Edexcel S2 2019 January — Question 3 8 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2019
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypeCalculate probabilities from CDF
DifficultyModerate -0.3 This is a straightforward S2 question requiring reading values from a CDF graph, finding PDF constants using f(x) = F'(x) (which reduces to finding gradients of linear segments), and computing E(X) using standard integration. All steps are routine applications of well-practiced techniques with no conceptual challenges or novel problem-solving required.
Spec5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles

3. Figure 1 shows an accurate graph of the cumulative distribution function, \(\mathrm { F } ( x )\), for the continuous random variable \(X\) \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{17296edc-9ab4-4f81-ae68-c76190986fd1-08_535_1152_354_342} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure}
  1. Find \(\mathrm { P } ( 3 < X < 7 )\) The probability density function of \(X\) is given by $$\mathrm { f } ( x ) = \begin{cases} a & 2 \leqslant x < 4 \\ b & 4 \leqslant x < 6 \\ c & 6 \leqslant x \leqslant 8 \\ 0 & \text { otherwise } \end{cases}$$ where \(a\), \(b\) and \(c\) are constants.
  2. Find the value of \(a\), the value of \(b\) and the value of \(c\)
  3. Find \(\mathrm { E } ( X )\)

Question 3:
Part (a)
AnswerMarks Guidance
WorkingMark Guidance
\(P(3 < X < 7) = F(7) - F(3) = 0.7 - 0.2\)M1 For writing or using \(F(7) - F(3)\); implied by answer 0.5
\(= \mathbf{0.5}\)A1
Part (b)
AnswerMarks Guidance
WorkingMark Guidance
\(a = \frac{0.4-0}{4-2}\), \(b = \frac{0}{6-4}\), \(c = \frac{1-0.4}{8-6}\)M1 For attempt to find gradient of at least 1 line segment
\(\underline{a=0.2}\), \(\underline{b=0}\), \(\underline{c=0.3}\)A1 A1 1st A1: two values correct; 2nd A1: all three correct
Part (c)
AnswerMarks Guidance
WorkingMark Guidance
\(E(X) = \int_2^4 "0.2"\,x\,dx + \int_4^6 "0"\,x\,dx + \int_6^8 "0.3"\,x\,dx\)M1 For correct expression \(\int xf(x)\,dx\) (ignore limits) using their \(a\), \(b\), \(c\); and attempt to integrate (\(x^n \to x^{n+1}\)); no need to see \(\int_4^6 "0"\,x\,dx\) if \(b=0\), otherwise must be present
\(E(X) = \left[\frac{0.2x^2}{2}\right]_2^4 + \left[\frac{0.3x^2}{2}\right]_6^8\)A1ft Correct integration with correct limits (ft their \(a\), \(b\), \(c\))
\(= \mathbf{5.4}\)A1
Alternative:
AnswerMarks Guidance
WorkingMark Guidance
\(E(X) = 3 \times p + 7 \times 2 \times "c"\) or \(3 \times 2 \times "a" + 7 \times p\) where \(0 < p < 1\)M1 Must have 3, 7 and one half correct ft their \(a/c\)
\(= 3 \times "0.4" + 7 \times "0.6"\)A1ft \(3 \times 2 \times "a" + 7 \times 2 \times "c"\)
\(= \mathbf{5.4}\)A1
# Question 3:

## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $P(3 < X < 7) = F(7) - F(3) = 0.7 - 0.2$ | M1 | For writing or using $F(7) - F(3)$; implied by answer 0.5 |
| $= \mathbf{0.5}$ | A1 | |

## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $a = \frac{0.4-0}{4-2}$, $b = \frac{0}{6-4}$, $c = \frac{1-0.4}{8-6}$ | M1 | For attempt to find gradient of at least 1 line segment |
| $\underline{a=0.2}$, $\underline{b=0}$, $\underline{c=0.3}$ | A1 A1 | 1st A1: two values correct; 2nd A1: all three correct |

## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| $E(X) = \int_2^4 "0.2"\,x\,dx + \int_4^6 "0"\,x\,dx + \int_6^8 "0.3"\,x\,dx$ | M1 | For correct expression $\int xf(x)\,dx$ (ignore limits) using their $a$, $b$, $c$; and attempt to integrate ($x^n \to x^{n+1}$); no need to see $\int_4^6 "0"\,x\,dx$ if $b=0$, otherwise must be present |
| $E(X) = \left[\frac{0.2x^2}{2}\right]_2^4 + \left[\frac{0.3x^2}{2}\right]_6^8$ | A1ft | Correct integration with correct limits (ft their $a$, $b$, $c$) |
| $= \mathbf{5.4}$ | A1 | |

**Alternative:**
| Working | Mark | Guidance |
|---------|------|----------|
| $E(X) = 3 \times p + 7 \times 2 \times "c"$ or $3 \times 2 \times "a" + 7 \times p$ where $0 < p < 1$ | M1 | Must have 3, 7 and one half correct ft their $a/c$ |
| $= 3 \times "0.4" + 7 \times "0.6"$ | A1ft | $3 \times 2 \times "a" + 7 \times 2 \times "c"$ |
| $= \mathbf{5.4}$ | A1 | |

---
3. Figure 1 shows an accurate graph of the cumulative distribution function, $\mathrm { F } ( x )$, for the continuous random variable $X$

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{17296edc-9ab4-4f81-ae68-c76190986fd1-08_535_1152_354_342}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( 3 < X < 7 )$

The probability density function of $X$ is given by

$$\mathrm { f } ( x ) = \begin{cases} a & 2 \leqslant x < 4 \\ b & 4 \leqslant x < 6 \\ c & 6 \leqslant x \leqslant 8 \\ 0 & \text { otherwise } \end{cases}$$

where $a$, $b$ and $c$ are constants.
\item Find the value of $a$, the value of $b$ and the value of $c$
\item Find $\mathrm { E } ( X )$
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2019 Q3 [8]}}