| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Multiple unknowns from expectation and variance |
| Difficulty | Standard +0.3 This is a standard S1 question requiring systematic application of probability axioms (sum to 1), cumulative distribution function definition, and a simple transformation. All steps are routine textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 5.02a Discrete probability distributions: general5.03a Continuous random variables: pdf and cdf |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | Sum of probabilities \(= 1\) gives \(\frac{a+b}{60} + \frac{2a+b}{60} + \frac{3a+b}{60} + \frac{4a+b}{60} = 1\) | M1 |
| e.g. \(\frac{10a+4b}{60} = 1\) leading to \(5a + 2b = 30*\) | A1cso | A1cso for fully correct solution with no errors or omissions seen and at least one intermediate line of working seen |
| (2) | ||
| (b) | 1st M1 for use of \(\sum P(X = i) = \frac{13}{20}\) or \(P(X = 4) = \frac{7}{20}\) to form a 2nd equation in \(a\) and \(b\) | M1 |
| \(\frac{6a+3b}{60} = \frac{13}{20}\) or \(\frac{4a+b}{60} = \frac{7}{20}\) | ||
| e.g. \(\frac{(6a + 3b) = 39}{(5a + 2b = 30) \times 3}\) leading to \(3a = 12\) | dM1 | 1st A1 for a correct 3 term 2nd equation in \(a\) and \(b\) with \(a\) and \(b\) terms collected. 2nd dM1 dependent on 1st M1 for solving 2 relevant linear equations i.e. eliminating \(a\) or \(b\) leading to a linear equation in 1 variable. Allow 1 numerical or sign slip. |
| \(\underline{a = 4 \text{ and } b = 5}\) | A1 | 2nd A1 for both \(a = 4\) and \(b = 5\) (Correct answer only can score all 4 marks) |
| (4) | ||
| (c) | \([y]\) | \([<1]\) |
| \([F(y)]\) | \([0]\) | \(\frac{9}{60}\left(=\frac{3}{20}\right)\) |
| [10 marks] | ||
| Probability distribution of X | ||
| \(x\) | 1 | 2 |
| \(P(X=x)\) | \(\frac{9}{60}\) | \(\frac{11}{60}\) |
| NB \(F(y) = \frac{2y + 7\sqrt{y}}{60}\) for \(y = 1,4,9,16\) (o.e.) Is OK for all marks only with \(y\) values given | ||
| Notes | ||
| (a) | 1st M1 for use of sum of probabilities \(= 1\) to form a linear equation in \(a\) and \(b\) (4 terms seen). A1cso for fully correct solution with no errors or omissions seen and at least one intermediate line of working seen. | |
| (b) | 1st M1 for use of \(\sum_{i=1}^{3} P(X = i) = \frac{13}{20}\) or \(P(X = 4) = \frac{7}{20}\) to form a 2nd equation in \(a\) and \(b\). 1st A1 for a correct 3 term 2nd equation in \(a\) and \(b\) with \(a\) and \(b\) terms collected. 2nd dM1 dependent on 1st M1 for solving 2 relevant linear equations i.e. eliminating \(a\) or \(b\) leading to a linear equation in 1 variable. Allow 1 numerical or sign slip. 2nd A1 for both \(a = 4\) and \(b = 5\) (Correct answer only can score all 4 marks). | |
| (c) | 1st B1 for all y-values, can allow label of \(x^2\) (accept 1, 4, 9 and 16 or 1, \(2^2\), \(3^2\), \(4^2\)). 2nd B1cao for Fr(1) \(= \frac{9}{60}\) oe but must be clearly labelled as cdf linked to \(Y = 1\) but not for P(Y = y) or P(Y = 1). M1 for a correct method to find Fr(4) or Fr(9) if their \(a\) and \(b\) [dep' on correct y-values seen]. A1 for fully correct cumulative distribution function allow F(1) = \(\frac{9}{60}\), F(4) = \(\frac{22}{60}\), F(9) = \(\frac{39}{60}\), F(16)\(= 1\). |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | Sum of probabilities $= 1$ gives $\frac{a+b}{60} + \frac{2a+b}{60} + \frac{3a+b}{60} + \frac{4a+b}{60} = 1$ | M1 | 1st M1 for use of sum of probabilities $= 1$ to form a linear equation in $a$ and $b$ (4 terms seen) |
| | e.g. $\frac{10a+4b}{60} = 1$ leading to $5a + 2b = 30*$ | A1cso | A1cso for fully correct solution with no errors or omissions seen and at least one intermediate line of working seen |
