Edexcel S1 2002 November — Question 6 15 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2002
SessionNovember
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeMultiple unknowns from expectation and variance
DifficultyModerate -0.8 This is a standard S1 textbook exercise testing routine application of discrete probability distribution formulas. Part (a) requires solving simultaneous equations using ΣP=1 and E(X) definition, parts (b-e) are direct formula applications with no problem-solving insight needed. The multi-part structure and 15 marks reflect thoroughness rather than difficulty.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance

The discrete random variable \(X\) has the following probability distribution.
\(x\)\(-2\)\(-1\)\(0\)\(1\)\(2\)
\(P(X = x)\)\(\alpha\)\(0.2\)\(0.1\)\(0.2\)\(\beta\)
  1. Given that \(E(X) = -0.2\), find the value of \(\alpha\) and the value of \(\beta\). [6]
  2. Write down \(F(0.8)\). [1]
  3. Evaluate \(\text{Var}(X)\). [4]
Find the value of
  1. \(E(3X - 2)\), [2]
  2. \(\text{Var}(2X + 6)\). [2]

(a)
AnswerMarks
\(\alpha + \beta = 0.5\)B1
\(-2\alpha + 2\beta = -0.2\)M1
\(\alpha = 0.3, \beta = 0.2\)M1 A1; A1
(6 marks)
(b)
AnswerMarks
\(F(0.8) = 0.6\)B1 ft
(1 mark)
(c)
AnswerMarks
\(E(X^2) = (4 \times 0.3) + \ldots + (4 \times 0.2) = 2.4\)M1, A1
\(\text{Var}(X) = 2.4 - (-0.2)^2 = 2.36\)M1, A1
(4 marks)
(d)
AnswerMarks
\(E(3X - 2) = 3E(X) - 2 = -2.6\)M1, A1 ft
(2 marks)
(e)
AnswerMarks
\(\text{Var}(2X + 6) = 4 \text{Var}(X) = 9.44\)M1, A1 ft
(2 marks)
Total: 15 marks
## (a)
$\alpha + \beta = 0.5$ | B1 |
$-2\alpha + 2\beta = -0.2$ | M1 |
$\alpha = 0.3, \beta = 0.2$ | M1 A1; A1 |
| (6 marks) |

## (b)
$F(0.8) = 0.6$ | B1 ft |
| (1 mark) |

## (c)
$E(X^2) = (4 \times 0.3) + \ldots + (4 \times 0.2) = 2.4$ | M1, A1 |
$\text{Var}(X) = 2.4 - (-0.2)^2 = 2.36$ | M1, A1 |
| (4 marks) |

## (d)
$E(3X - 2) = 3E(X) - 2 = -2.6$ | M1, A1 ft |
| (2 marks) |

## (e)
$\text{Var}(2X + 6) = 4 \text{Var}(X) = 9.44$ | M1, A1 ft |
| (2 marks) |

**Total: 15 marks**

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The discrete random variable $X$ has the following probability distribution.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & $-2$ & $-1$ & $0$ & $1$ & $2$ \\
\hline
$P(X = x)$ & $\alpha$ & $0.2$ & $0.1$ & $0.2$ & $\beta$ \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}[label=(\alph*)]
\item Given that $E(X) = -0.2$, find the value of $\alpha$ and the value of $\beta$. [6]

\item Write down $F(0.8)$. [1]

\item Evaluate $\text{Var}(X)$. [4]
\end{enumerate}

Find the value of

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item $E(3X - 2)$, [2]

\item $\text{Var}(2X + 6)$. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2002 Q6 [15]}}