| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2016 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Standard +0.3 This is a standard S3 probability density function question covering routine techniques: finding k by integration, sketching a symmetric function, calculating variance using E(X²), finding quartiles by solving ∫f(x)dx = 0.75, and applying CLT for sample means. All parts follow textbook methods with no novel insight required, though part (iv) involves slightly more algebraic manipulation than typical. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(k\int_{-1}^{1}(1-x^2)dx = 1 \;\left(\rightarrow k\left[x - \frac{x^3}{3}\right]_{-1}^{1} = 1\right)\) | M1 | Correct integral including limits |
| \(\rightarrow \frac{4k}{3} = 1\) | M1 | \((\text{const})\times k = 1\) |
| \(\rightarrow k = \frac{3}{4}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Graph of \(f(x)\): general shape between \(-1\) and \(+1\) | B1 | |
| Axes labelled with scales and intercepts (FT their \(k\)) | B1 | |
| Nothing outside \( | x | < 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 0 \rightarrow V(X) = E(X^2)\) | B1 | for \(E(X) = 0\) |
| \(V(X) = \frac{3}{4}\int_{-1}^{1}(x^2 - x^4)dx = \frac{3}{4}\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_{-1}^{1}\) | M1 | for correct integral including limits |
| \(= \frac{1}{5}\) | A1 | cao (ignore mistakes in working) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3}{4}\int_0^q(1-x^2)dx = \frac{1}{4}\) | M1 | Correct limits and equality |
| integration \(= \frac{3}{4}\left[x - \frac{x^3}{3}\right]\) | B1 | f/t their \(k\) |
| \(\rightarrow q - \frac{q^3}{3} = \frac{1}{3}\) or \(\rightarrow q^3 - 3q + 1 = 0\) | A1 | any correct simplified (3-term) cubic |
| \(g(0.345) = 0.006\); \(g(0.355) = -0.02\) | M1 | (allow correct alternative); If solving using calculator: state all three solutions |
| Change of sign \(\rightarrow 0.345 < q < 0.355\); So upper quartile \(= 0.35\) to 2 dp | E1 | must be explained clearly; If solving by calculator: explain why only one works |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum X_i > 5 \rightarrow \bar{X}_{40} > 0.125n\) large and so can use Central Limit Theorem | B1 | both (or \(\bar{X}\) normal) |
| \(\bar{X}_{40} \sim N\left(0, \frac{0.2}{40}\right)\) | M1 | for \(\frac{\text{Var}}{40}\) or \(\frac{\text{Var}}{\sqrt{40}}\) |
| A1 | correct mean and variance (ft any positive variance from iii) | |
| \(P(\bar{X}_{40} > 5) = P\left(Z > \frac{0.125-0}{0.0707}\right) = P(Z > 1.768)\) | ||
| \(P(\bar{X}_{40} > 0.125) = P\left(Z > \frac{0.125-0}{0.0707}\right) = P(Z > 1.768) = 0.0385\) | A1 | cao 0.0385 or 0.0386 |
# Question 3:
## Part i:
$k\int_{-1}^{1}(1-x^2)dx = 1 \;\left(\rightarrow k\left[x - \frac{x^3}{3}\right]_{-1}^{1} = 1\right)$ | M1 | Correct integral including limits
$\rightarrow \frac{4k}{3} = 1$ | M1 | $(\text{const})\times k = 1$
$\rightarrow k = \frac{3}{4}$ | A1 | cao
## Part ii:
Graph of $f(x)$: general shape between $-1$ and $+1$ | B1 |
Axes labelled with scales and intercepts (FT their $k$) | B1 |
Nothing outside $|x| < 1$ | B1 |
## Part iii:
$E(X) = 0 \rightarrow V(X) = E(X^2)$ | B1 | for $E(X) = 0$
$V(X) = \frac{3}{4}\int_{-1}^{1}(x^2 - x^4)dx = \frac{3}{4}\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_{-1}^{1}$ | M1 | for correct integral including limits
$= \frac{1}{5}$ | A1 | cao (ignore mistakes in working)
## Part iv:
$\frac{3}{4}\int_0^q(1-x^2)dx = \frac{1}{4}$ | M1 | Correct limits and equality
integration $= \frac{3}{4}\left[x - \frac{x^3}{3}\right]$ | B1 | f/t their $k$
$\rightarrow q - \frac{q^3}{3} = \frac{1}{3}$ or $\rightarrow q^3 - 3q + 1 = 0$ | A1 | any correct simplified (3-term) cubic
$g(0.345) = 0.006$; $g(0.355) = -0.02$ | M1 | (allow correct alternative); If solving using calculator: state all three solutions
Change of sign $\rightarrow 0.345 < q < 0.355$; So upper quartile $= 0.35$ to 2 dp | E1 | must be explained clearly; If solving by calculator: explain why only one works
## Part v:
$\sum X_i > 5 \rightarrow \bar{X}_{40} > 0.125n$ large and so can use Central Limit Theorem | B1 | both (or $\bar{X}$ normal)
$\bar{X}_{40} \sim N\left(0, \frac{0.2}{40}\right)$ | M1 | for $\frac{\text{Var}}{40}$ or $\frac{\text{Var}}{\sqrt{40}}$
| A1 | correct mean and variance (ft any positive variance from iii)
$P(\bar{X}_{40} > 5) = P\left(Z > \frac{0.125-0}{0.0707}\right) = P(Z > 1.768)$ | |
$P(\bar{X}_{40} > 0.125) = P\left(Z > \frac{0.125-0}{0.0707}\right) = P(Z > 1.768) = 0.0385$ | A1 | cao 0.0385 or 0.0386
---3 The random variable $X$ has the following probability density function:
$$\mathrm { f } ( x ) = \begin{cases} k \left( 1 - x ^ { 2 } \right) & - 1 \leqslant x \leqslant 1 \\ 0 & \text { elsewhere } \end{cases}$$
where $k$ is a positive constant.\\
(i) Calculate the value of $k$.\\
(ii) Sketch the probability density function.\\
(iii) Calculate $\operatorname { Var } ( X )$.\\
(iv) Find a cubic equation satisfied by the upper quartile $q$, and hence verify that $q = 0.35$ to 2 decimal places.\\
(v) A random sample of 40 values of $X$ is taken. Using a suitable approximating distribution, calculate the probability that the mean of these values is greater than 0.125 . Justify your choice of distribution.
\hfill \mbox{\textit{OCR MEI S3 2016 Q3 [18]}}