Edexcel S2 — Question 5 10 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeMultiple observations or trials
DifficultyModerate -0.3 This is a straightforward application of continuous uniform distribution with standard probability calculations. Parts (a) and (b) require basic probability density function evaluation, while part (c) involves a simple binomial probability calculation. The question is slightly easier than average as it's a direct application of formulas with no conceptual challenges or problem-solving required.
Spec5.02b Expectation and variance: discrete random variables5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf

5. In a party game, a bottle is spun and whoever it points to when it stops has to play next. The acute angle, in degrees, that the bottle makes with the side of the room is modelled by a rectangular distribution over the interval [0,90]. Find the probability that on one spin this angle is
  1. between \(25 ^ { \circ }\) and \(38 ^ { \circ }\),
  2. \(45 ^ { \circ }\) to the nearest degree. The bottle is spun ten times.
  3. Find the probability that the acute angle it makes with the side of the room is less than \(10 ^ { \circ }\) more than twice.

AnswerMarks Guidance
(a) \(13 \times \frac{1}{90} = \frac{13}{90}\) or 0.1444 (4sf)M1 A1
(b) \(P(44.5° \text{ to } 45.5°) \therefore \frac{1}{90}\)M1 A1
(c) \(P(< 10°) = 10 \times \frac{1}{90} = \frac{1}{9}\); let X = no. of times \(< 10° \therefore X \sim B(10, \frac{1}{9})\); \(P(X > 2) = 1 - P(X \le 2) = 1 - [(\frac{8}{9})^{10} + 10(\frac{1}{9})(\frac{8}{9})^9 + \frac{10\cdot9}{2}(\frac{1}{9})^2(\frac{8}{9})^8] = 1 - 0.9094 = 0.0906\) (3sf)A1 M1 M1 M1 A1 (10 marks total)
(a) $13 \times \frac{1}{90} = \frac{13}{90}$ or 0.1444 (4sf) | M1 A1 |

(b) $P(44.5° \text{ to } 45.5°) \therefore \frac{1}{90}$ | M1 A1 |

(c) $P(< 10°) = 10 \times \frac{1}{90} = \frac{1}{9}$; let X = no. of times $< 10° \therefore X \sim B(10, \frac{1}{9})$; $P(X > 2) = 1 - P(X \le 2) = 1 - [(\frac{8}{9})^{10} + 10(\frac{1}{9})(\frac{8}{9})^9 + \frac{10\cdot9}{2}(\frac{1}{9})^2(\frac{8}{9})^8] = 1 - 0.9094 = 0.0906$ (3sf) | A1 M1 M1 M1 A1 | (10 marks total)

---
5. In a party game, a bottle is spun and whoever it points to when it stops has to play next. The acute angle, in degrees, that the bottle makes with the side of the room is modelled by a rectangular distribution over the interval [0,90].

Find the probability that on one spin this angle is
\begin{enumerate}[label=(\alph*)]
\item between $25 ^ { \circ }$ and $38 ^ { \circ }$,
\item $45 ^ { \circ }$ to the nearest degree.

The bottle is spun ten times.
\item Find the probability that the acute angle it makes with the side of the room is less than $10 ^ { \circ }$ more than twice.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q5 [10]}}