| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Normal approximation to binomial |
| Difficulty | Moderate -0.3 This is a straightforward two-part question on binomial distribution and normal approximation. Part (i) requires direct calculation using B(15, 0.65) for P(X > 12), which is routine. Part (ii) applies the standard normal approximation to B(250, 0.65) with continuity correction—a textbook application requiring no insight beyond recognizing when the approximation is valid (np and nq both > 5). Both parts are standard S1 exercises slightly easier than average A-level difficulty. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((P > 12) = P(13, 14, 15)\) | M1 | Binomial term of form \(^{15}C_x p^x (1-p)^{15-x}\), \(0 < p < 1\) any \(p\), \(x \neq 15, 0\) |
| \(= {}^{15}C_{13}(0.65)^{13}(0.35)^2 + {}^{15}C_{14}(0.65)^{14}(0.35)^1 + (0.65)^{15}\) | A1 | Correct unsimplified answer |
| \(= 0.0617\) | A1 | SC if use np and npq with justification give \((12.5 - 9.75)/\sqrt{3.41}\) M1 \(1-F(1.489)\) A1 0.0681 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| mean \(= 250 \times 0.65 = 162.5\), variance \(= 250 \times 0.65 \times 0.35 = 56.875\) | B1 | Correct unsimplified \(np\) and \(npq\) |
| \(P(< 179) = P\left(z < \frac{178.5 - 162.5}{\sqrt{56.875}}\right) = P(z < 2.122)\) | M1 | Substituting their \(\mu\) and \(\sigma\) (condone \(\sigma^2\)) into the Standardisation Formula with a numerical value for '\(178.5\)'. Continuity correct not required for this M1. Condone \(\pm\) standardisation formula |
| Using continuity correction 178.5 or 179.5 | M1 | |
| \(= 0.983\) | A1 | Correct final answer |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(P > 12) = P(13, 14, 15)$ | M1 | Binomial term of form $^{15}C_x p^x (1-p)^{15-x}$, $0 < p < 1$ any $p$, $x \neq 15, 0$ |
| $= {}^{15}C_{13}(0.65)^{13}(0.35)^2 + {}^{15}C_{14}(0.65)^{14}(0.35)^1 + (0.65)^{15}$ | A1 | Correct unsimplified answer |
| $= 0.0617$ | A1 | SC if use np and npq with justification give $(12.5 - 9.75)/\sqrt{3.41}$ M1 $1-F(1.489)$ A1 0.0681 A0 |
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## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| mean $= 250 \times 0.65 = 162.5$, variance $= 250 \times 0.65 \times 0.35 = 56.875$ | B1 | Correct unsimplified $np$ and $npq$ |
| $P(< 179) = P\left(z < \frac{178.5 - 162.5}{\sqrt{56.875}}\right) = P(z < 2.122)$ | M1 | Substituting their $\mu$ and $\sigma$ (condone $\sigma^2$) into the Standardisation Formula with a numerical value for '$178.5$'. Continuity correct not required for this M1. Condone $\pm$ standardisation formula |
| Using continuity correction 178.5 or 179.5 | M1 | |
| $= 0.983$ | A1 | Correct final answer |
---5 In a certain country the probability that a child owns a bicycle is 0.65 .\\
(i) A random sample of 15 children from this country is chosen. Find the probability that more than 12 own a bicycle.\\
(ii) A random sample of 250 children from this country is chosen. Use a suitable approximation to find the probability that fewer than 179 own a bicycle.\\
\hfill \mbox{\textit{CAIE S1 2019 Q5 [7]}}