CAIE S2 2015 June — Question 1 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeGeometric/graphical PDF with k
DifficultyModerate -0.8 This is a straightforward PDF question requiring use of the fundamental property that total area equals 1 to find the constant, followed by a standard expectation calculation using the given linear function. The geometric shape (trapezoid) makes integration simple, and the answer is provided to verify the calculation. This is below average difficulty as it's purely procedural with no problem-solving insight required.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration
1 \includegraphics[max width=\textwidth, alt={}, center]{cfffe79d-91c9-48b8-a3e6-887d7891441d-2_478_691_260_724} The random variable \(X\) has probability density function, f , as shown in the diagram, where \(a\) is a constant. Find the value of \(a\) and hence show that \(\mathrm { E } ( X ) = 0.943\) correct to 3 significant figures. [5]
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{1}{2}a^2 = 1\)M1 or \(\int_0^a x\,dx = 1\)
\(a = \sqrt{2}\)A1 Allow 1.41 or better
\(\int_0^{\sqrt{2}} x^2\,dx\)M1 ignore limits
\(= \left[\frac{x^3}{3}\right]_0^{\sqrt{2}}\)A1f correct integral and limits, but ft their \(a\)
\(= \frac{(\sqrt{2})^3}{3} = \frac{2^{1.5}}{3}\) or \(\frac{2.83}{3}\) or 0.9428 \((= 0.943)\) AGA1 [5] must see this numerical expression, or equiv. SR: Equating \(\int x\,f(x)\) to 0.943 scores M1. Solving to find \(a = 1.41\) scores A1 
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}a^2 = 1$ | M1 | or $\int_0^a x\,dx = 1$ |
| $a = \sqrt{2}$ | A1 | Allow 1.41 or better |
| $\int_0^{\sqrt{2}} x^2\,dx$ | M1 | ignore limits |
| $= \left[\frac{x^3}{3}\right]_0^{\sqrt{2}}$ | A1f | correct integral and limits, but ft their $a$ |
| $= \frac{(\sqrt{2})^3}{3} = \frac{2^{1.5}}{3}$ or $\frac{2.83}{3}$ or 0.9428 $(= 0.943)$ **AG** | A1 [5] | must see this numerical expression, or equiv. SR: Equating $\int x\,f(x)$ to 0.943 scores M1. Solving to find $a = 1.41$ scores A1 |

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1\\
\includegraphics[max width=\textwidth, alt={}, center]{cfffe79d-91c9-48b8-a3e6-887d7891441d-2_478_691_260_724}

The random variable $X$ has probability density function, f , as shown in the diagram, where $a$ is a constant. Find the value of $a$ and hence show that $\mathrm { E } ( X ) = 0.943$ correct to 3 significant figures. [5]

\hfill \mbox{\textit{CAIE S2 2015 Q1 [5]}}
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