| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Calculate probabilities from CDF |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring direct application of CDF properties: part (a) uses P(X>a)=1-F(a), part (b) applies conditional probability formula with CDF values, and part (c) requires finding the pdf by differentiation then integrating xf(x). All steps are standard procedures with no conceptual challenges, making it slightly easier than average. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(1 - F(4) = 1 - \frac{1}{16}(4-1)^2 = \frac{7}{16}\) | M1 A1 (2) | M1 for writing or using \(1 - F(4)\); A1 for \(\frac{7}{16}\) oe (allow 0.4375 or 0.438) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
\([P(X>3 \mid 2| M1, dM1, A1 (3) |
1st M1 for writing or using \(F(4)-F(3)\) (may be implied by \(\frac{5}{16}\) or 0.3125); 2nd dM1 (dep on 1st M1) for a ratio of probabilities with \(F(4)-F(2)\) in denominator (may be implied by \(\frac{1}{2}\)); do not award 2nd M1 if numerator > denominator; A1 for \(\frac{5}{8}\) or 0.625 |
|
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(f(x) = \frac{d}{dx}F(x) = \frac{1}{8}(x-1)\) | M1 | 1st M1 for differentiating \(F(x)\) to find \(f(x)\) (at least one \(x^n \to x^{n-1}\)) |
| \(E(X) = \int_1^5 \frac{1}{8}x(x-1)\,dx\) | dM1 | 2nd dM1 (dep on 1st M1) for multiplying \(x \cdot f(x)\) and integrating (at least one \(x^n \to x^{n+1}\)) |
| \(E(X) = \left[\frac{1}{24}x^3 - \frac{1}{16}x^2\right]_1^5 = \left(\frac{5^3}{24}-\frac{5^2}{16}\right) - \left(\frac{1}{24}-\frac{1}{16}\right) = \dfrac{11}{3}\) | dM1 A1 (4) | 3rd dM1 (dep on 2nd M1) for substitution of correct limits; may be implied by \(\frac{175}{48}-\left(-\frac{1}{48}\right)\); A1 for \(\frac{11}{3}\) or awrt 3.67 |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $1 - F(4) = 1 - \frac{1}{16}(4-1)^2 = \frac{7}{16}$ | M1 A1 (2) | M1 for writing or using $1 - F(4)$; A1 for $\frac{7}{16}$ oe (allow 0.4375 or 0.438) |
## Part (b)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $[P(X>3 \mid 2<X<4)=] \dfrac{F(4)-F(3)}{F(4)-F(2)} = \dfrac{\frac{9}{16}-\frac{4}{16}}{\frac{9}{16}-\frac{1}{16}} = \dfrac{5}{8}$ | M1, dM1, A1 (3) | 1st M1 for writing or using $F(4)-F(3)$ (may be implied by $\frac{5}{16}$ or 0.3125); 2nd dM1 (dep on 1st M1) for a ratio of probabilities with $F(4)-F(2)$ in denominator (may be implied by $\frac{1}{2}$); do not award 2nd M1 if numerator > denominator; A1 for $\frac{5}{8}$ or 0.625 |
## Part (c)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $f(x) = \frac{d}{dx}F(x) = \frac{1}{8}(x-1)$ | M1 | 1st M1 for differentiating $F(x)$ to find $f(x)$ (at least one $x^n \to x^{n-1}$) |
| $E(X) = \int_1^5 \frac{1}{8}x(x-1)\,dx$ | dM1 | 2nd dM1 (dep on 1st M1) for multiplying $x \cdot f(x)$ and integrating (at least one $x^n \to x^{n+1}$) |
| $E(X) = \left[\frac{1}{24}x^3 - \frac{1}{16}x^2\right]_1^5 = \left(\frac{5^3}{24}-\frac{5^2}{16}\right) - \left(\frac{1}{24}-\frac{1}{16}\right) = \dfrac{11}{3}$ | dM1 A1 (4) | 3rd dM1 (dep on 2nd M1) for substitution of correct limits; may be implied by $\frac{175}{48}-\left(-\frac{1}{48}\right)$; A1 for $\frac{11}{3}$ or awrt 3.67 |
**Total: 9 marks**\begin{enumerate}
\item A continuous random variable $X$ has cumulative distribution function
\end{enumerate}
$$\mathrm { F } ( x ) = \left\{ \begin{array} { l r }
0 & x < 1 \\
\frac { 1 } { 16 } ( x - 1 ) ^ { 2 } & 1 \leqslant x \leqslant 5 \\
1 & x > 5
\end{array} \right.$$
(a) Find $\mathrm { P } ( X > 4 )$\\
(b) Find $\mathrm { P } ( X > 3 \mid 2 < X < 4 )$\\
(c) Find the exact value of $\mathrm { E } ( X )$\\
\hfill \mbox{\textit{Edexcel S2 2018 Q1 [9]}}