or $\frac{P(Y=15)}{P(Y<20)}$; allow $\frac{P(1.5<X<3)}{P(X>1.5)}$ or correct ratio of probabilities ft $P(Y<20)$ |
| $\frac{2}{9}$ | A1 A1 | For correct numerator; for $\frac{2}{9}$ oe (allow decimal 3sf or better e.g. 0.222) |
---\begin{enumerate}
\item The discrete random variable $X$ has the following probability distribution.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 10 }$ & $\frac { 1 } { 5 }$ & $\frac { 3 } { 10 }$ & $\frac { 2 } { 5 }$ \\
\hline
\end{tabular}
\end{center}
(a) Show that $\mathrm { E } \left( \frac { 1 } { X } \right) = \frac { 2 } { 5 }$\\
(b) Find $\operatorname { Var } \left( \frac { 1 } { X } \right)$
The random variable $Y = \frac { 30 } { X }$\\
(c) Find\\
(i) $\mathrm { E } ( Y )$\\
(ii) $\operatorname { Var } ( Y )$\\
(d) Find $\mathrm { P } ( X < 3 \mid Y < 20 )$
\hfill \mbox{\textit{Edexcel S1 2023 Q4 [12]}}