Edexcel S2 2013 January — Question 4 14 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2013
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeDerive or verify variance formula
DifficultyStandard +0.3 This is a straightforward S2 uniform distribution question requiring standard techniques: reading off the mean, calculating probabilities using the rectangular distribution, and performing routine integration to verify E(Y²). Part (f) involves equating two probabilities but follows directly from the uniform distribution formula. All parts are textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration
4. The continuous random variable \(X\) is uniformly distributed over the interval \([ - 4,6 ]\).
  1. Write down the mean of \(X\).
  2. Find \(\mathrm { P } ( X \leqslant 2.4 )\)
  3. Find \(\mathrm { P } ( - 3 < X - 5 < 3 )\) The continuous random variable \(Y\) is uniformly distributed over the interval \([ a , 4 a ]\).
  4. Use integration to show that \(\mathrm { E } \left( Y ^ { 2 } \right) = 7 a ^ { 2 }\)
  5. Find \(\operatorname { Var } ( Y )\).
  6. Given that \(\mathrm { P } \left( X < \frac { 8 } { 3 } \right) = \mathrm { P } \left( Y < \frac { 8 } { 3 } \right)\), find the value of \(a\).
Question 4:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Mean \(= 1\)B1
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(X \leq 2.4) = (2.4 - {-4}) \times \frac{1}{10}\)M1 \((2.4--4)\times\frac{1}{10}\) or \(1-(6-2.4)\times\frac{1}{10}\) o.e.
\(= 0.64\) or \(\frac{16}{25}\)A1
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(-3 < X - 5 < 3) = P(2 < X < 6)\)M1 Finding \(P(2 < X < 6)\) or \(P(X > 2)\) or \(1 - P(X < 2)\). May be implied by correct answer if no incorrect working. NB if distribution changed to U[-9,1] then M1 is for finding \(P(-3 < X < 1)\) or \(P(X > -3)\) or \(1 - P(X < -3)\)
\(= 0.4\)A1
Part (d):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\int_a^{4a} \frac{y^2}{4a-a}\,dy = \left[\frac{y^3}{9a}\right]_a^{4a}\)M1, M1dep 1st M1: writing or using \(\int_a^{4a} y^2 f(y)\,dy\) with correct limits used at some point; 2nd M1 dep: attempting to integrate \(y^n \to \frac{y^{n+1}}{n+1}\)
\(= \frac{64a^3 - a^3}{9a}\)A1 1st A1: correct expression with correct limits substituted
\(= 7a^2\) *AGA1cso Answer given — must show working
Part (e):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\text{Var}(Y) = \frac{1}{12}(4a-a)^2\) or \(\text{Var}(Y) = 7a^2 - \left(\frac{5}{2}a\right)^2\)M1 Either use of \(\frac{(b-a)^2}{12}\) or \(E(Y^2) - [E(Y)]^2\) — may use part (d) for \(E(Y^2)\)
\(= \frac{3}{4}a^2\)A1cso
Part (f):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{2}{3} = \frac{1}{3a}\!\left(\frac{8}{3} - a\right)\)M1, A1 M1: using \(\frac{1}{3a}\!\left(\frac{8}{3}-a\right) =\) a probability or \(\frac{1}{3a}\!\left(4a - \frac{8}{3}\right) =\) a probability
\(a = \frac{8}{9}\)A1 An answer of \(\frac{8}{9}\) with no incorrect working gains M1A1A1 
## Question 4:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean $= 1$ | B1 | |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X \leq 2.4) = (2.4 - {-4}) \times \frac{1}{10}$ | M1 | $(2.4--4)\times\frac{1}{10}$ or $1-(6-2.4)\times\frac{1}{10}$ o.e. |
| $= 0.64$ or $\frac{16}{25}$ | A1 | |

### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(-3 < X - 5 < 3) = P(2 < X < 6)$ | M1 | Finding $P(2 < X < 6)$ or $P(X > 2)$ or $1 - P(X < 2)$. May be implied by correct answer if no incorrect working. NB if distribution changed to U[-9,1] then M1 is for finding $P(-3 < X < 1)$ or $P(X > -3)$ or $1 - P(X < -3)$ |
| $= 0.4$ | A1 | |

### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_a^{4a} \frac{y^2}{4a-a}\,dy = \left[\frac{y^3}{9a}\right]_a^{4a}$ | M1, M1dep | 1st M1: writing or using $\int_a^{4a} y^2 f(y)\,dy$ with correct limits used at some point; 2nd M1 dep: attempting to integrate $y^n \to \frac{y^{n+1}}{n+1}$ |
| $= \frac{64a^3 - a^3}{9a}$ | A1 | 1st A1: correct expression with correct limits substituted |
| $= 7a^2$ *AG | A1cso | Answer given — must show working |

### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(Y) = \frac{1}{12}(4a-a)^2$ or $\text{Var}(Y) = 7a^2 - \left(\frac{5}{2}a\right)^2$ | M1 | Either use of $\frac{(b-a)^2}{12}$ or $E(Y^2) - [E(Y)]^2$ — may use part (d) for $E(Y^2)$ |
| $= \frac{3}{4}a^2$ | A1cso | |

### Part (f):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{3} = \frac{1}{3a}\!\left(\frac{8}{3} - a\right)$ | M1, A1 | M1: using $\frac{1}{3a}\!\left(\frac{8}{3}-a\right) =$ a probability **or** $\frac{1}{3a}\!\left(4a - \frac{8}{3}\right) =$ a probability |
| $a = \frac{8}{9}$ | A1 | An answer of $\frac{8}{9}$ with no incorrect working gains M1A1A1 |

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4. The continuous random variable $X$ is uniformly distributed over the interval $[ - 4,6 ]$.
\begin{enumerate}[label=(\alph*)]
\item Write down the mean of $X$.
\item Find $\mathrm { P } ( X \leqslant 2.4 )$
\item Find $\mathrm { P } ( - 3 < X - 5 < 3 )$

The continuous random variable $Y$ is uniformly distributed over the interval $[ a , 4 a ]$.
\item Use integration to show that $\mathrm { E } \left( Y ^ { 2 } \right) = 7 a ^ { 2 }$
\item Find $\operatorname { Var } ( Y )$.
\item Given that $\mathrm { P } \left( X < \frac { 8 } { 3 } \right) = \mathrm { P } \left( Y < \frac { 8 } { 3 } \right)$, find the value of $a$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2013 Q4 [14]}}
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