| | | (2) | |
| (b) | 1st M1 for use of $\sum P(X = i) = \frac{13}{20}$ or $P(X = 4) = \frac{7}{20}$ to form a 2nd equation in $a$ and $b$ | M1 | |
| | $\frac{6a+3b}{60} = \frac{13}{20}$ or $\frac{4a+b}{60} = \frac{7}{20}$ | | |
| | e.g. $\frac{(6a + 3b) = 39}{(5a + 2b = 30) \times 3}$ leading to $3a = 12$ | dM1 | 1st A1 for a correct 3 term 2nd equation in $a$ and $b$ with $a$ and $b$ terms collected. 2nd dM1 dependent on 1st M1 for solving 2 relevant linear equations i.e. eliminating $a$ or $b$ leading to a linear equation in 1 variable. Allow 1 numerical or sign slip. |
| | $\underline{a = 4 \text{ and } b = 5}$ | A1 | 2nd A1 for both $a = 4$ and $b = 5$ (Correct answer only can score all 4 marks) |
| | | (4) | |
| (c) | $[y]$ | $[<1]$ | $1[\leq y < 4]$ | $4[\leq y < 9]$ | $9[\leq y < 16]$ | $[\geq ] 16$ | B1, B1cao, M1, A1 | 1st B1 for all y-values, can allow label of $x^2$ (accept 1, 4, 9 and 16 or 1, $2^2$, $3^2$, $4^2$). 2nd B1cao for Fr(1) $= \frac{9}{60}$ oe but must be clearly labelled as cdf linked to $Y = 1$ but not for P(Y = y) or P(Y = 1). M1 for a correct method to find Fr(4) or Fr(9) if their $a$ and $b$ [dep' on correct y-values seen] A1 for fully correct cumulative distribution function allow F(1) = $\frac{9}{60}$, F(4) = $\frac{22}{60}$, F(9) = $\frac{39}{60}$, F(16)$= 1$. |
| | $[F(y)]$ | $[0]$ | $\frac{9}{60}\left(=\frac{3}{20}\right)$ | $\frac{22}{60}\left(=\frac{11}{30}\right)$ | $\frac{39}{60}\left(=\frac{13}{20}\right)$ | $\frac{60}{60}(=1)$ | | |
| | | | [10 marks] | | |
| | **Probability distribution of X** | | | | | | |
| | $x$ | 1 | 2 | 3 | 4 | | | |
| | $P(X=x)$ | $\frac{9}{60}$ | $\frac{11}{60}$ | $\frac{17}{60}$ | $\frac{21}{60}$ | | | |
| | NB $F(y) = \frac{2y + 7\sqrt{y}}{60}$ for $y = 1,4,9,16$ (o.e.) Is OK for all marks only with $y$ values given | | | |
| **Notes** | | | | | | | |
| (a) | 1st M1 for use of sum of probabilities $= 1$ to form a linear equation in $a$ and $b$ (4 terms seen). A1cso for fully correct solution with no errors or omissions seen and at least one intermediate line of working seen. | | |
| (b) | 1st M1 for use of $\sum_{i=1}^{3} P(X = i) = \frac{13}{20}$ or $P(X = 4) = \frac{7}{20}$ to form a 2nd equation in $a$ and $b$. 1st A1 for a correct 3 term 2nd equation in $a$ and $b$ with $a$ and $b$ terms collected. 2nd dM1 dependent on 1st M1 for solving 2 relevant linear equations i.e. eliminating $a$ or $b$ leading to a linear equation in 1 variable. Allow 1 numerical or sign slip. 2nd A1 for both $a = 4$ and $b = 5$ (Correct answer only can score all 4 marks). | | |
| (c) | 1st B1 for all y-values, can allow label of $x^2$ (accept 1, 4, 9 and 16 or 1, $2^2$, $3^2$, $4^2$). 2nd B1cao for Fr(1) $= \frac{9}{60}$ oe but must be clearly labelled as cdf linked to $Y = 1$ but not for P(Y = y) or P(Y = 1). M1 for a correct method to find Fr(4) or Fr(9) if their $a$ and $b$ [dep' on correct y-values seen]. A1 for fully correct cumulative distribution function allow F(1) = $\frac{9}{60}$, F(4) = $\frac{22}{60}$, F(9) = $\frac{39}{60}$, F(16)$= 1$. | | |\begin{enumerate}
\item The discrete random variable $X$ can take only the values $1,2,3$ and 4 . For these values, the probability function is given by
\end{enumerate}
$$\mathrm { P } ( X = x ) = \frac { a x + b } { 60 } \quad \text { for } x = 1,2,3,4$$
where $a$ and $b$ are constants.\\
(a) Show that $5 a + 2 b = 30$
Given that $\mathrm { F } ( 3 ) = \frac { 13 } { 20 }$\\
(b) find the value of $a$ and the value of $b$
Given also that $Y = X ^ { 2 }$\\
(c) find the cumulative distribution function of $Y$\\
\hfill \mbox{\textit{Edexcel S1 2017 Q7 [10]}}