| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with k |
| Difficulty | Standard +0.3 This is a standard S2 piecewise PDF question covering routine techniques: finding k by integration (∫f=1), sketching, identifying mode by inspection, computing variance using Var(X)=E(X²)-[E(X)]², finding CDF constants by continuity, and solving F(x)=0.75 for upper quartile. All steps are textbook exercises requiring careful algebra but no novel insight or problem-solving. Slightly easier than average due to being methodical and guided. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int_1^3 \frac{1}{16}x^2\,dx + \int_3^4 k(4-x)\,dx = 1\) | M1 | Equating sum of two expressions for area to 1 (ignore limits) |
| \(\left[\frac{x^3}{48}\right]_1^3 + \left[k\left(4x - \frac{x^2}{2}\right)\right]_3^4 = 1\) | M1 | Correct use of limits to obtain a linear equation in \(k\) |
| \(\left(\frac{27}{48} - \frac{1}{48}\right) + k\left((16-8)-(12-\frac{9}{2})\right) = 1\) | ||
| \(k = \frac{11}{12}\) | A1cao | For \(k=\frac{11}{12}\) with correct integration and no incorrect working seen |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Correct shape for \(1 \leq x < 3\) | B1 | Positive quadratic starting above \(x\)-axis with correct curvature |
| Correct shape for \(3 \leq x \leq 4\) | B1 | Line with negative gradient starting above the quadratic, finishing on \(x\)-axis; labels 1, 3 and 4 on \(x\)-axis; ignore sketch outside \(1 \leq x \leq 4\) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([X =]\, 3\) | B1 | |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\text{E}(X^2) = \int_1^3 \frac{1}{16}x^4\,dx + \int_3^4 \frac{11}{12}(4x^2 - x^3)\,dx\) | M1 | Attempt to find \(\text{E}(X^2)\) by multiplying \(f(x)\) by \(x^2\); attempt to integrate \(x^n \to x^{n+1}\) |
| \(\text{E}(X^2) = \left[\frac{1}{80}x^5\right]_1^3 + \left[\frac{11}{12}\left(\frac{4}{3}x^3 - \frac{1}{4}x^4\right)\right]_3^4\) | dM1 | Correct integration with correct limits (condone ft on \(k\)) |
| \(\text{Var}(X) = \frac{5863}{720} - \left(\frac{25}{9}\right)^2 = \frac{2767}{6480} \approx 0.427\ldots\) | M1, A1 | Use of \(\text{Var}(X) = \text{E}(X^2) - \left(\frac{25}{9}\right)^2\); awrt 0.427 |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(c = -1\) | B1cao | |
| (see (3)) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(F(4) = 1\) | ||
| \(\frac{11}{12}(4(4) - \frac{1}{2}(4^2) + d) = 1\) | M1 | Use of \(F(4)=1\) [or use of \(F(3) = \frac{13}{24}\)] (condone ft on \(k\)) |
| \(d = -\frac{76}{11}\) | A1cao | |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{11}{12}\left(4x - \frac{1}{2}x^2 - \frac{76}{11}\right) = 0.75\) | M1 | Equating their third line of CDF to 0.75 (condone ft on \(k\)) |
| \(11x^2 - 88x + 170 = 0\) | ||
| \(x = 3.26\) only | A1 | |
| (2) | ||
| Total 15 |
# Question 7:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $\int_1^3 \frac{1}{16}x^2\,dx + \int_3^4 k(4-x)\,dx = 1$ | M1 | Equating sum of two expressions for area to 1 (ignore limits) |
| $\left[\frac{x^3}{48}\right]_1^3 + \left[k\left(4x - \frac{x^2}{2}\right)\right]_3^4 = 1$ | M1 | Correct use of limits to obtain a linear equation in $k$ |
