| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution probability then binomial/normal approximation on sample |
| Difficulty | Standard +0.3 Part (i) is a standard normal distribution calculation requiring two z-score conversions and subtraction. Part (ii) applies normal approximation to binomial with continuity correction—a routine S1 technique. Both parts are straightforward applications of well-practiced methods with no conceptual challenges beyond standard curriculum content. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z_1 = \pm\frac{4.1-5.7}{0.8} = -2 \quad z_2 = \pm\frac{5-5.7}{0.8} = -0.875\) | M1 | At least one standardising; no cc, no sq rt, no sq; using 5.7 and 0.8 and either 4.1 or 5 |
| \(P(\text{Toffee Apple}) = P(d < 5.0) - P(d < 4.1) = P(z < -0.875) - P(z < -2) = \Phi(-0.875) - \Phi(-2) = \Phi(2) - \Phi(0.875)\) | M1 | Correct area \(\Phi - \Phi\) legitimately obtained; need 2 negative \(z\)-values or 2 positives, not one of each |
| \(= 0.9772 - 0.8092 = 0.168\) (or \(0.1908 - 0.0228\)) | A1 | Correct final answer |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(np = 250 \times 0.168 = 42, \quad npq = 34.944\) | B1ft | Correct unsimplified mean and var — ft their prob for (i) providing \(0 < p < 1\); implied by \(\sigma = \sqrt{34.944} = 5.911\) |
| \(P(<50) = P\!\left(z < \frac{49.5 - 42}{\sqrt{34.944}}\right) = P(z < 1.2687)\) | M1 | \(\pm\) Standardising using 50, their mean and sd; must have sq rt |
| M1 | 49.5 or 50.5 seen as a cc | |
| \(= \Phi(1.2687)\) | M1 | Correct area \(\Phi(>0.5\) for \(+z\) and \(<0.5\) for \(-z)\) in their final answer |
| \(= 0.898\) | A1 | Correct final answer |
| Total: 5 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z_1 = \pm\frac{4.1-5.7}{0.8} = -2 \quad z_2 = \pm\frac{5-5.7}{0.8} = -0.875$ | M1 | At least one standardising; no cc, no sq rt, no sq; using 5.7 and 0.8 and either 4.1 or 5 |
| $P(\text{Toffee Apple}) = P(d < 5.0) - P(d < 4.1) = P(z < -0.875) - P(z < -2) = \Phi(-0.875) - \Phi(-2) = \Phi(2) - \Phi(0.875)$ | M1 | Correct area $\Phi - \Phi$ legitimately obtained; need 2 negative $z$-values or 2 positives, not one of each |
| $= 0.9772 - 0.8092 = 0.168$ (or $0.1908 - 0.0228$) | A1 | Correct final answer |
| **Total: 3** | | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $np = 250 \times 0.168 = 42, \quad npq = 34.944$ | B1ft | Correct unsimplified mean and var — ft their prob for (i) providing $0 < p < 1$; implied by $\sigma = \sqrt{34.944} = 5.911$ |
| $P(<50) = P\!\left(z < \frac{49.5 - 42}{\sqrt{34.944}}\right) = P(z < 1.2687)$ | M1 | $\pm$ Standardising using 50, their mean and sd; must have sq rt |
| | M1 | 49.5 or 50.5 seen as a cc |
| $= \Phi(1.2687)$ | M1 | Correct area $\Phi(>0.5$ for $+z$ and $<0.5$ for $-z)$ in their final answer |
| $= 0.898$ | A1 | Correct final answer |
| **Total: 5** | | |6 The diameters of apples in an orchard have a normal distribution with mean 5.7 cm and standard deviation 0.8 cm . Apples with diameters between 4.1 cm and 5 cm can be used as toffee apples.\\
(i) Find the probability that an apple selected at random can be used as a toffee apple.\\
(ii) 250 apples are chosen at random. Use a suitable approximation to find the probability that fewer than 50 can be used as toffee apples.\\
\hfill \mbox{\textit{CAIE S1 2018 Q6 [8]}}