| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Moderate -0.3 Part (i) is a straightforward binomial probability calculation with n=10, requiring P(X≥8) = P(X=8) + P(X=9) + P(X=10). Part (ii) is a standard normal approximation to binomial with continuity correction for n=150, p=0.66. Both parts are routine applications of textbook methods with no conceptual challenges, though the calculation in (ii) requires proper application of continuity correction and standardization, making it slightly easier than average overall. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(8,9,10) = {}^{10}C_8\ 0.66^8\ 0.34^2 + {}^{10}C_9\ 0.66^9\ 0.34^1 + 0.66^{10}\) | M1 | Correct binomial term, \({}^{10}C_a\ 0.66^a(1-0.66)^b\), \(a+b=10\), \(0 < a,b < 10\) |
| A1 | Correct unsimplified expression | |
| \(0.284\) | B1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(np = 0.66\times150 = 99\); \(npq = 0.66\times(1-0.66)\times150 = 33.66\) | B1 | Accept evaluated or unsimplified \(\mu\), \(\sigma^2\) numerical expressions, condone \(\sigma = \sqrt{33.66} = 5.8017\) or \(5.802\). CAO |
| \(P(X>84) = P\left(Z > \frac{84.5-99}{\sqrt{33.66}}\right)\) | M1 | \(\pm\) Standardise, \(\frac{x - their\ 99}{\sqrt{their\ 33.66}}\), condone \(\sigma^2\), \(x\) a value |
| M1 | \(84.5\) or \(83.5\) used in *their* standardisation formula | |
| \((= P(Z > -2.499))\) | M1 | Correct final area |
| \(0.994\) | A1 | Final answer (accept \(0.9938\)). SC if no standardisation formula seen, B2 \(P(Z > -2.499) = 0.994\) |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(8,9,10) = {}^{10}C_8\ 0.66^8\ 0.34^2 + {}^{10}C_9\ 0.66^9\ 0.34^1 + 0.66^{10}$ | M1 | Correct binomial term, ${}^{10}C_a\ 0.66^a(1-0.66)^b$, $a+b=10$, $0 < a,b < 10$ |
| | A1 | Correct unsimplified expression |
| $0.284$ | B1 | CAO |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $np = 0.66\times150 = 99$; $npq = 0.66\times(1-0.66)\times150 = 33.66$ | B1 | Accept evaluated or unsimplified $\mu$, $\sigma^2$ numerical expressions, condone $\sigma = \sqrt{33.66} = 5.8017$ or $5.802$. CAO |
| $P(X>84) = P\left(Z > \frac{84.5-99}{\sqrt{33.66}}\right)$ | M1 | $\pm$ Standardise, $\frac{x - their\ 99}{\sqrt{their\ 33.66}}$, condone $\sigma^2$, $x$ a value |
| | M1 | $84.5$ or $83.5$ used in *their* standardisation formula |
| $(= P(Z > -2.499))$ | M1 | Correct final area |
| $0.994$ | A1 | Final answer (accept $0.9938$). SC if no standardisation formula seen, B2 $P(Z > -2.499) = 0.994$ |
---4 In Quarendon, $66 \%$ of households are satisfied with the speed of their wifi connection.\\
(i) Find the probability that, out of 10 households chosen at random in Quarendon, at least 8 are satisfied with the speed of their wifi connection.\\
(ii) A random sample of 150 households in Quarendon is chosen. Use a suitable approximation to find the probability that more than 84 are satisfied with the speed of their wifi connection. [5]\\
\hfill \mbox{\textit{CAIE S1 2019 Q4 [8]}}