| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Find cumulative distribution F(x) |
| Difficulty | Moderate -0.3 This is a standard S1 cumulative distribution function question requiring routine techniques: finding k from F(5)=1, converting F(x) to P(X=x), calculating probabilities from the table, applying variance transformation rules (Var(aX+b)=a²Var(X)), and solving an inequality involving the linear transformation. All steps are algorithmic with no novel insight required, making it slightly easier than average for A-level. |
| 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 | 5 |
| \(\mathrm {~F} ( x )\) | \(\frac { 3 k } { 2 }\) | \(4 k\) | \(\frac { 15 k } { 2 }\) | \(12 k\) | \(\frac { 35 k } { 2 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(k = \frac{2}{35}\) | B1 | May be implied by a correct probability |
| M1, A1 | M1: need x values with a prob and at least one correct prob. (Allow probs in terms of k) | |
| (3) | A1: all values correct – accept decimals 3sf or better | |
| \(P(X = x)\) table: \(x\) = 1, 2, 3, 4, 5 with values \(\frac{3}{35}, \frac{5}{35}\) or \(\frac{1}{7}, \frac{7}{35}\) or \(\frac{1}{5}, \frac{9}{35}, \frac{11}{35}\) | ||
| (b) \(\frac{5}{35} + \frac{7}{35} + \frac{12}{35}\) | M1 A1ft | M1: "their P(X = 2)" + "their P(X = 3)". A1ft: ft providing < 1. Allow answer in [0.3428, 0.343] or 6k |
| (2) | ||
| (c) \(E(X) = 1 \times \frac{3}{35} + 2 \times \frac{5}{35} + 3 \times \frac{7}{35} + 4 \times \frac{9}{35} + 5 \times \frac{11}{35} = \left[\frac{25}{7}\right]\) | M1 | 1st M1: using \(\sum xP(X = x)\) or \(\frac{25}{7}\) or \(\frac{125}{k}\) [or \(\sum yP(Y = y)\) or \(– 13\) (> 4 correct terms or ft)] |
| M1 | 2nd M1: using \(\sum x^2P(X = x)\) or \(\frac{101}{7}\) or \(\frac{505}{k}\) [or \(\sum y^2P(Y = y)\) (> 4 correct terms or ft)] | |
| \(E(X^2) = 1 \times \frac{3}{35} + 4 \times \frac{5}{35} + 9 \times \frac{7}{35} + 16 \times \frac{9}{35} + 25 \times \frac{11}{35} = \left[\frac{101}{7}\right]\) | ||
| \(\text{Var}(X) = \frac{101}{7} - \left(\frac{25}{7}\right)^2 ; = \frac{82}{49}\) (allow 1.67~1.674) | M1; A1 | M1: using \(\text{Var}(X) = E(X^2) - [E(X)]^2\) or \(\text{Var}(Y) = E(Y^2) - [E(Y)]^2\) (or for E(Y²) = 251) |
| (6) | A1: for a correct answer (allow 3sf) | |
| \(\text{Var}(12 - 7X) = 7^2 \times \frac{82}{49} ; = 82\) | M1; A1 | 4th M1: \(49 \times \text{Var}(X)\). 2nd A1: for 82 only |
| (d) \(4X \le | Y | \) when \(X = 1, 4\) or \(5\), so probability \(= "{\frac{3}{35}}" + "{\frac{9}{35}}" + "{\frac{11}{35}}" = \frac{23}{35}\) |
| (3) | ||
| Total 14 |
**(a)** $k = \frac{2}{35}$ | B1 | May be implied by a correct probability
| | M1, A1 | M1: need x values with a prob and at least one correct prob. (Allow probs in terms of k)
| | (3) | A1: all values correct – accept decimals 3sf or better
| | | $P(X = x)$ table: $x$ = 1, 2, 3, 4, 5 with values $\frac{3}{35}, \frac{5}{35}$ or $\frac{1}{7}, \frac{7}{35}$ or $\frac{1}{5}, \frac{9}{35}, \frac{11}{35}$ |
**(b)** $\frac{5}{35} + \frac{7}{35} + \frac{12}{35}$ | M1 A1ft | M1: "their P(X = 2)" + "their P(X = 3)". A1ft: ft providing < 1. Allow answer in [0.3428, 0.343] or 6k
| | (2) |
**(c)** $E(X) = 1 \times \frac{3}{35} + 2 \times \frac{5}{35} + 3 \times \frac{7}{35} + 4 \times \frac{9}{35} + 5 \times \frac{11}{35} = \left[\frac{25}{7}\right]$ | M1 | 1st M1: using $\sum xP(X = x)$ or $\frac{25}{7}$ or $\frac{125}{k}$ [or $\sum yP(Y = y)$ or $– 13$ (> 4 correct terms or ft)]
| | M1 | 2nd M1: using $\sum x^2P(X = x)$ or $\frac{101}{7}$ or $\frac{505}{k}$ [or $\sum y^2P(Y = y)$ (> 4 correct terms or ft)]
| $E(X^2) = 1 \times \frac{3}{35} + 4 \times \frac{5}{35} + 9 \times \frac{7}{35} + 16 \times \frac{9}{35} + 25 \times \frac{11}{35} = \left[\frac{101}{7}\right]$ | |
| $\text{Var}(X) = \frac{101}{7} - \left(\frac{25}{7}\right)^2 ; = \frac{82}{49}$ (allow 1.67~1.674) | M1; A1 | M1: using $\text{Var}(X) = E(X^2) - [E(X)]^2$ or $\text{Var}(Y) = E(Y^2) - [E(Y)]^2$ (or for E(Y²) = 251)
| | (6) | A1: for a correct answer (allow 3sf)
| $\text{Var}(12 - 7X) = 7^2 \times \frac{82}{49} ; = 82$ | M1; A1 | 4th M1: $49 \times \text{Var}(X)$. 2nd A1: for 82 only
**(d)** $4X \le |Y|$ when $X = 1, 4$ or $5$, so probability $= "{\frac{3}{35}}" + "{\frac{9}{35}}" + "{\frac{11}{35}}" = \frac{23}{35}$ | M1;A1ft; A1 | M1: for $X = 1, 4$ or $5$ [or $Y = 5, -16, -23$] and at least one correct ft probability. A1ft: their "$\frac{3}{35}$"+their "$\frac{9}{35}$"+their "$\frac{11}{35}$" providing sum is < 1 (allow in terms of k). A1: cao (allow $\frac{23}{k}$)
| | (3) |
| | **Total 14** |
---\begin{enumerate}
\item The discrete random variable $X$ is defined by the cumulative distribution function
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm {~F} ( x )$ & $\frac { 3 k } { 2 }$ & $4 k$ & $\frac { 15 k } { 2 }$ & $12 k$ & $\frac { 35 k } { 2 }$ \\
\hline
\end{tabular}
\end{center}
where $k$ is a constant.\\
(a) Find the probability distribution of $X$.\\
(b) Find $\mathrm { P } ( 1.5 < X \leqslant 3.5 )$
The random variable $Y = 12 - 7 X$\\
(c) Calculate Var(Y)\\
(d) Calculate $\mathrm { P } ( 4 X \leqslant | Y | )$
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel S1 2018 Q5 [14]}}