| 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 | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Moderate -0.3 This is a straightforward two-part question testing standard binomial distribution calculations and normal approximation. Part (i) requires direct binomial probability computation (P(X=4)+P(X=5)+P(X=6)), while part (ii) is a textbook application of normal approximation with continuity correction. Both parts involve routine procedures with no conceptual challenges beyond recognizing when to apply the approximation (np and nq both >5). Slightly easier than average due to clear structure and standard technique application. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(4,5,6) = ^{15}C_4(0.22)^4(0.78)^{11} + ^{15}C_5(0.22)^5(0.78)^{10} + ^{15}C_6(0.22)^6(0.78)^9\) | M1 | One binomial term \(^{15}C_x p^x (1-p)^{15-x}\), \(0 < p < 1\) |
| A1 | Correct unsimplified expression | |
| \(= 0.398\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu = 145 \times 0.22 = 31.9\), \(\sigma^2 = 145 \times 0.22 \times 0.78 = 24.882\) | B1 | Correct unsimplified mean and variance |
| \(P(x > 26) = P\left(z > \frac{26.5 - 31.9}{\sqrt{24.882}}\right) = P(z > -1.08255)\) | M1 | Standardising must have sq rt |
| M1 | 25.5 or 26.5 seen as a cc | |
| \(= \Phi(1.08255)\) | M1 | Correct area \(\Phi\), must agree with their \(\mu\) |
| \(= 0.861\) | A1 | Correct final answer, accept 0.861, or 0.860 from 0.8604 not from 0.8599 |
## Question 5(i):
$P(4,5,6) = ^{15}C_4(0.22)^4(0.78)^{11} + ^{15}C_5(0.22)^5(0.78)^{10} + ^{15}C_6(0.22)^6(0.78)^9$ | M1 | One binomial term $^{15}C_x p^x (1-p)^{15-x}$, $0 < p < 1$ |
| A1 | Correct unsimplified expression |
$= 0.398$ | A1 | Correct answer |
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## Question 5(ii):
$\mu = 145 \times 0.22 = 31.9$, $\sigma^2 = 145 \times 0.22 \times 0.78 = 24.882$ | B1 | Correct unsimplified mean and variance |
$P(x > 26) = P\left(z > \frac{26.5 - 31.9}{\sqrt{24.882}}\right) = P(z > -1.08255)$ | M1 | Standardising must have sq rt |
| M1 | 25.5 or 26.5 seen as a cc |
$= \Phi(1.08255)$ | M1 | Correct area $\Phi$, must agree with their $\mu$ |
$= 0.861$ | A1 | Correct final answer, accept 0.861, or 0.860 from 0.8604 not from 0.8599 |
---5 In Pelmerdon 22\% of families own a dishwasher.\\
(i) Find the probability that, of 15 families chosen at random from Pelmerdon, between 4 and 6 inclusive own a dishwasher.\\
(ii) A random sample of 145 families from Pelmerdon is chosen. Use a suitable approximation to find the probability that more than 26 families own a dishwasher.\\
\hfill \mbox{\textit{CAIE S1 2018 Q5 [8]}}