Edexcel S1 — Question 3 9 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicUniform Distribution
TypeCalculate basic probabilities
DifficultyModerate -0.8 This is a straightforward S1 question testing basic discrete uniform distribution formulas. Part (a) requires direct application of standard formulas for expectation and variance, part (b) is simple counting of favorable outcomes, and part (c) involves solving a linear equation. All parts are routine calculations with no conceptual challenges or problem-solving required.
Spec5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution
3. The random variable \(X\) has the discrete uniform distribution over the set of consecutive integers \(\{ - 7 , - 6 , \ldots , 10 \}\).
Calculate (a) the expectation and variance of \(X\),
(b) \(\mathrm { P } ( X > 7 )\),
(c) the value of \(n\) for which \(\mathrm { P } ( - n \leq X \leq n ) = \frac { 7 } { 18 }\).
AnswerMarks Guidance
(a) Mean = \(\frac{10 + (-7)}{2} = 1.5\)M1 A1
Var\((X) = \text{Var}(X + 8) = \frac{(18^2 - 1)}{12} = 26\frac{11}{12}\) or \(26.9\)M1 M1 A1
(b) \(P(X > 7) = \frac{3}{18} - \frac{1}{6}\)M1 A1
(c) 7 consecutive integers centred on 0 are \(-3\) to \(3\), so \(x = 3\)M1 A1 Total: 9 marks 
(a) Mean = $\frac{10 + (-7)}{2} = 1.5$ | M1 A1 |

Var$(X) = \text{Var}(X + 8) = \frac{(18^2 - 1)}{12} = 26\frac{11}{12}$ or $26.9$ | M1 M1 A1 |

(b) $P(X > 7) = \frac{3}{18} - \frac{1}{6}$ | M1 A1 |

(c) 7 consecutive integers centred on 0 are $-3$ to $3$, so $x = 3$ | M1 A1 | **Total: 9 marks**
3. The random variable $X$ has the discrete uniform distribution over the set of consecutive integers $\{ - 7 , - 6 , \ldots , 10 \}$.\\
Calculate (a) the expectation and variance of $X$,\\
(b) $\mathrm { P } ( X > 7 )$,\\
(c) the value of $n$ for which $\mathrm { P } ( - n \leq X \leq n ) = \frac { 7 } { 18 }$.\\

\hfill \mbox{\textit{Edexcel S1  Q3 [9]}}
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