| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Relationship between two random variables |
| Difficulty | Moderate -0.3 This is a straightforward S1 question testing standard discrete probability concepts: finding probabilities from a uniform distribution, linear transformations of random variables, and variance calculations. All parts follow routine procedures with no novel problem-solving required, though part (c) requires careful enumeration of cases and part (d)(ii) involves more arithmetic than typical. Slightly easier than average due to the simple uniform distribution and direct application of formulas. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Scheme | Marks |
| (a) | \(\frac{2}{5}\) | B1 (1) |
| (b) | \(E(W) = 3\) | B1 |
| \(E(5 - 2W) = 5 - 2E(W)\) or \(\frac{1}{5}(3 + 1 + \ldots + -5)\) | M1 | M1 use of \(E(5 - 2W) = 5 - 2E(W)\) or \(\frac{1}{5}(3 + 1 + \ldots + -5)\) Condone use of \(X\) instead of \(W\) |
| \(E(X) = -1\) | A1 (3) | A1 cao and labelled \(E(X)\) |
| (c) | \(P(X < W) = P(5 - 2W < W) = P(W > \frac{5}{3})\) or \(P(W \geq 2)\) | M1 |
| \(= \frac{4}{5}\) | A1 (2) | A1 o.e. |
| (d)(i) | B1 (3) | |
| \(\begin{array}{c\ | ccccc} y & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \hline p & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} \end{array}\) | |
| (ii) | \(E(Y) = \frac{1}{5}\left(1 + \frac{1}{2} + \ldots + \frac{1}{5}\right)\) or \(\frac{1}{5} + \frac{1}{10} + \frac{1}{15} + \frac{1}{20} + \frac{1}{25} = \left[\frac{137}{300} = 0.4566\ldots\right]\) | M1 |
| \(E(Y^2) = \frac{1}{5}\left[1^2 + \left(\frac{1}{2}\right)^2 + \ldots + \left(\frac{1}{5}\right)^2\right]\) or \(\frac{1}{5} + \frac{1}{20} + \frac{1}{45} + \frac{1}{80} + \frac{1}{125} = \left[\frac{5269}{18000} = 0.2927\ldots\right]\) | M1 | M1 attempt at expression for \(E(Y^2)\) using their values of \(y\) and \(p\) (at least 2 terms seen) or awrt 0.293 (0.2885 if have 0.3 rather than 1/3) Condone incorrect labelling |
| \(\text{Var}(Y) = '0.2927\ldots' - ('0.4566\ldots')^2\) | M1 A1 (5) | M1 A1 For use of "\(E(Y^2)\)"–"\((E(Y))^2\)" it their values for \(E(Y^2)\) and \(E(Y)\). A1 awrt 0.0842 or \(\frac{947}{11250}\) |
| awrt 0.758 | A1ft (2) | A1ft M1 for use of \((-3)^2 \text{Var}(Y)\) with their \(\text{Var}(Y) > 0\) condone \((3)^2 \text{Var}(Y)\) A1ft \(\frac{947}{1250}\) or \(9 \times\) "their part (d) > 0" evaluated correctly to 3sf or exact fraction |
| (e) | \(\text{Var}(2 - 3Y) = (-3)^2 \text{Var}(Y)\) |
| Part | Scheme | Marks | Guidance |
|------|--------|-------|----------|
| (a) | $\frac{2}{5}$ | B1 (1) | B1 o.e. |
| (b) | $E(W) = 3$ | B1 | B1 sight of $E(W) = 3$ or the $x$ values 3, 1, –1, –3, –5 (they may be added) |
| | $E(5 - 2W) = 5 - 2E(W)$ or $\frac{1}{5}(3 + 1 + \ldots + -5)$ | M1 | M1 use of $E(5 - 2W) = 5 - 2E(W)$ or $\frac{1}{5}(3 + 1 + \ldots + -5)$ Condone use of $X$ instead of $W$ |
| | $E(X) = -1$ | A1 (3) | A1 cao and labelled $E(X)$ |
| (c) | $P(X < W) = P(5 - 2W < W) = P(W > \frac{5}{3})$ or $P(W \geq 2)$ | M1 | M1 for identifying $W > \frac{5}{3}$ or $W \geq 2$ e.g. $1 - P(W = 1) \geq 2$ or $1 - P(W \leq 1) \geq 2$ |
| | $= \frac{4}{5}$ | A1 (2) | A1 o.e. |
| (d)(i) | | B1 (3) | B1 Correct distribution (probabilities may be implied by correct use). May be seen in any part |
| | $\begin{array}{c\|ccccc} y & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \hline p & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} \end{array}$ | | |
| (ii) | $E(Y) = \frac{1}{5}\left(1 + \frac{1}{2} + \ldots + \frac{1}{5}\right)$ or $\frac{1}{5} + \frac{1}{10} + \frac{1}{15} + \frac{1}{20} + \frac{1}{25} = \left[\frac{137}{300} = 0.4566\ldots\right]$ | M1 | M1 attempt at expression for $E(Y)$ using their values of $y$ and $p$ (at least 2 terms seen) or awrt 0.457 (0.45 if have 0.3 rather than 1/3) Condone incorrect labelling |
| | $E(Y^2) = \frac{1}{5}\left[1^2 + \left(\frac{1}{2}\right)^2 + \ldots + \left(\frac{1}{5}\right)^2\right]$ or $\frac{1}{5} + \frac{1}{20} + \frac{1}{45} + \frac{1}{80} + \frac{1}{125} = \left[\frac{5269}{18000} = 0.2927\ldots\right]$ | M1 | M1 attempt at expression for $E(Y^2)$ using their values of $y$ and $p$ (at least 2 terms seen) or awrt 0.293 (0.2885 if have 0.3 rather than 1/3) Condone incorrect labelling |
| | $\text{Var}(Y) = '0.2927\ldots' - ('0.4566\ldots')^2$ | M1 A1 (5) | M1 A1 For use of "$E(Y^2)$"–"$(E(Y))^2$" it their values for $E(Y^2)$ and $E(Y)$. A1 awrt 0.0842 or $\frac{947}{11250}$ |
| | awrt **0.758** | A1ft (2) | A1ft M1 for use of $(-3)^2 \text{Var}(Y)$ with their $\text{Var}(Y) > 0$ condone $(3)^2 \text{Var}(Y)$ A1ft $\frac{947}{1250}$ or $9 \times$ "their part (d) > 0" evaluated correctly to 3sf or exact fraction |
| (e) | $\text{Var}(2 - 3Y) = (-3)^2 \text{Var}(Y)$ | | |
---\begin{enumerate}
\item The random variable $W$ has a discrete uniform distribution where
\end{enumerate}
$$\mathrm { P } ( W = w ) = \frac { 1 } { 5 } \quad \text { for } w = 1,2,3,4,5$$
(a) Find $\mathrm { P } ( 2 \leqslant W < 3.5 )$
The discrete random variable $X = 5 - 2 W$\\
(b) Find $\mathrm { E } ( X )$\\
(c) Find $\mathrm { P } ( X < W )$
The discrete random variable $\mathrm { Y } = \frac { 1 } { W }$\\
(d) Find\\
(i) the probability distribution of $Y$\\
(ii) $\operatorname { Var } ( Y )$, showing your working.\\
(e) Find $\operatorname { Var } ( 2 - 3 Y )$\\
\hfill \mbox{\textit{Edexcel S1 2022 Q4 [13]}}