Edexcel S1 — Question 5 13 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeSimple algebraic expression for P(X=x)
DifficultyModerate -0.8 This is a straightforward S1 probability distribution question requiring standard techniques: summing probabilities to find k, calculating E(X) using the definition, applying probability and expectation rules, and computing variance. All parts follow textbook procedures with no problem-solving insight needed, making it easier than average but not trivial due to the arithmetic involved across multiple parts.
Spec2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables

The discrete random variable \(X\) has the probability function shown below. $$P(X = x) = \begin{cases} kx, & x = 2, 3, 4, 5, 6, \\ 0, & \text{otherwise}. \end{cases}$$
  1. Find the value of \(k\). [2 marks]
  2. Show that E\((X) = \frac{9}{2}\). [3 marks]
Find
  1. P\([X > \text{E}(X)]\), [2 marks]
  2. E\((2X - 5)\), [2 marks]
  3. Var\((X)\). [4 marks]

AnswerMarks Guidance
(a)M1 A1 \(2k + 3k + 4k + 5k + 6k = 1\); \(k = \frac{1}{30}\)
(b)M2 A1 \(\sum xP(x) = \frac{1}{30}(4 + 9 + 16 + 25 + 36) = \frac{9}{2}\)
(c)M1 A1 \(= \mathrm{P}(X > \frac{9}{2}) = \frac{5}{20} + \frac{6}{20} = \frac{11}{20}\)
(d)M1 A1 \((2 \times \frac{9}{2}) - 5 = 4\)
(e)M1 A1 \(E(X^2) = \sum x^2P(x) = \frac{1}{30}(8 + 27 + 64 + 125 + 216) = 22\)
M1 A1, (13)\(\mathrm{Var}(X) = 22 - (\frac{9}{2})^2 = \frac{7}{4}\)
(a) | M1 A1 | $2k + 3k + 4k + 5k + 6k = 1$; $k = \frac{1}{30}$

(b) | M2 A1 | $\sum xP(x) = \frac{1}{30}(4 + 9 + 16 + 25 + 36) = \frac{9}{2}$

(c) | M1 A1 | $= \mathrm{P}(X > \frac{9}{2}) = \frac{5}{20} + \frac{6}{20} = \frac{11}{20}$

(d) | M1 A1 | $(2 \times \frac{9}{2}) - 5 = 4$

(e) | M1 A1 | $E(X^2) = \sum x^2P(x) = \frac{1}{30}(8 + 27 + 64 + 125 + 216) = 22$

| M1 A1, (13) | $\mathrm{Var}(X) = 22 - (\frac{9}{2})^2 = \frac{7}{4}$

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The discrete random variable $X$ has the probability function shown below.
$$P(X = x) = \begin{cases} kx, & x = 2, 3, 4, 5, 6, \\ 0, & \text{otherwise}. \end{cases}$$

\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$. [2 marks]
\item Show that E$(X) = \frac{9}{2}$. [3 marks]
\end{enumerate}

Find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item P$[X > \text{E}(X)]$, [2 marks]
\item E$(2X - 5)$, [2 marks]
\item Var$(X)$. [4 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1  Q5 [13]}}