AQA Further AS Paper 2 Statistics 2018 June — Question 6 6 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Statistics (Further AS Paper 2 Statistics)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeCalculate Var(X) from probability function
DifficultyStandard +0.3 This is a straightforward variance calculation requiring finding k from ΣP=1, computing E(Y) and E(Y²), then applying Var(aY+b)=a²Var(Y). All steps are routine applications of standard formulas with simple arithmetic (summing 2k(1+2+3+4)=1 gives k=1/20, then basic calculations). Slightly easier than average as it's purely mechanical with no problem-solving required.
Spec5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance

6 The discrete random variable \(Y\) has the probability function $$\mathrm { P } ( Y = y ) = \begin{cases} 2 k y & y = 1,2,3,4 \\ 0 & \text { otherwise } \end{cases}$$ where \(k\) is a constant. Show that \(\operatorname { Var } ( 5 Y - 2 ) = 25\) \includegraphics[max width=\textwidth, alt={}, center]{313cd5ce-07ff-4781-a134-565b8b221145-07_2488_1716_219_153}

Question 6:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2k + 4k + 6k + 8k = 1 \Rightarrow 20k = 1 \Rightarrow k = \frac{1}{20}\) or \(0.05\)B1 Uses sum of probabilities \(= 1\)
\(E(Y) = 1 \times 2k + 2 \times 4k + 3 \times 6k + 4 \times 8k = 60k\) or \(3\)M1 Calculates \(E(Y)\) or \(E(Y^2)\) correctly
\(E(Y^2) = 1^2 \times 2k + 2^2 \times 4k + 3^2 \times 6k + 4^2 \times 8k = 200k\) or \(10\)A1 Calculates both \(E(Y)\) and \(E(Y^2)\) correctly
\(\text{Var}(Y) = E(Y^2) - [E(Y)]^2 = 200k - (60k)^2\) or \(10 - 3^2 = 200k - 3600k^2\) or \(1\)M1 Calculates \(\text{Var}(Y)\)
\(\text{Var}(5Y-2) = 5^2\,\text{Var}(Y) = 25 \times 1 = 25\)M1 Uses formula for \(\text{Var}(aY \pm b)\)
\(\text{Var}(5Y-2) = 25\) AGR1 Fully correct logical argument
## Question 6:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2k + 4k + 6k + 8k = 1 \Rightarrow 20k = 1 \Rightarrow k = \frac{1}{20}$ or $0.05$ | B1 | Uses sum of probabilities $= 1$ |
| $E(Y) = 1 \times 2k + 2 \times 4k + 3 \times 6k + 4 \times 8k = 60k$ or $3$ | M1 | Calculates $E(Y)$ or $E(Y^2)$ correctly |
| $E(Y^2) = 1^2 \times 2k + 2^2 \times 4k + 3^2 \times 6k + 4^2 \times 8k = 200k$ or $10$ | A1 | Calculates both $E(Y)$ and $E(Y^2)$ correctly |
| $\text{Var}(Y) = E(Y^2) - [E(Y)]^2 = 200k - (60k)^2$ or $10 - 3^2 = 200k - 3600k^2$ or $1$ | M1 | Calculates $\text{Var}(Y)$ |
| $\text{Var}(5Y-2) = 5^2\,\text{Var}(Y) = 25 \times 1 = 25$ | M1 | Uses formula for $\text{Var}(aY \pm b)$ |
| $\text{Var}(5Y-2) = 25$ AG | R1 | Fully correct logical argument |

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6 The discrete random variable $Y$ has the probability function

$$\mathrm { P } ( Y = y ) = \begin{cases} 2 k y & y = 1,2,3,4 \\ 0 & \text { otherwise } \end{cases}$$

where $k$ is a constant.

Show that $\operatorname { Var } ( 5 Y - 2 ) = 25$\\

\includegraphics[max width=\textwidth, alt={}, center]{313cd5ce-07ff-4781-a134-565b8b221145-07_2488_1716_219_153}

\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2018 Q6 [6]}}