Edexcel S2 2005 June — Question 2 11 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2005
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeCumulative distribution function
DifficultyEasy -1.3 This is a straightforward application of standard uniform distribution formulas with no problem-solving required. All parts involve direct recall and substitution into well-known formulas (pdf = 1/(b-a), E(X) = (a+b)/2, Var(X) = (b-a)²/12, CDF integration, probability calculation). This is easier than average A-level questions which typically require multi-step reasoning or technique combination.
Spec5.02e Discrete uniform distribution5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration

2. The continuous random variable \(X\) is uniformly distributed over the interval \([ 2,6 ]\).
  1. Write down the probability density function \(\mathrm { f } ( x )\). Find
  2. \(\mathrm { E } ( X )\),
  3. \(\operatorname { Var } ( X )\),
  4. the cumulative distribution function of \(X\), for all \(x\),
  5. \(\mathrm { P } ( 2.3 < X < 3.4 )\).

2(a)
AnswerMarks Guidance
\(f(x) = \frac{1}{4}\) for \(2 \le x \le 6\)B1 \(\frac{1}{4}\) and range
\(f(x) = 0\) otherwiseB1 0 and range
2(b)
AnswerMarks
\(E(X) = 4\) by symmetry or formulaB1
2(c)
AnswerMarks Guidance
\(\text{Var}(X) = \frac{(6-2)^2}{12}\)M1 Use of formula
\(= \frac{4}{3}\)A1 \(1.3\) or \(1\frac{1}{3}\) or \(\frac{4}{3}\) or \(1.33\)
2(d)
AnswerMarks Guidance
\(F(x) = \int_2^x \frac{1}{4} dt = \left[\frac{1}{4}t\right]_2^x\)M1 Use of \(\int f(x)dx\)
\(= \frac{1}{4}(x-2)\)A1 \(\frac{1}{4}(x-2)\) or equiv.
\(F(x) = \frac{1}{4}(x-2), 2 \le x \le 6\)B1, B1t \(\frac{1}{4}(x-2)\) and range, ends and ranges
\(F(x) = 1\) for \(x > 6\); \(F(x) = 0\) for \(x < 2\)B1
2(e)
AnswerMarks Guidance
\(P(2.3 < X < 3.4) = \frac{1}{4}(3.4 - 2.3)\)M1 Use of area or F(x)
\(= 0.275\)A1 \(0.275\) or \(\frac{11}{40}\)
**2(a)**
| $f(x) = \frac{1}{4}$ for $2 \le x \le 6$ | B1 | $\frac{1}{4}$ and range |
| $f(x) = 0$ otherwise | B1 | 0 and range |

**2(b)**
| $E(X) = 4$ by symmetry or formula | B1 | |

**2(c)**
| $\text{Var}(X) = \frac{(6-2)^2}{12}$ | M1 | Use of formula |
| $= \frac{4}{3}$ | A1 | $1.3$ or $1\frac{1}{3}$ or $\frac{4}{3}$ or $1.33$ |

**2(d)**
| $F(x) = \int_2^x \frac{1}{4} dt = \left[\frac{1}{4}t\right]_2^x$ | M1 | Use of $\int f(x)dx$ |
| $= \frac{1}{4}(x-2)$ | A1 | $\frac{1}{4}(x-2)$ or equiv. |
| $F(x) = \frac{1}{4}(x-2), 2 \le x \le 6$ | B1, B1t | $\frac{1}{4}(x-2)$ and range, ends and ranges |
| $F(x) = 1$ for $x > 6$; $F(x) = 0$ for $x < 2$ | B1 | |

**2(e)**
| $P(2.3 < X < 3.4) = \frac{1}{4}(3.4 - 2.3)$ | M1 | Use of area or F(x) |
| $= 0.275$ | A1 | $0.275$ or $\frac{11}{40}$ |

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2. The continuous random variable $X$ is uniformly distributed over the interval $[ 2,6 ]$.
\begin{enumerate}[label=(\alph*)]
\item Write down the probability density function $\mathrm { f } ( x )$.

Find
\item $\mathrm { E } ( X )$,
\item $\operatorname { Var } ( X )$,
\item the cumulative distribution function of $X$, for all $x$,
\item $\mathrm { P } ( 2.3 < X < 3.4 )$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2005 Q2 [11]}}