Edexcel S1 — Question 4 11 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeSimple algebraic expression for P(X=x)
DifficultyModerate -0.3 This is a straightforward S1 probability distribution question requiring standard techniques: summing probabilities to find k, calculating P(X<0) by addition, and deriving a cumulative distribution function. While part (c) involves algebraic manipulation to show the quadratic form, all steps follow routine procedures with no novel problem-solving required. Slightly easier than average due to being a standard textbook-style exercise.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.03e Find cdf: by integration

The discrete random variable \(X\) has probability function P\((X = x) = k(x + 4)\). Given that \(X\) can take any of the values \(-3, -2, -1, 0, 1, 2, 3, 4\),
  1. find the value of the constant \(k\). [3 marks]
  2. Find P\((X < 0)\). [2 marks]
  3. Show that the cumulative distribution F\((x)\) is given by $$\text{F}(x) = \lambda(x + 4)(x + 5)$$ where \(\lambda\) is a constant to be found. [6 marks]

AnswerMarks
(a) \(k(1 + 2 + \ldots + 8) = 1\)M1 M1 A1
\(36k = 1\), so \(k = \frac{1}{36}\)
(b) \(P(X < 0) = \frac{5}{36} \times \frac{1}{6}\)M1 A1
(c) \(F(X) = \frac{1}{36} + \frac{2}{36} + \ldots + \frac{x+4}{36} = \frac{1}{36}(1 + 2 + \ldots [x+4])\)M1 A1
\(= \frac{1}{2} \times \frac{1}{36}(x+4)(x+5) + 1) = \frac{1}{72}(x+4)(x+5)\)M1 M1 A1 A1
11 marks total
(a) $k(1 + 2 + \ldots + 8) = 1$ | M1 M1 A1 |
$36k = 1$, so $k = \frac{1}{36}$ |

(b) $P(X < 0) = \frac{5}{36} \times \frac{1}{6}$ | M1 A1 |

(c) $F(X) = \frac{1}{36} + \frac{2}{36} + \ldots + \frac{x+4}{36} = \frac{1}{36}(1 + 2 + \ldots [x+4])$ | M1 A1 |
$= \frac{1}{2} \times \frac{1}{36}(x+4)(x+5) + 1) = \frac{1}{72}(x+4)(x+5)$ | M1 M1 A1 A1 |
| 11 marks total |
The discrete random variable $X$ has probability function P$(X = x) = k(x + 4)$.

Given that $X$ can take any of the values $-3, -2, -1, 0, 1, 2, 3, 4$,

\begin{enumerate}[label=(\alph*)]
\item find the value of the constant $k$. [3 marks]
\item Find P$(X < 0)$. [2 marks]
\item Show that the cumulative distribution F$(x)$ is given by
$$\text{F}(x) = \lambda(x + 4)(x + 5)$$
where $\lambda$ is a constant to be found. [6 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1  Q4 [11]}}