| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(aX+b) transformations |
| Difficulty | Moderate -0.3 This is a straightforward S1 question testing standard discrete probability distribution calculations. Parts (a)-(c) involve routine application of expectation and variance formulas including the linear transformation Var(aX+b) = a²Var(X). Parts (d)-(f) require basic probability calculations with independent variables. All techniques are standard textbook exercises with no novel problem-solving 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 |
| \(x\) | 5 | 6 | 7 | 8 |
| \(\mathrm { P } ( X = x )\) | 0.13 | 0.21 | 0.29 | 0.37 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \([E(X) =] 5 \times 0.13 + 6 \times 0.21 + 7 \times 0.29 + 8 \times 0.37 = \underline{6.9}\) | M1, A1 |
| (2) | ||
| (b) | \([E(X^2) =] 5^2 \times 0.13 + 6^2 \times 0.21 + 7^2 \times 0.29 + 8^2 \times 0.37 [= 48.7]\) | M1 |
| \(\text{Var}(X) = 48.7 - '6.9'^2\) | M1, A1 | 2nd M1 for correct use of \(\text{Var}(X) = E(X^2) - [E(X)]^2\) f.t. their \(E(X)\). A1 for 1.09 (Correct answer only is M1M1A1) |
| \(= \underline{1.09}\) | ||
| (3) | ||
| (c) | \(\text{Var}(3-2X) = (-2)^2 \text{Var}(X)\) | M1, A1 |
| \(= \underline{4.36}\) | ||
| (2) | ||
| (d) | \([E(Y) =] \underline{6.5}\) or \(\frac{13}{2}\) (o.e.) | |
| (e) | \(P(X = Y) = \frac{1}{4} \times 0.13 + \frac{1}{4} \times 0.21 + \frac{1}{4} \times 0.29 + \frac{1}{4} \times 0.37 = \frac{1}{4}\) (oe) | M1, A1 |
| (2) | ||
| (f) | \(P(X > Y) = P(X = 6 \cap Y < 6) + P(X = 7 \cap Y < 7) + P(X = 8 \cap Y < 8)\) | M1 |
| \(= 0.21 \times 0.25 + 0.29 \times 0.50 + 0.37 \times 0.75\) | M1 | 2nd M1 for correct probability expression(i.e. correct values in formula) |
| \(= \underline{0.475}\) | A1 | A1 for 0.475 or \(\frac{19}{40}\). NB alternative expressions e.g. \(\frac{1}{4}(0.37 + 0.66 + 0.87)\) from listing \(Y < X\) rather than \(X > Y\). The 1st M1 may be implied by scoring the 2nd M1. |
| (3) | ||
| [13 marks] | ||
| SC/ | Only apply if they reach \([P(Y > X) = 0.13 \times \frac{3}{4} + 0.21 \times \frac{2}{4} + 0.29 \times \frac{1}{4} = ] 0.275\) | |
| (Y >X) |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $[E(X) =] 5 \times 0.13 + 6 \times 0.21 + 7 \times 0.29 + 8 \times 0.37 = \underline{6.9}$ | M1, A1 | M1 for a correct expression for $E(X)$ (Correct answer only is M1A1) |
| | | (2) | |
| (b) | $[E(X^2) =] 5^2 \times 0.13 + 6^2 \times 0.21 + 7^2 \times 0.29 + 8^2 \times 0.37 [= 48.7]$ | M1 | 1st M1 for attempting a correct expression for $E(X^2)$, sum of at least 3 correct products seen. The first M1 can be implied by 48.7. Stating $\text{Var}(X) =$ the expression for $E(X^2)$ can score M1M0A0 and may get M1 in (c). |
| | $\text{Var}(X) = 48.7 - '6.9'^2$ | M1, A1 | 2nd M1 for correct use of $\text{Var}(X) = E(X^2) - [E(X)]^2$ f.t. their $E(X)$. A1 for 1.09 (Correct answer only is M1M1A1) |
| | | $= \underline{1.09}$ | |
| | | (3) | |
| (c) | $\text{Var}(3-2X) = (-2)^2 \text{Var}(X)$ | M1, A1 | M1 for $(-2)^2 \text{Var}(X)$ or $(-2)^2 \times \text{'(b)'} [if \text{'(b)'} > 0]$ [condone no brackets if final answer is $\geq 0$] or a fully correct expr' for $\text{Var}(3-2X)$ based on $\begin{array}{c\|c\|c\|c\|c} 3-2x & -7 & -9 & -11 & -13 \\ \hline \text{Prob} & 0.13 & 0.21 & 0.29 & 0.37 \end{array}$. A1 for 4.36 (Correct answer only with no working scores M1A1) |
| | | $= \underline{4.36}$ | |
| | | (2) | |
| (d) | | $[E(Y) =] \underline{6.5}$ or $\frac{13}{2}$ (o.e.) | B1 | (1) |
| (e) | $P(X = Y) = \frac{1}{4} \times 0.13 + \frac{1}{4} \times 0.21 + \frac{1}{4} \times 0.29 + \frac{1}{4} \times 0.37 = \frac{1}{4}$ (oe) | M1, A1 | M1 for an expression for $P(X = Y)$ (at least 3 of the 4 products correct). May be implied by a correct answer. |
| | | (2) | |
| (f) | $P(X > Y) = P(X = 6 \cap Y < 6) + P(X = 7 \cap Y < 7) + P(X = 8 \cap Y < 8)$ | M1 | 1st M1 for a correct probability formula (as in scheme) or complete list of $X > Y$ [e.g. $X = 6$ and $Y=5; X = 7$ and $Y=5; X = 8$ and $Y = 7]$ |
| | $= 0.21 \times 0.25 + 0.29 \times 0.50 + 0.37 \times 0.75$ | M1 | 2nd M1 for correct probability expression(i.e. correct values in formula) |
| | | $= \underline{0.475}$ | A1 | A1 for 0.475 or $\frac{19}{40}$. NB alternative expressions e.g. $\frac{1}{4}(0.37 + 0.66 + 0.87)$ from listing $Y < X$ rather than $X > Y$. The 1st M1 may be implied by scoring the 2nd M1. |
| | | (3) | |
| | | [13 marks] | |
| **SC/** | Only apply if they reach $[P(Y > X) = 0.13 \times \frac{3}{4} + 0.21 \times \frac{2}{4} + 0.29 \times \frac{1}{4} = ] 0.275$ | | |
| **(Y >X)** | | | |\begin{enumerate}
\item In a game, the number of points scored by a player in the first round is given by the random variable $X$ with probability distribution
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 5 & 6 & 7 & 8 \\
\hline
$\mathrm { P } ( X = x )$ & 0.13 & 0.21 & 0.29 & 0.37 \\
\hline
\end{tabular}
\end{center}
Find\\
(a) $\mathrm { E } ( X )$\\
(b) $\operatorname { Var } ( X )$\\
(c) $\operatorname { Var } ( 3 - 2 X )$
The number of points scored by a player in the second round is given by the random variable $Y$ and is independent of the number of points scored in the first round.
The random variable $Y$ has probability function
$$\mathrm { P } ( Y = y ) = \frac { 1 } { 4 } \quad \text { for } y = 5,6,7,8$$
(d) Write down the value of $\mathrm { E } ( Y )$\\
(e) Find $\mathrm { P } ( X = Y )$\\
(f) Find the probability that the number of points scored by a player in the first round is greater than the number of points scored by the player in the second round.
\hfill \mbox{\textit{Edexcel S1 2017 Q4 [13]}}