Edexcel S1 2014 June — Question 1 9 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2014
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeOne unknown from sum constraint only
DifficultyEasy -1.2 This is a straightforward S1 question testing basic probability distribution properties. Part (a) uses the simple constraint that probabilities sum to 1 (0.4 + p + 0.05 + 0.15 + p = 1). Parts (b)-(d) are routine calculations of expectation, cumulative probability, and probability of an inequality. Part (e) applies the standard variance property Var(aX+b) = a²Var(X). All parts require only direct application of formulas with no problem-solving or insight needed.
Spec2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance

  1. The discrete random variable \(X\) has probability distribution
\(x\)- 4- 2135
\(\mathrm { P } ( X = x )\)0.4\(p\)0.050.15\(p\)
  1. Show that \(p = 0.2\) Find
  2. \(\mathrm { E } ( X )\)
  3. \(\mathrm { F } ( 0 )\)
  4. \(\mathrm { P } ( 3 X + 2 > 5 )\) Given that \(\operatorname { Var } ( X ) = 13.35\)
  5. find the possible values of \(a\) such that \(\operatorname { Var } ( a X + 3 ) = 53.4\)

\begin{enumerate}
  \item The discrete random variable $X$ has probability distribution
\end{enumerate}

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 4 & - 2 & 1 & 3 & 5 \\
\hline
$\mathrm { P } ( X = x )$ & 0.4 & $p$ & 0.05 & 0.15 & $p$ \\
\hline
\end{tabular}
\end{center}

(a) Show that $p = 0.2$

Find\\
(b) $\mathrm { E } ( X )$\\
(c) $\mathrm { F } ( 0 )$\\
(d) $\mathrm { P } ( 3 X + 2 > 5 )$

Given that $\operatorname { Var } ( X ) = 13.35$\\
(e) find the possible values of $a$ such that $\operatorname { Var } ( a X + 3 ) = 53.4$\\

\hfill \mbox{\textit{Edexcel S1 2014 Q1 [9]}}