Edexcel S1 — Question 4 12 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeFind cumulative distribution F(x)
DifficultyModerate -0.8 This is a straightforward S1 probability distribution question requiring standard techniques: solving a quadratic equation for p using the probability sum rule, calculating simple probabilities by addition, computing mean and variance using standard formulas, and constructing a cumulative distribution table. All steps are routine applications of basic probability concepts with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation needed.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.03e Find cdf: by integration

The discrete random variable \(X\) has the following probability distribution:
\(x\)012345
\(\text{P}(X = x)\)0.110.170.20.13\(p\)\(p^2\)
  1. Find the value of \(p\). [4 marks]
  2. Find
    1. \(\text{P}(0 < X \leq 2)\),
    2. \(\text{P}(X \geq 3)\).
    [3 marks]
  3. Find the mean and the variance of \(X\). [3 marks]
  4. Construct a table to represent the cumulative distribution function \(\text{F}(x)\). [2 marks]

AnswerMarks Guidance
(a) \(p^2 + p - 0·39 = 0\) → \((p + 1·3)(p - 0·3) = 0\) → \(p = 0·3\)M1 A1 M1 A1
(b)(i) \(P(0 < X \leq 2) = 0·17 + 0·2 = 0·37\)B1
(ii) \(P(X \geq 3) = 0·13 + 0·3 + 0·09 = 0·52\)M1 A1
(c) \(E(X) = 0·17 + 0·4 + 0·39 + 1·2 + 0·45 = 2·61\)B1
\(E(X^2) = 9·19\)M1 A1
\(\text{Var}(X) = 9·19 - 2·61^2 = 2·38\)
(d) Cumulative distribution table completedB2 12 marks total
\(x\)0 1
\(F(x)\)0·11 0·28
(a) $p^2 + p - 0·39 = 0$ → $(p + 1·3)(p - 0·3) = 0$ → $p = 0·3$ | M1 A1 M1 A1 |

(b)(i) $P(0 < X \leq 2) = 0·17 + 0·2 = 0·37$ | B1 |

(ii) $P(X \geq 3) = 0·13 + 0·3 + 0·09 = 0·52$ | M1 A1 |

(c) $E(X) = 0·17 + 0·4 + 0·39 + 1·2 + 0·45 = 2·61$ | B1 |

$E(X^2) = 9·19$ | M1 A1 |

$\text{Var}(X) = 9·19 - 2·61^2 = 2·38$ | |

(d) Cumulative distribution table completed | B2 | 12 marks total

| $x$ | 0 | 1 | 2 | 3 | 4 | 5 |
|-----|-------|-------|-------|-------|-------|------|
| $F(x)$ | 0·11 | 0·28 | 0·48 | 0·61 | 0·91 | 1 |
The discrete random variable $X$ has the following probability distribution:

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$x$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\text{P}(X = x)$ & 0.11 & 0.17 & 0.2 & 0.13 & $p$ & $p^2$ \\
\hline
\end{tabular}

\begin{enumerate}[label=(\alph*)]
\item Find the value of $p$. [4 marks]
\item Find \begin{enumerate}[label=(\roman*)] \item $\text{P}(0 < X \leq 2)$, \item $\text{P}(X \geq 3)$. \end{enumerate} [3 marks]
\item Find the mean and the variance of $X$. [3 marks]
\item Construct a table to represent the cumulative distribution function $\text{F}(x)$. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1  Q4 [12]}}