| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Standard +0.3 Part (i) is a standard inverse normal problem requiring use of tables and algebraic manipulation (z-score formula). Part (ii) requires finding a normal probability then applying binomial distribution with n=4. Both are routine S1 techniques with no novel insight needed, but the two-stage process and binomial calculation make it slightly above average difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(z = -1.282\) | B1 | \(\pm 1.282\) or \(\pm 1.281\) seen |
| \(\text{P}(x < 20) = P\left(z < \frac{20-\mu}{0.8}\right)\) \(-1.282 = \frac{20-\mu}{0.8}\) \(\mu = 21.0\) cm \((21.0256)\) | M1, A1 [3] | Standardising, no cc, must have \(0.8\), must be a z-value; Correct answer |
| (ii) \(\text{P}(21.5 < x < 22.5) = P\left(\frac{21.5-21.03}{0.8}\right) < z < \left(\frac{22.5-21.03}{0.8}\right)\) \(= \Phi(1.8375) - \Phi(0.5875) = 0.9670 - 0.7217 = 0.2453\) | M1, M1, A1 | 2 attempts at standardising with their mean, must have \(0.8\) oe; Subtracting 2 \(\Phi\)s fi their mean; Needn't be entirely accurate, rounding to \(0.24\) or \(0.25\) |
| \(\text{P}(< 2) = \text{P}(0) + \text{P}(1) = (0.7547)^4 + (0.2453)'(0.7547)^3 {}^4C_1\) | M1, M1 | Binomial term with \({}^nC_rp^r(1-p)^{n-r}\) seen \(r = 0\), any \(p < 1\); Bin expression for P(0) + P(1), any \(p < 1\) |
| \(= 0.746\) | A1 [6] | Accept 3sf rounding to \(0.75\) |
**(i)** $z = -1.282$ | B1 | $\pm 1.282$ or $\pm 1.281$ seen
$\text{P}(x < 20) = P\left(z < \frac{20-\mu}{0.8}\right)$ $-1.282 = \frac{20-\mu}{0.8}$ $\mu = 21.0$ cm $(21.0256)$ | M1, A1 [3] | Standardising, no cc, must have $0.8$, must be a z-value; Correct answer
**(ii)** $\text{P}(21.5 < x < 22.5) = P\left(\frac{21.5-21.03}{0.8}\right) < z < \left(\frac{22.5-21.03}{0.8}\right)$ $= \Phi(1.8375) - \Phi(0.5875) = 0.9670 - 0.7217 = 0.2453$ | M1, M1, A1 | 2 attempts at standardising with their mean, must have $0.8$ oe; Subtracting 2 $\Phi$s fi their mean; Needn't be entirely accurate, rounding to $0.24$ or $0.25$
$\text{P}(< 2) = \text{P}(0) + \text{P}(1) = (0.7547)^4 + (0.2453)'(0.7547)^3 {}^4C_1$ | M1, M1 | Binomial term with ${}^nC_rp^r(1-p)^{n-r}$ seen $r = 0$, any $p < 1$; Bin expression for P(0) + P(1), any $p < 1$
$= 0.746$ | A1 [6] | Accept 3sf rounding to $0.75$6 The lengths, in centimetres, of drinking straws produced in a factory have a normal distribution with mean $\mu$ and variance 0.64 . It is given that $10 \%$ of the straws are shorter than 20 cm .\\
(i) Find the value of $\mu$.\\
(ii) Find the probability that, of 4 straws chosen at random, fewer than 2 will have a length between 21.5 cm and 22.5 cm .
\hfill \mbox{\textit{CAIE S1 2011 Q6 [9]}}