| Exam Board | OCR |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2018 |
| Session | March |
| Marks | 9 |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Moderate -0.3 This is a standard Further Statistics 1 question requiring two simultaneous equations (probabilities sum to 1, expectation equals 2.55) to find x and y, followed by routine application of variance and linear transformation formulas. While it involves multiple steps, all techniques are direct textbook applications 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 |
| Score, \(N\) | 1 | 2 | 3 | 4 | 5 |
| Probability | 0.3 | 0.2 | 0.2 | \(x\) | \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.7 + x + y = 1\) | M1\* | 3.1b |
| \(1.3 + 4x + 5y = 2.55\) | M1\* | 1.1a |
| Solve simultaneously to get \(x = 0.25, y = 0.05\) | DepM1 | 1.1a |
| \(\sum n^2 p (= 8.15)\) | M1\* | 1.1a |
| \(-2.55^2\) | depM1 | 1.1a |
| \(= \frac{659}{400}\) or \(1.6475\) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(3 \times 2.55 + 2 = \frac{193}{20}\) or \(9.65\) | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(9 \times\) their (i) \(= [14.83]\) | B1FT | 1.1 |
### Part (i)
$0.7 + x + y = 1$ | **M1\*** | 3.1b | Equation $\sum p = 1$ seen or implied
$1.3 + 4x + 5y = 2.55$ | **M1\*** | 1.1a | $\sum np = 2.55$ seen or implied
Solve simultaneously to get $x = 0.25, y = 0.05$ | **DepM1** | 1.1a | Or BC
$\sum n^2 p (= 8.15)$ | **M1\*** | 1.1a | Use $\sum n^2 p$
$-2.55^2$ | **depM1** | 1.1a |
$= \frac{659}{400}$ or $1.6475$ | **A1** | 1.1 | Exact or awrt 1.65, www
### Part (ii)
$3 \times 2.55 + 2 = \frac{193}{20}$ or $9.65$ | **B1** | 1.1 | Exact only
### Part (iii)
$9 \times$ their (i) $= [14.83]$ | **B1FT** | 1.1 | FT on their variance
---4 A spinner has edges numbered $1,2,3,4$ and 5 . When the spinner is spun, the number of the edge on which it lands is the score. The probability distribution of the score, $N$, is given in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
Score, $N$ & 1 & 2 & 3 & 4 & 5 \\
\hline
Probability & 0.3 & 0.2 & 0.2 & $x$ & $y$ \\
\hline
\end{tabular}
\end{center}
It is known that $\mathrm { E } ( N ) = 2.55$.\\
(i) Find $\operatorname { Var } ( N )$.\\
(ii) Find $\mathrm { E } ( 3 N + 2 )$.\\
(iii) Find $\operatorname { Var } ( 3 N + 2 )$.
\hfill \mbox{\textit{OCR FS1 AS 2018 Q4 [9]}}