Edexcel FS2 2023 June — Question 7 9 marks

Exam BoardEdexcel
ModuleFS2 (Further Statistics 2)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeE(g(X)) and Var(g(X)) by integration
DifficultyStandard +0.3 This is a straightforward application of E(g(X)) for a uniform distribution. Part (a) is trivial recall, part (b) is a guided 'show that' requiring integration of r² over [2,10], and part (c) requires integration of r³. All steps are routine for FS2 students with no conceptual challenges beyond applying the standard formula E(g(X)) = ∫g(x)f(x)dx. Slightly easier than average due to the simple uniform pdf and polynomial functions.
Spec5.02a Discrete probability distributions: general5.03c Calculate mean/variance: by integration
  1. The random variable \(R\) has a continuous uniform distribution over the interval \([ 2,10 ]\)
    1. Write down the probability density function \(\mathrm { f } ( r )\) of \(R\)
    A sphere of radius \(R \mathrm {~cm}\) is formed.
    The surface area of the sphere, \(S \mathrm {~cm} ^ { 2 }\), is given by \(S = 4 \pi R ^ { 2 }\)
  2. Show that \(\mathrm { E } ( S ) = \frac { 496 \pi } { 3 }\) The volume of the sphere, \(V \mathrm {~cm} ^ { 3 }\), is given by \(V = \frac { 4 } { 3 } \pi R ^ { 3 }\)
  3. Find, using algebraic integration, the expected value of \(V\)
Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(r) = \frac{1}{8}\) for \(2 \leq r \leq 10\) [0 otherwise]B1 1.1b - correct pdf with limits; allow any letter but must be consistent
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(4\pi R^2) = 4\pi E(R^2)\); \(\int_2^{10} 4\pi r^2 \cdot \frac{1}{8}\, dr\)M1 1.1b - use of \(E(aR) = aE(R)\); correct integral with their \(f(r)\)
\(E(R) = 6\); \(= \left[\frac{1}{2}\pi\left(\frac{r^3}{3}\right)\right]_2^{10}\)M1 1.1b - \(E(R)=6\) seen or implied; use of integration
\(E(4\pi R^2) = 4\pi(\text{Var}(R) + [E(R)]^2) = 4\pi\left(\frac{16}{3} + 36\right)\); \(= \frac{1}{2}\pi\left(\frac{10^3}{3} - \frac{2^3}{3}\right)\)M1 2.1 - use of \(E(R^2) = \text{Var}(R) + [E(R)]^2\)
\(= \frac{496\pi}{3}\)A1* 2.2a - exact value from correct working
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E\!\left(\frac{4}{3}\pi R^3\right) = \int_2^{10} \frac{4}{3}\pi r^3 \cdot \frac{1}{8}\, dr\)B1ft 1.1b - correct integral for \(E(V)\) including limits
\(= \left[\frac{1}{6}\pi\left(\frac{r^4}{4}\right)\right]_2^{10}\)M1 2.1 - integration of their pdf to find \(E(V)\)
\(= \frac{1}{6}\pi\left(\frac{10^4}{4} - \frac{2^4}{4}\right)\)dM1 1.1b - dep on previous M1; use of correct limits
\(= 416\pi\)A1 2.2a - allow awrt 1310 to 3 sf 
# Question 7:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(r) = \frac{1}{8}$ for $2 \leq r \leq 10$ [0 otherwise] | B1 | 1.1b - correct pdf with limits; allow any letter but must be consistent |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(4\pi R^2) = 4\pi E(R^2)$; $\int_2^{10} 4\pi r^2 \cdot \frac{1}{8}\, dr$ | M1 | 1.1b - use of $E(aR) = aE(R)$; correct integral with their $f(r)$ |
| $E(R) = 6$; $= \left[\frac{1}{2}\pi\left(\frac{r^3}{3}\right)\right]_2^{10}$ | M1 | 1.1b - $E(R)=6$ seen or implied; use of integration |
| $E(4\pi R^2) = 4\pi(\text{Var}(R) + [E(R)]^2) = 4\pi\left(\frac{16}{3} + 36\right)$; $= \frac{1}{2}\pi\left(\frac{10^3}{3} - \frac{2^3}{3}\right)$ | M1 | 2.1 - use of $E(R^2) = \text{Var}(R) + [E(R)]^2$ |
| $= \frac{496\pi}{3}$ | A1* | 2.2a - exact value from correct working |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E\!\left(\frac{4}{3}\pi R^3\right) = \int_2^{10} \frac{4}{3}\pi r^3 \cdot \frac{1}{8}\, dr$ | B1ft | 1.1b - correct integral for $E(V)$ including limits |
| $= \left[\frac{1}{6}\pi\left(\frac{r^4}{4}\right)\right]_2^{10}$ | M1 | 2.1 - integration of their pdf to find $E(V)$ |
| $= \frac{1}{6}\pi\left(\frac{10^4}{4} - \frac{2^4}{4}\right)$ | dM1 | 1.1b - dep on previous M1; use of correct limits |
| $= 416\pi$ | A1 | 2.2a - allow awrt 1310 to 3 sf |

---
\begin{enumerate}
  \item The random variable $R$ has a continuous uniform distribution over the interval $[ 2,10 ]$\\
(a) Write down the probability density function $\mathrm { f } ( r )$ of $R$
\end{enumerate}

A sphere of radius $R \mathrm {~cm}$ is formed.\\
The surface area of the sphere, $S \mathrm {~cm} ^ { 2 }$, is given by $S = 4 \pi R ^ { 2 }$\\
(b) Show that $\mathrm { E } ( S ) = \frac { 496 \pi } { 3 }$

The volume of the sphere, $V \mathrm {~cm} ^ { 3 }$, is given by $V = \frac { 4 } { 3 } \pi R ^ { 3 }$\\
(c) Find, using algebraic integration, the expected value of $V$

\hfill \mbox{\textit{Edexcel FS2 2023 Q7 [9]}}
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