| $\left(\frac{27}{48} - \frac{1}{48}\right) + k\left((16-8)-(12-\frac{9}{2})\right) = 1$ | | |
| $k = \frac{11}{12}$ | A1cao | For $k=\frac{11}{12}$ with correct integration and no incorrect working seen |
| | **(3)** | |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| Correct shape for $1 \leq x < 3$ | B1 | Positive quadratic starting above $x$-axis with correct curvature |
| Correct shape for $3 \leq x \leq 4$ | B1 | Line with negative gradient starting above the quadratic, finishing on $x$-axis; labels 1, 3 and 4 on $x$-axis; ignore sketch outside $1 \leq x \leq 4$ |
| | **(2)** | |
## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| $[X =]\, 3$ | B1 | |
| | **(1)** | |
## Part (d)
| Working | Mark | Guidance |
|---------|------|----------|
| $\text{E}(X^2) = \int_1^3 \frac{1}{16}x^4\,dx + \int_3^4 \frac{11}{12}(4x^2 - x^3)\,dx$ | M1 | Attempt to find $\text{E}(X^2)$ by multiplying $f(x)$ by $x^2$; attempt to integrate $x^n \to x^{n+1}$ |
| $\text{E}(X^2) = \left[\frac{1}{80}x^5\right]_1^3 + \left[\frac{11}{12}\left(\frac{4}{3}x^3 - \frac{1}{4}x^4\right)\right]_3^4$ | dM1 | Correct integration with correct limits (condone ft on $k$) |
| $\text{Var}(X) = \frac{5863}{720} - \left(\frac{25}{9}\right)^2 = \frac{2767}{6480} \approx 0.427\ldots$ | M1, A1 | Use of $\text{Var}(X) = \text{E}(X^2) - \left(\frac{25}{9}\right)^2$; awrt **0.427** |
| | **(4)** | |
## Part (e)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $c = -1$ | B1cao | |
| | **(see (3))** | |
## Part (e)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $F(4) = 1$ | | |
| $\frac{11}{12}(4(4) - \frac{1}{2}(4^2) + d) = 1$ | M1 | Use of $F(4)=1$ [or use of $F(3) = \frac{13}{24}$] (condone ft on $k$) |
| $d = -\frac{76}{11}$ | A1cao | |
| | **(3)** | |
## Part (f)
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{11}{12}\left(4x - \frac{1}{2}x^2 - \frac{76}{11}\right) = 0.75$ | M1 | Equating their third line of CDF to 0.75 (condone ft on $k$) |
| $11x^2 - 88x + 170 = 0$ | | |
| $x = 3.26$ only | A1 | |
| | **(2)** | |
| | **Total 15** | |\begin{enumerate}
\item The continuous random variable $X$ has probability density function $\mathrm { f } ( x )$ given by
\end{enumerate}
$$f ( x ) = \begin{cases} \frac { 1 } { 16 } x ^ { 2 } & 1 \leqslant x < 3 \\ k ( 4 - x ) & 3 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
(a) Show that $k = \frac { 11 } { 12 }$\\
(b) Sketch $\mathrm { f } ( x )$ for $1 \leqslant x \leqslant 4$\\
(c) Write down the mode of $X$
Given that $\mathrm { E } ( X ) = \frac { 25 } { 9 }$\\
(d) use algebraic integration to find $\operatorname { Var } ( X )$, giving your answer to 3 significant figures.
The cumulative distribution function of $X$ is given by
$$\mathrm { F } ( x ) = \left\{ \begin{array} { l r }
0 & x < 1 \\
\frac { 1 } { 48 } \left( x ^ { 3 } + c \right) & 1 \leqslant x < 3 \\
\frac { 11 } { 12 } \left( 4 x - \frac { 1 } { 2 } x ^ { 2 } + d \right) & 3 \leqslant x \leqslant 4 \\
1 & x > 4
\end{array} \right.$$
(e) (i) Find the exact value of $C$\\
(ii) Find the exact value of $d$\\
(f) Calculate, to 3 significant figures, the upper quartile of $X$
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\hfill \mbox{\textit{Edexcel S2 2018 Q7 [15]}}