| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Symmetry property of PDF |
| Difficulty | Standard +0.3 This is a standard S2 question testing routine PDF/CDF properties. Parts (a)-(d) involve basic sketching and expectation calculations. Parts (e)-(g) use symmetry to find quartiles with minimal computation. Part (h) requires solving F(x) equations but is straightforward. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cap\) shape which does not go below the \(x\)-axis [condone missing patios] | B1 | |
| Graph must end at points \((1,0)\) and \((5,0)\) with points labelled at 1 and 5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = 3\) (by symmetry) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X^2) = \int x^2 f(x)\,dx = \frac{3}{32}\int(6x^3 - x^4 - 5x^2)\,dx\) | M1 | Showing intention of doing \(\int x^2 f(x)\) and attempt to multiply out bracket |
| \(= \frac{3}{32}\left[\frac{6x^4}{4} - \frac{x^5}{5} - \frac{5x^3}{3}\right]_1^5\) | A1 | Correct integration, cao, ignore limits |
| \(= \frac{3}{32}\!\left(\!\left[\frac{6 \times 625}{4} - 625 - \frac{625}{3}\right] - \left[\frac{6}{4} - \frac{1}{5} - \frac{5}{3}\right]\!\right) = 9.8\) (*) | M1 A1 cso | \(2^\text{nd}\) M1 for use of correct limits; \(2^\text{nd}\) A1 cso leading to 9.8; do not ignore subsequent working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{s.d.} = \sqrt{9.8 - E(X)^2}\) | M1 | Must include \(\sqrt{\ldots}\) |
| \(= 0.8944\ldots\) awrt 0.894 | A1 | Allow awrt 0.894, \(\sqrt{0.8}\), \(\frac{2\sqrt{5}}{5}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(1) = 0 \Rightarrow \frac{1}{32}(a - 15 + 9 - 1) = 0\), leading to \(\underline{a = 7}\) | M1 A1 | M1 for correct method to find \(a\), e.g. \(F(5)=1\) or \(\int_1^5 f(x)=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(2.29) = 0.2449\ldots\), \(F(2.31) = 0.2515\ldots\) | M1 A1 | M1 for attempt at \(F(2.29)\) or \(F(2.31)\) or put \(F(x)=0.25\); A1 both values seen awrt 0.245 and 0.252 |
| Since \(F(q_1) = 0.25\) and these values are either side of 0.25 then \(2.29 < q_1 < 2.31\) | A1 | \(2^\text{nd}\) A1 for comparison with 0.25 and stating \(Q_1\) lies between 2.29 and 2.31 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Since the distribution is symmetric \(q_3 = 5 - 1.3 = \underline{3.7}\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| We know \(P(q_1 = 2.3 < X < 3.7 = q_3) = 0.5\), so \(k\sigma = 0.7\) | M1 | For \(k\sigma = \) awrt 0.7 |
| \(k = \frac{0.7}{0.894\ldots} = 0.7826\ldots =\) awrt 0.78 | A1 | Allow awrt 0.78; NB a correct awrt 0.78 gains M1 A1 |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cap$ shape which does not go below the $x$-axis [condone missing patios] | B1 | |
| Graph must end at points $(1,0)$ and $(5,0)$ with points labelled at 1 and 5 | B1 | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 3$ (by symmetry) | B1 | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X^2) = \int x^2 f(x)\,dx = \frac{3}{32}\int(6x^3 - x^4 - 5x^2)\,dx$ | M1 | Showing intention of doing $\int x^2 f(x)$ and attempt to multiply out bracket |
| $= \frac{3}{32}\left[\frac{6x^4}{4} - \frac{x^5}{5} - \frac{5x^3}{3}\right]_1^5$ | A1 | Correct integration, cao, ignore limits |
| $= \frac{3}{32}\!\left(\!\left[\frac{6 \times 625}{4} - 625 - \frac{625}{3}\right] - \left[\frac{6}{4} - \frac{1}{5} - \frac{5}{3}\right]\!\right) = 9.8$ (*) | M1 A1 cso | $2^\text{nd}$ M1 for use of correct limits; $2^\text{nd}$ A1 cso leading to 9.8; do not ignore subsequent working |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{s.d.} = \sqrt{9.8 - E(X)^2}$ | M1 | Must include $\sqrt{\ldots}$ |
| $= 0.8944\ldots$ awrt 0.894 | A1 | Allow awrt 0.894, $\sqrt{0.8}$, $\frac{2\sqrt{5}}{5}$ oe |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(1) = 0 \Rightarrow \frac{1}{32}(a - 15 + 9 - 1) = 0$, leading to $\underline{a = 7}$ | M1 A1 | M1 for correct method to find $a$, e.g. $F(5)=1$ or $\int_1^5 f(x)=1$ |
### Part (f):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(2.29) = 0.2449\ldots$, $F(2.31) = 0.2515\ldots$ | M1 A1 | M1 for attempt at $F(2.29)$ or $F(2.31)$ or put $F(x)=0.25$; A1 both values seen awrt 0.245 and 0.252 |
| Since $F(q_1) = 0.25$ and these values are either side of 0.25 then $2.29 < q_1 < 2.31$ | A1 | $2^\text{nd}$ A1 for comparison with 0.25 and stating $Q_1$ lies between 2.29 and 2.31 |
### Part (g):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Since the distribution is symmetric $q_3 = 5 - 1.3 = \underline{3.7}$ | B1 | cao |
### Part (h):
| Answer/Working | Marks | Guidance |
|---|---|---|
| We know $P(q_1 = 2.3 < X < 3.7 = q_3) = 0.5$, so $k\sigma = 0.7$ | M1 | For $k\sigma = $ awrt 0.7 |
| $k = \frac{0.7}{0.894\ldots} = 0.7826\ldots =$ **awrt 0.78** | A1 | Allow awrt 0.78; NB a correct awrt 0.78 gains M1 A1 |
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\item The continuous random variable $X$ has probability density function given by
\end{enumerate}
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { 3 } { 32 } ( x - 1 ) ( 5 - x ) & 1 \leqslant x \leqslant 5 \\
0 & \text { otherwise }
\end{array} \right.$$
(a) Sketch $\mathrm { f } ( x )$ showing clearly the points where it meets the $x$-axis.\\
(b) Write down the value of the mean, $\mu$, of $X$.\\
(c) Show that $\mathrm { E } \left( X ^ { 2 } \right) = 9.8$\\
(d) Find the standard deviation, $\sigma$, of $X$.
The cumulative distribution function of $X$ is given by
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c }
0 & x < 1 \\
\frac { 1 } { 32 } \left( a - 15 x + 9 x ^ { 2 } - x ^ { 3 } \right) & 1 \leqslant x \leqslant 5 \\
1 & x > 5
\end{array} \right.$$
where $a$ is a constant.\\
(e) Find the value of $a$.\\
(f) Show that the lower quartile of $X , q _ { 1 }$, lies between 2.29 and 2.31\\
(g) Hence find the upper quartile of $X$, giving your answer to 1 decimal place.\\
(h) Find, to 2 decimal places, the value of $k$ so that
$$\mathrm { P } ( \mu - k \sigma < X < \mu + k \sigma ) = 0.5$$
\hfill \mbox{\textit{Edexcel S2 2011 Q7 [17]}}