Edexcel S2 2011 January — Question 3 11 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2011
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeMultiple observations or trials
DifficultyModerate -0.3 Parts (a)-(d) are direct formula applications for uniform distribution (E(X)=midpoint, Var(X)=range²/12, standard results). Part (e) requires binomial approximation to normal, which is S2-standard but involves multiple steps. Overall slightly easier than average due to routine nature of all components.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration

The continuous random variable \(X\) is uniformly distributed over the interval \([-1,3]\). Find
  1. E(\(X\)) [1]
  2. Var(\(X\)) [2]
  3. E(\(X^2\)) [2]
  4. P(\(X < 1.4\)) [1]
A total of 40 observations of \(X\) are made.
  1. Find the probability that at least 10 of these observations are negative. [5]

AnswerMarks Guidance
(a) \(E(X) = \frac{3-1}{2} = 1\)B1cao (1 mark)
(b) \(\text{Var}(X) = \frac{(3+1)^2}{12} = \frac{4}{3}\) or equivalentM1A1 (2 marks)
(c) \(E(X^2) = \frac{4}{3} + 1, = \frac{7}{3}\) or equivalentM1, A1 (2 marks)
(d) \(P(X < 1.4) = 0.6\)B1cao (1 mark)
(e) \(P(X < 0) = 0.25\); \(Y\) is number of values less than 0; \(Y \sim \text{Bin}(40, 0.25)\); \(P(Y \geq 10) = 1 - P(Y \leq 9) = 1 - 0.4395 = 0.5605\)B1, M1A1, M1, A1 (5 marks)
Notes:
- (b) M1 \(\frac{(3-1)^2}{12}\) or \(\frac{(3+1)^2}{12}\) or \(\frac{(3-(-1))^2}{12}\); A1 awrt 1.33.
- (c) M1 "their(b)" + ["their(a)"]² or \(\int_{-1}^{4} \frac{x^2}{4} dx\); A1 awrt 2.33.
- (e) B1 For writing or using the probability of a negative = 0.25; M1 Writing or use of \(B(40, p)\); A1 Writing or use of \(B(40, 0.25)\); M1 Writing or using \(1 - P(Y \leq 9)\); A1 awrt 0.561 or 0.560.
[11 marks total]
**(a)** $E(X) = \frac{3-1}{2} = 1$ | B1cao | (1 mark)

**(b)** $\text{Var}(X) = \frac{(3+1)^2}{12} = \frac{4}{3}$ or equivalent | M1A1 | (2 marks)

**(c)** $E(X^2) = \frac{4}{3} + 1, = \frac{7}{3}$ or equivalent | M1, A1 | (2 marks)

**(d)** $P(X < 1.4) = 0.6$ | B1cao | (1 mark)

**(e)** $P(X < 0) = 0.25$; $Y$ is number of values less than 0; $Y \sim \text{Bin}(40, 0.25)$; $P(Y \geq 10) = 1 - P(Y \leq 9) = 1 - 0.4395 = 0.5605$ | B1, M1A1, M1, A1 | (5 marks)

**Notes:**
- (b) M1 $\frac{(3-1)^2}{12}$ or $\frac{(3+1)^2}{12}$ or $\frac{(3-(-1))^2}{12}$; A1 awrt 1.33.
- (c) M1 "their(b)" + ["their(a)"]² or $\int_{-1}^{4} \frac{x^2}{4} dx$; A1 awrt 2.33.
- (e) B1 For writing or using the probability of a negative = 0.25; M1 Writing or use of $B(40, p)$; A1 Writing or use of $B(40, 0.25)$; M1 Writing or using $1 - P(Y \leq 9)$; A1 awrt 0.561 or 0.560.

**[11 marks total]**

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The continuous random variable $X$ is uniformly distributed over the interval $[-1,3]$. Find

\begin{enumerate}[label=(\alph*)]
\item E($X$)
[1]

\item Var($X$)
[2]

\item E($X^2$)
[2]

\item P($X < 1.4$)
[1]
\end{enumerate}

A total of 40 observations of $X$ are made.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Find the probability that at least 10 of these observations are negative.
[5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2011 Q3 [11]}}