Edexcel S2 — Question 4 13 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeStandard applied PDF calculations
DifficultyStandard +0.3 This is a straightforward S2 probability density function question requiring standard techniques: finding k by integration (∫k(10-t)dt = 1), computing E(T) = ∫tf(t)dt, and solving ∫f(t)dt = 0.95 for the percentile. The linear pdf makes all integrations simple, and part (d) requires only basic interpretation. Slightly easier than average due to the simple functional form and routine application of standard methods.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles

The waiting time, in minutes, at a dentist is modelled by the continuous random variable \(T\) with probability density function $$f(t) = k(10 - t) \quad 0 \leq t \leq 10,$$ $$f(t) = 0 \quad \text{otherwise}.$$
  1. Sketch the graph of \(f(t)\) and find the value of \(k\). [4 marks]
  2. Find the mean value of \(T\). [4 marks]
  3. Find the 95th percentile of \(T\). [3 marks]
  4. State whether you consider this function to be a sensible model for \(T\) and suggest how it could be modified to provide a better model. [2 marks]

AnswerMarks Guidance
(a) Graph: straight line from \((0, 10k)\) to \((10, 0)\); on x-axis otherwiseB2
\(\frac{1}{2} \times 10 \times 10k = 1\) \(k = \frac{1}{50}\)M1 A1
(b) \(E(T) = \int_0^{10} f(t) \, dt = \frac{1}{50} \int_0^{10} 10t - t^2 \, dt = \frac{1}{50}\left[5t^2 - \frac{t^3}{3}\right]_0^{10} = 3\frac{1}{3}\)M1 A1 M1 A1
(c) From graph, \(\frac{1}{2}(10-p)\left(\frac{10-p}{50}\right) = \frac{5}{100}\) \((10-p)^2 = 5\) \(p = 7.76\)M1 A1 M1 A1
(d) Some wait more than 10 mins; more gradual slope neededB1 B1 13 marks total
(a) Graph: straight line from $(0, 10k)$ to $(10, 0)$; on x-axis otherwise | B2 |

$\frac{1}{2} \times 10 \times 10k = 1$ $k = \frac{1}{50}$ | M1 A1 |

(b) $E(T) = \int_0^{10} f(t) \, dt = \frac{1}{50} \int_0^{10} 10t - t^2 \, dt = \frac{1}{50}\left[5t^2 - \frac{t^3}{3}\right]_0^{10} = 3\frac{1}{3}$ | M1 A1 M1 A1 |

(c) From graph, $\frac{1}{2}(10-p)\left(\frac{10-p}{50}\right) = \frac{5}{100}$ $(10-p)^2 = 5$ $p = 7.76$ | M1 A1 M1 A1 |

(d) Some wait more than 10 mins; more gradual slope needed | B1 B1 | 13 marks total
The waiting time, in minutes, at a dentist is modelled by the continuous random variable $T$ with probability density function
$$f(t) = k(10 - t) \quad 0 \leq t \leq 10,$$
$$f(t) = 0 \quad \text{otherwise}.$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $f(t)$ and find the value of $k$. [4 marks]
\item Find the mean value of $T$. [4 marks]
\item Find the 95th percentile of $T$. [3 marks]
\item State whether you consider this function to be a sensible model for $T$ and suggest how it could be modified to provide a better model. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q4 [13]}}