AQA Further Paper 3 Statistics Specimen — Question 2 1 marks

Exam BoardAQA
ModuleFurther Paper 3 Statistics (Further Paper 3 Statistics)
SessionSpecimen
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeCalculate probability P(X in interval)
DifficultyEasy -1.2 This is a straightforward application of the definition of cumulative distribution function: P(Y > 4) = 1 - F(4) = 1 - 16/36 = 5/9. Requires only direct substitution into a given formula with no problem-solving or conceptual insight, making it easier than average even for Further Maths statistics.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf

2 The continuous random variable \(Y\) has cumulative distribution function defined by $$\mathrm { F } ( y ) = \left\{ \begin{array} { c c } 0 & y < 0 \\ \frac { y ^ { 2 } } { 36 } & 0 \leq y \leq 6 \\ 1 & y > 6 \end{array} \right.$$ Find the value of \(\mathrm { P } ( Y > 4 )\) Circle your answer. \(\frac { 4 } { 9 }\) \(\frac { 5 } { 9 }\) \(\frac { 16 } { 27 }\) \(\frac { 11 } { 27 }\)

Question 2:
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{5}{9}\)B1 Circles correct answer
Total: 1 mark
## Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{5}{9}$ | B1 | Circles correct answer |

**Total: 1 mark**

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2 The continuous random variable $Y$ has cumulative distribution function defined by

$$\mathrm { F } ( y ) = \left\{ \begin{array} { c c } 
0 & y < 0 \\
\frac { y ^ { 2 } } { 36 } & 0 \leq y \leq 6 \\
1 & y > 6
\end{array} \right.$$

Find the value of $\mathrm { P } ( Y > 4 )$\\
Circle your answer.\\
$\frac { 4 } { 9 }$\\
$\frac { 5 } { 9 }$\\
$\frac { 16 } { 27 }$\\
$\frac { 11 } { 27 }$

\hfill \mbox{\textit{AQA Further Paper 3 Statistics  Q2 [1]}}