Edexcel S2 Specimen — Question 7 20 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
SessionSpecimen
Marks20
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeInterquartile range calculation
DifficultyStandard +0.3 This is a standard S2 continuous probability distribution question requiring routine application of well-practiced techniques: sketching a pdf, calculating E(X) and Var(X) using standard integrals, finding the CDF by integration, and solving quartile equations. All steps are procedural with no novel insight required, making it slightly easier than average for A-level.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.03a 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

The continuous random variable \(X\) has probability density function f(\(x\)) given by $$\text{f}(x) = \begin{cases} \frac{1}{20}x^3, & 1 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases}$$
  1. Sketch f(\(x\)) for all values of \(x\). [3]
  2. Calculate E(\(X\)). [3]
  3. Show that the standard deviation of \(X\) is 0.459 to 3 decimal places. [3]
  4. Show that for \(1 \leq x \leq 3\), P(\(X \leq x\)) is given by \(\frac{1}{80}(x^4 - 1)\) and specify fully the cumulative distribution function of \(X\). [5]
  5. Find the interquartile range for the random variable \(X\). [4]
Some statisticians use the following formula to estimate the interquartile range: $$\text{interquartile range} = \frac{4}{3} \times \text{standard deviation}.$$
  1. Use this formula to estimate the interquartile range in this case, and comment. [2]

Part (a)
AnswerMarks Guidance
Graph showing curve starting at \(\left(\frac{1}{20}, \frac{1}{20}\right)\), passing through \((3, \frac{27}{20})\)B1, B1 B1 (3 marks)
Part (b)
AnswerMarks Guidance
\(E(X) = \int_1^3 \frac{1}{20}x^3 dx = \left[\frac{x^5}{100}\right]_1^3 = \frac{242}{100} = 2.42\)M1 [M1] A1 (3 marks)
Part (c)
\(\sigma^2 = \int_0^3 \frac{1}{20}x^5 dx - \mu^2 = \left[\frac{x^6}{120}\right]_1^3 - \mu^2 = \frac{728}{120} - (2.42)^2 = 0.21026\)
AnswerMarks Guidance
\(\therefore \sigma = 0.459\)M1 [M1] A1 cso (3 marks)
Part (d)
AnswerMarks Guidance
\(P(X \leq x) = \int_1^x \frac{1}{20}t^3 dt = \left[\frac{t^4}{80}\right]_1^x = \frac{x^4}{80} - \frac{1}{80}\)M1 [M1] A1 cso
\[F(x) = \begin{cases} 0 & x \leq 1 \\ \frac{1}{80}(x^4 - 1) & 1 < x < 3 \\ 1 & x \geq 3 \end{cases}\]B1 ft, centre B1 ends (5 marks)
Part (e)
\(F(p) = 0.25 = \frac{1}{80}(p^4 - 1) = \frac{1}{4} \therefore p^4 = 21 \Rightarrow p = 2.14\ldots\)
\(F(q) = 0.75 = \frac{1}{80}(q^4 - 1) = \frac{3}{4} \therefore q^4 = 61 \Rightarrow q = 2.79\ldots\)
AnswerMarks Guidance
\(\text{IQR} = 0.65\)M1 A1 A1 A1 ft (4 marks)
Part (f)
\(\text{IQR} \approx \frac{4}{3} \times 0.459 = 0.612\)
AnswerMarks Guidance
Sensible comment, e.g. reasonable approximation or slight underestimateB1 B1 (2 marks)
Total: 20 marks
## Part (a)
Graph showing curve starting at $\left(\frac{1}{20}, \frac{1}{20}\right)$, passing through $(3, \frac{27}{20})$ | B1, B1 B1 | (3 marks)

## Part (b)
$E(X) = \int_1^3 \frac{1}{20}x^3 dx = \left[\frac{x^5}{100}\right]_1^3 = \frac{242}{100} = 2.42$ | M1 [M1] A1 | (3 marks)

## Part (c)
$\sigma^2 = \int_0^3 \frac{1}{20}x^5 dx - \mu^2 = \left[\frac{x^6}{120}\right]_1^3 - \mu^2 = \frac{728}{120} - (2.42)^2 = 0.21026$

$\therefore \sigma = 0.459$ | M1 [M1] A1 cso | (3 marks)

## Part (d)
$P(X \leq x) = \int_1^x \frac{1}{20}t^3 dt = \left[\frac{t^4}{80}\right]_1^x = \frac{x^4}{80} - \frac{1}{80}$ | M1 [M1] A1 cso

$$F(x) = \begin{cases} 0 & x \leq 1 \\ \frac{1}{80}(x^4 - 1) & 1 < x < 3 \\ 1 & x \geq 3 \end{cases}$$ | B1 ft, centre B1 ends | (5 marks)

## Part (e)
$F(p) = 0.25 = \frac{1}{80}(p^4 - 1) = \frac{1}{4} \therefore p^4 = 21 \Rightarrow p = 2.14\ldots$

$F(q) = 0.75 = \frac{1}{80}(q^4 - 1) = \frac{3}{4} \therefore q^4 = 61 \Rightarrow q = 2.79\ldots$

$\text{IQR} = 0.65$ | M1 A1 A1 A1 ft | (4 marks)

## Part (f)
$\text{IQR} \approx \frac{4}{3} \times 0.459 = 0.612$

Sensible comment, e.g. reasonable approximation or slight underestimate | B1 B1 | (2 marks)

**Total: 20 marks**
The continuous random variable $X$ has probability density function f($x$) given by

$$\text{f}(x) = \begin{cases}
\frac{1}{20}x^3, & 1 \leq x \leq 3 \\
0, & \text{otherwise}
\end{cases}$$

\begin{enumerate}[label=(\alph*)]
\item Sketch f($x$) for all values of $x$. [3]
\item Calculate E($X$). [3]
\item Show that the standard deviation of $X$ is 0.459 to 3 decimal places. [3]
\item Show that for $1 \leq x \leq 3$, P($X \leq x$) is given by $\frac{1}{80}(x^4 - 1)$ and specify fully the cumulative distribution function of $X$. [5]
\item Find the interquartile range for the random variable $X$. [4]
\end{enumerate}

Some statisticians use the following formula to estimate the interquartile range:

$$\text{interquartile range} = \frac{4}{3} \times \text{standard deviation}.$$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{5}
\item Use this formula to estimate the interquartile range in this case, and comment. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q7 [20]}}