| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Direct variance calculation from pdf |
| Difficulty | Moderate -0.3 This is a standard S2 continuous probability distribution question requiring routine integration techniques: finding k from normalization, calculating E(Y) and Var(Y), and a probability calculation. All steps follow textbook procedures with straightforward polynomial integration, making it slightly easier than average A-level difficulty. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^2 k(4-y^2)dy = [1]\) or attempt F(y) | M1 | |
| \(k\left[4y - \frac{y^3}{3}\right]_0^2 = [1]\) | A1 | |
| \(k\left[4 \times 2 - \frac{2^3}{3}\right] = 1\) or must use F(2) = 1 | M1d, A1cso | |
| \(k = \frac{3}{16}\) (*) | A1cso | (4) |
| \(E(Y) = \frac{3}{16}\int_0^2(4y - y^3)dy\) | M1 | |
| \(= \frac{3}{16}\left[2y^2 - \frac{y^4}{4}\right]_0^2\), \(= \frac{12}{16}\) or 0.75 | A1, A1 | |
| \(= 750\) (kg) | A1cao | (4) |
| \(E(Y^2) = \frac{3}{16}\int_0^2 4y^2 - y^5 dy\) | M1 | |
| \(= \frac{3}{16}\left[\frac{4y^3}{3} - \frac{y^5}{5}\right]_0^2\) \((= 0.8)\) | A1 | |
| Var(Y) = \(0.8 - 0.75^2 = 0.2375\) | M1, A1 | |
| Standard deviation \(= 0.48734...\) or 487 (kg) | A1, B1 | (5) |
| \(P(Y > 1.5) = \frac{3}{16}\int_{1.5}^2(4-y^2)dy\) or \(1 - \frac{3}{16}\left[4y - \frac{y^3}{3}\right]_0^{1.5}\) | M1 | |
| \(= \frac{3}{16}\left[4y - \frac{y^3}{3}\right]_{1.5}\) or \(1 - \frac{3}{16}\left[4y - \frac{y^3}{3}\right]_0^{1.5}\) | A1 | |
| \(= 0.0859\) or \(\frac{11}{128}\) | A1 | (3) |
| [16] |
| Answer | Marks |
|---|---|
| Item | Guidance |
| (a) | 1st M1: attempting to integrate \(f(y)\), (at least one term \(y^n \to y^{n+1}\)). Ignore limits. |
| 1st A1: fully correct integration. Ignore limits and accept any letters. | |
| 2nd M1: dep on 1st M1. Subst in correct limits – condone not seeing 0 substituted. | |
| NB: An "= 1" must appear somewhere before the line \(\frac{16k}{3} = 1\) | |
| (b) | 1st M1: Attempting to integrate \(y f(y)\), (at least one term \(y^n \to y^{n+1}\)). Ignore limits |
| 1st A1: correct integration which must be shown. No integration loses all 4 marks | |
| 2nd A1: 0.75 or any equivalent. May be implied by a correct ans. of 750 (kg) | |
| 3rd A1cao: 750 only. Condone missing "kg" | |
| (c) | 1st M1: Attempting to integrate \(y^2 f(y)\) (at least one term \(y^n \to y^{n+1}\)). Ignore limits. Condone in \(\sqrt{\ }\) |
| 1st A1: correct integration. Condone inside \(\sqrt{\ }\). May be implied by sight of 0.8 | |
| 2nd M1: using E(Y²) – [E(Y)]² follow through their \(E(Y^2)\) and \(E(Y)^2\). Must see values used | |
| 2nd A1: 0.2375 may be implied by correct sd. Allow \(\frac{19}{80}\) or exact equivalent | |
| 3rd A1: awrt 0.487 or awrt 487. (no fractions) | |
| (d) | B1: using 1.5 in an integral or \(1 - F(1.5)\). Must be part of a correct expression. |
| M1: Correct integration and at least intention to use correct limits so 1.5, 2 or 0, 1.5 seen | |
| A1: awrt 0.0859 or \(\frac{11}{128}\) or exact equivalent |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^2 k(4-y^2)dy = [1]$ or attempt F(y) | M1 | |
