| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Measurement error modeling |
| Difficulty | Standard +0.3 This is a straightforward S2 uniform distribution question requiring understanding of measurement error (part a), standard formula application for uniform distribution variance (part b), and routine normal approximation to binomial (part c). All techniques are standard bookwork with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| VIAN SIHI NI IIIHM ION OC | VIUV SIHILNI JMAMALONOO | VI4V SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(a = 4 \times (-0.5)\) and \(b = 4 \times 0.5\) | B1 (1) |
| (b)(i) | \(\sqrt{\frac{(2-(-2))^2}{12} - \frac{2\sqrt{3}}{3}} = 1.1547\ldots\) awrt \(\mathbf{1.15}\) | M1 A1 (i) awrt 1.15 (allow 1.155) |
| (ii) | \([\text{P}(-\frac{2\sqrt{3}}{3} < W < \frac{2\sqrt{3}}{3})] = [\frac{\frac{2\sqrt{3}}{3}-(-\frac{2\sqrt{3}}{3})}{\sqrt{3}}{2-(-2)}] = 0.57735\ldots\) awrt \(\mathbf{0.577}\) | M1 A1 (4) |
| (c) | \(P(W > 1.9) = \frac{2-1.9}{2-(-2)} = [0.025]\) \(X \sim \text{B}(100, 0.025)\) \(\to \text{Po}(2.5)\) \(P(X \geq 5) = 1 - P(X \leq 4) = 1 - 0.8912 = 0.1088\) awrt \(\mathbf{0.109}\) | M1 M1 M1 A1 (4) Total 9 |
(a) | $a = 4 \times (-0.5)$ and $b = 4 \times 0.5$ | B1 (1) |
(b)(i) | $\sqrt{\frac{(2-(-2))^2}{12} - \frac{2\sqrt{3}}{3}} = 1.1547\ldots$ awrt $\mathbf{1.15}$ | M1 A1 (i) awrt **1.15** (allow 1.155) | M1 for use of correct formula with square root / A1 awrt 1.15
(ii) | $[\text{P}(-\frac{2\sqrt{3}}{3} < W < \frac{2\sqrt{3}}{3})] = [\frac{\frac{2\sqrt{3}}{3}-(-\frac{2\sqrt{3}}{3})}{\sqrt{3}}{2-(-2)}] = 0.57735\ldots$ awrt $\mathbf{0.577}$ | M1 A1 (4) | M1 for a correct follow through ($2 \times \frac{1}{4} \times$ their (b)(i)) / A1 awrt 0.577
(c) | $P(W > 1.9) = \frac{2-1.9}{2-(-2)} = [0.025]$ $X \sim \text{B}(100, 0.025)$ $\to \text{Po}(2.5)$ $P(X \geq 5) = 1 - P(X \leq 4) = 1 - 0.8912 = 0.1088$ awrt $\mathbf{0.109}$ | M1 M1 M1 A1 (4) Total 9 | 1st M1 for a correct expression for $P(W > 1.9)$ / 2nd M1 for Binomial distribution B(100, $P(W > 1.9)$) / 3rd M1 for Poisson approximation with mean $100 \times P(W > 1.9)$ and $1 - P(X \leq 4)$ / Note: using binomial distribution gives 0.106... |
---\begin{enumerate}
\item One side of a square is measured to the nearest centimetre and this measurement is multiplied by 4 to estimate the perimeter of the square. The random variable, $W \mathrm {~cm}$, represents the estimated perimeter of the square minus the true perimeter of the square.\\
$W$ is uniformly distributed over the interval $[ a , b ]$\\
(a) Explain why $a = - 2$ and $b = 2$
\end{enumerate}
The standard deviation of $W$ is $\sigma$\\
(b) (i) Find $\sigma$\\
(ii) Find the probability that the estimated perimeter of the square is within $\sigma$ of the true perimeter of the square.
One side of each of 100 squares are now measured. Using a suitable approximation,\\
(c) find the probability that $W$ is greater than 1.9 for at least 5 of these squares.
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VIAN SIHI NI IIIHM ION OC & VIUV SIHILNI JMAMALONOO & VI4V SIHI NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel S2 2018 Q6 [9]}}