Edexcel S1 — Question 4 14 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicUniform Distribution
TypeVariance of linear transformation
DifficultyModerate -0.8 This question tests basic discrete probability distributions with straightforward calculations. Part (a) is simple counting, (b) requires recognizing a discrete uniform distribution and applying standard formulas, (c) involves constructing a simple probability table, and (d) applies the variance transformation rule Var(aX+b) = a²Var(X). All parts are routine textbook exercises requiring recall and direct application of formulas with no problem-solving insight needed.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02e Discrete uniform distribution

A pack of 52 cards contains 4 cards bearing each of the integers from 1 to 13. A card is selected at random. The random variable \(X\) represents the number on the card.
  1. Find \(P(X \leq 5)\). [1 mark]
  2. Name the distribution of \(X\) and find the expectation and variance of \(X\). [4 marks]
A hand of 12 cards consists of three 2s, four 3s, two 4s, two 5s and one 6. The random variable \(Y\) represents the number on a card chosen at random from this hand.
  1. Draw up a table to show the probability distribution of \(Y\). [3 marks]
  2. Calculate \(\text{Var}(3Y - 2)\). [6 marks]

AnswerMarks Guidance
(a) \(P(X \leq 5) = \frac{20}{52} = \frac{5}{13}\)\(B1\)
(b) Discrete uniform dist. on \(\{1, \ldots, 13\}\) \(E(X) = 7\), \(\text{Var}(X) = 14\)\(B1\) \(B1\) \(M1\) \(A1\)
(c)\(y\) 2
\(P(Y = y)\)\(\frac{1}{4}\) \(\frac{1}{3}\)
(d) \(E(Y) = 3.5\)\(M1\) \(A1\) \(A1\)
\(E(Y^2) = 1 + 3 + \frac{8}{3} + \frac{25}{6} + 3 = 13.83\)
\(\text{Var}(Y) = 1.58\)\(M1\) \(A1\) \(A1\)
\(\text{Var}(3Y - 2) = 9 \text{ Var}(Y) = 14.25\)
Total: 14 marks
**(a)** $P(X \leq 5) = \frac{20}{52} = \frac{5}{13}$ | $B1$ |

**(b)** Discrete uniform dist. on $\{1, \ldots, 13\}$ $E(X) = 7$, $\text{Var}(X) = 14$ | $B1$ $B1$ $M1$ $A1$ |

**(c)** | $y$ | 2 | 3 | 4 | 5 | 6 | |
| $P(Y = y)$ | $\frac{1}{4}$ | $\frac{1}{3}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{12}$ | $M1$ $A1$ $A1$ |

**(d)** $E(Y) = 3.5$ | $M1$ $A1$ $A1$ |
$E(Y^2) = 1 + 3 + \frac{8}{3} + \frac{25}{6} + 3 = 13.83$ | |
$\text{Var}(Y) = 1.58$ | $M1$ $A1$ $A1$ |
$\text{Var}(3Y - 2) = 9 \text{ Var}(Y) = 14.25$ | |

**Total: 14 marks**
A pack of 52 cards contains 4 cards bearing each of the integers from 1 to 13. A card is selected at random. The random variable $X$ represents the number on the card.

\begin{enumerate}[label=(\alph*)]
\item Find $P(X \leq 5)$. [1 mark]
\item Name the distribution of $X$ and find the expectation and variance of $X$. [4 marks]
\end{enumerate}

A hand of 12 cards consists of three 2s, four 3s, two 4s, two 5s and one 6. The random variable $Y$ represents the number on a card chosen at random from this hand.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Draw up a table to show the probability distribution of $Y$. [3 marks]
\item Calculate $\text{Var}(3Y - 2)$. [6 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1  Q4 [14]}}