| $k\left[4y - \frac{y^3}{3}\right]_0^2 = [1]$ | A1 | |
| $k\left[4 \times 2 - \frac{2^3}{3}\right] = 1$ or must use F(2) = 1 | M1d, A1cso | |
| $k = \frac{3}{16}$ (*) | A1cso | (4) |
| $E(Y) = \frac{3}{16}\int_0^2(4y - y^3)dy$ | M1 | |
| $= \frac{3}{16}\left[2y^2 - \frac{y^4}{4}\right]_0^2$, $= \frac{12}{16}$ or 0.75 | A1, A1 | |
| $= 750$ (kg) | A1cao | (4) |
| $E(Y^2) = \frac{3}{16}\int_0^2 4y^2 - y^5 dy$ | M1 | |
| $= \frac{3}{16}\left[\frac{4y^3}{3} - \frac{y^5}{5}\right]_0^2$ $(= 0.8)$ | A1 | |
| Var(Y) = $0.8 - 0.75^2 = 0.2375$ | M1, A1 | |
| Standard deviation $= 0.48734...$ or 487 (kg) | A1, B1 | (5) |
| $P(Y > 1.5) = \frac{3}{16}\int_{1.5}^2(4-y^2)dy$ or $1 - \frac{3}{16}\left[4y - \frac{y^3}{3}\right]_0^{1.5}$ | M1 | |
| $= \frac{3}{16}\left[4y - \frac{y^3}{3}\right]_{1.5}$ or $1 - \frac{3}{16}\left[4y - \frac{y^3}{3}\right]_0^{1.5}$ | A1 | |
| $= 0.0859$ or $\frac{11}{128}$ | A1 | (3) |
| | [16] | |
**Notes:**
| Item | Guidance |
|---|---|
| (a) | 1st M1: attempting to integrate $f(y)$, (at least one term $y^n \to y^{n+1}$). Ignore limits. |
| | 1st A1: fully correct integration. Ignore limits and accept any letters. |
| | 2nd M1: dep on 1st M1. Subst in correct limits – condone not seeing 0 substituted. |
| | NB: An "= 1" must appear somewhere **before** the line $\frac{16k}{3} = 1$ |
| (b) | 1st M1: Attempting to integrate $y f(y)$, (at least one term $y^n \to y^{n+1}$). Ignore limits |
| | 1st A1: correct integration which must be shown. **No integration loses all 4 marks** |
| | 2nd A1: 0.75 or any equivalent. May be implied by a correct ans. of 750 (kg) |
| | 3rd A1cao: 750 only. Condone missing "kg" |
| (c) | 1st M1: Attempting to integrate $y^2 f(y)$ (at least one term $y^n \to y^{n+1}$). Ignore limits. Condone in $\sqrt{\ }$ |
| | 1st A1: correct integration. Condone inside $\sqrt{\ }$. May be implied by sight of 0.8 |
| | 2nd M1: using E(Y²) – [E(Y)]² follow through their $E(Y^2)$ and $E(Y)^2$. Must see values **used** |
| | 2nd A1: 0.2375 may be implied by correct sd. Allow $\frac{19}{80}$ or exact equivalent |
| | 3rd A1: awrt 0.487 or awrt 487. (no fractions) |
| (d) | B1: using 1.5 in an integral or $1 - F(1.5)$. Must be part of a correct expression. |
| | M1: Correct integration and at least intention to use correct limits so 1.5, 2 or 0, 1.5 seen |
| | A1: awrt 0.0859 or $\frac{11}{128}$ or exact equivalent |
---2. The amount of flour used by a factory in a week is $Y$ thousand kg where $Y$ has probability density function
$$\mathrm { f } ( y ) = \left\{ \begin{array} { c c }
k \left( 4 - y ^ { 2 } \right) & 0 \leqslant y \leqslant 2 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Show that the value of $k$ is $\frac { 3 } { 16 }$
Use algebraic integration to find
\item the mean number of kilograms of flour used by the factory in a week,
\item the standard deviation of the number of kilograms of flour used by the factory in a week,
\item the probability that more than 1500 kg of flour will be used by the factory next week.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q2 [16]}}