| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Direct variance calculation from pdf |
| Difficulty | Standard +0.3 This is a standard S2 continuous probability distribution question requiring routine integration techniques: finding k by normalizing the pdf, computing E(X) and Var(X) using standard formulas, finding the CDF by integration, and evaluating a probability. All steps are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks |
|---|---|
| (a) \(k\int_0^1 x^2 - x^3 \, dx = 1\) | M1 A1 |
| \(k\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 1\) | (working shown) |
| \(k = 12\) | A1 |
| Graph sketched: parabola, vertex upwards, through (0, 0), (1, 0) | B2 |
| (b) \(E(X) = 12\int_0^1 x^3 - x^4 \, dx = 12\left[\frac{1}{4} - \frac{1}{5}\right] = 0.6\) | M1 A1 A1 |
| (c) \(\text{Var}(X) = 12\int_0^1 x^4 - x^5 \, dx - 0.6^2 = 12\left[\frac{1}{5} - \frac{1}{6}\right] - 0.36 = 0.04\) | M1 A1 A1 |
| (d) \(F(x) = 0\) (\(x < 0\)), \(F(x) = 4x^3 - 3x^4\) (\(0 \leq x \leq 1\)), \(F(x) = 1\) (\(x > 1\)) | B1 M1 A1 B1 |
| (e) \(P(x \leq 0.85) = F(0.85) = 4(0.85)^3 - 3(0.85)^4 = 0.89\), so cloud cover is \(\leq 85\%\) for 89% of the time | M1 A1 A1 |
(a) $k\int_0^1 x^2 - x^3 \, dx = 1$ | M1 A1 |
$k\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 1$ | (working shown) |
$k = 12$ | A1 |
Graph sketched: parabola, vertex upwards, through (0, 0), (1, 0) | B2 |
(b) $E(X) = 12\int_0^1 x^3 - x^4 \, dx = 12\left[\frac{1}{4} - \frac{1}{5}\right] = 0.6$ | M1 A1 A1 |
(c) $\text{Var}(X) = 12\int_0^1 x^4 - x^5 \, dx - 0.6^2 = 12\left[\frac{1}{5} - \frac{1}{6}\right] - 0.36 = 0.04$ | M1 A1 A1 |
(d) $F(x) = 0$ ($x < 0$), $F(x) = 4x^3 - 3x^4$ ($0 \leq x \leq 1$), $F(x) = 1$ ($x > 1$) | B1 M1 A1 B1 |
(e) $P(x \leq 0.85) = F(0.85) = 4(0.85)^3 - 3(0.85)^4 = 0.89$, so cloud cover is $\leq 85\%$ for 89% of the time | M1 A1 A1 |
**Total: 17 marks**7. The fraction of sky covered by cloud is modelled by the random variable $X$ with probability density function
$$\begin{array} { l l }
\mathrm { f } ( x ) = 0 & x < 0 \\
\mathrm { f } ( x ) = k x ^ { 2 } ( 1 - x ) & 0 \leq x \leq 1 , \\
\mathrm { f } ( x ) = 0 & x > 1 .
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find $k$ and sketch the graph of $\mathrm { f } ( x )$.
\item Find the mean and the variance of $X$.
\item Find the cumulative distribution function $\mathrm { F } ( x )$.
\item Given that flying is prohibited when $85 \%$ of the sky is covered by cloud, show that cloud conditions allow flying nearly $90 \%$ of the time.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [17]}}