| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Calculate probabilities from CDF |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics 2 question testing standard CDF operations: reading probabilities from the CDF formula, finding the median by solving F(x)=0.5, differentiating to get the pdf for E(X²), and sketching. All parts follow routine procedures with no novel insight required, though the E(X²) integration and skewness interpretation add slight complexity beyond basic recall. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(F(5) + (1 - F(8))\) giving \(\left[\frac{3}{4} + \left(1 - \frac{15}{16}\right)\right]\) | M1 | Equivalent correct probability statement, e.g. \(1-(F(8)-F(5))\) |
| \(= \frac{13}{16}\) | A1 | \(\frac{13}{16}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(F(m) = 0.5\) giving \(\left[1.25 - \frac{2.5}{m} = 0.5\right]\) | M1 | Use of \(F(m) = 0.5\) |
| \(m = \frac{10}{3}\) | A1 | \(\frac{10}{3}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(f(x) \left[= \frac{d}{dx}(F(x))\right] = 2.5x^{-2}\) | M1 | Realising that \(f(x)\) is required and attempting to differentiate \(F(x)\) |
| \(E(X^2)\left[= \int_2^{10} x^2 f(x)\, dx\right] = \int_2^{10} 2.5\, dx\) | M1 | Use of \(\int_2^{10} x^2 f(x)\, dx\) |
| \(= 20\) | A1 | 20 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Correct shape (decreasing curve) | M1 | Correct shape |
| 2 and 10 correctly labelled on horizontal axis | A1 | Correct labels |
| Therefore positive skew | A1 | Positive skew provided M1 scored |
# Question 3:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $F(5) + (1 - F(8))$ giving $\left[\frac{3}{4} + \left(1 - \frac{15}{16}\right)\right]$ | M1 | Equivalent correct probability statement, e.g. $1-(F(8)-F(5))$ |
| $= \frac{13}{16}$ | A1 | $\frac{13}{16}$ oe |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $F(m) = 0.5$ giving $\left[1.25 - \frac{2.5}{m} = 0.5\right]$ | M1 | Use of $F(m) = 0.5$ |
| $m = \frac{10}{3}$ | A1 | $\frac{10}{3}$ oe |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $f(x) \left[= \frac{d}{dx}(F(x))\right] = 2.5x^{-2}$ | M1 | Realising that $f(x)$ is required and attempting to differentiate $F(x)$ |
| $E(X^2)\left[= \int_2^{10} x^2 f(x)\, dx\right] = \int_2^{10} 2.5\, dx$ | M1 | Use of $\int_2^{10} x^2 f(x)\, dx$ |
| $= 20$ | A1 | 20 cao |
## Part (d)(i) and (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Correct shape (decreasing curve) | M1 | Correct shape |
| 2 and 10 correctly labelled on horizontal axis | A1 | Correct labels |
| Therefore positive skew | A1 | Positive skew provided M1 scored |
---\begin{enumerate}
\item The continuous random variable $X$ has cumulative distribution function given by
\end{enumerate}
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c r }
0 & x < 2 \\
1.25 - \frac { 2.5 } { x } & 2 \leqslant x \leqslant 10 \\
1 & x > 10
\end{array} \right.$$
(a) Find $\mathrm { P } ( \{ X < 5 \} \cup \{ X > 8 \} )$\\
(b) Find the median of $X$.\\
(c) Find $\mathrm { E } \left( X ^ { 2 } \right)$\\
(d) (i) Sketch the probability density function of $X$.\\
(ii) Describe the skewness of the distribution of $X$.
\hfill \mbox{\textit{Edexcel FS2 2021 Q3 [10]}}