| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Multinomial probability calculation |
| Difficulty | Standard +0.3 This is a straightforward application of binomial probability and normal approximation with clearly stated probabilities. Part (i) requires direct binomial calculation for small n, and part (ii) is a standard normal approximation with continuity correction. Both are routine S1 techniques with no conceptual challenges beyond recognizing which distribution to use. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(O\ \text{given}\ +) = \frac{0.37}{0.83}\ (0.4458)\) | B1 | \(0.83\) seen or implied |
| A1 | Attempt to find \(P(O\ \text{given}\ +)\) using conditional probability fraction | |
| M1 | Binomial term \({}^9C_r p^r(1-p)^{9-r},\ r \neq 0\) or \(9\) | |
| \(P(0,1,2) = (0.4458)^0(0.5542)^9 +\ {}^9C_1(0.4458)^1(0.5542)^8 +\ {}^9C_2(0.4458)^2(0.5542)^7\) | M1 | Binomial expression \(P(0,1,2)\) or \(P(0,1,2,3)\) powers summing to 9 any \(0 < p < 1\) |
| A1 | Correct unsimplified expression | |
| \(= 0.156\) | A1 [6] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mu = 150 \times 0.35 = 52.5\) | B1 | \(150 \times 0.35\ (52.5)\) and \(150 \times 0.35 \times 0.65\ (34.125)\) seen |
| \(\sigma^2 = 150 \times 0.35 \times 0.65 = 34.125\) | M1 | Standardising, using sd not variance |
| M1 | Using continuity correction, \(59.5\) or \(60.5\) | |
| \(P(>60.5) = P\!\left(z > \pm\frac{60.5-52.5}{\sqrt{34.125}}\right)\) | M1 | Correct area (\(< 0.5\), for mean \(<\) their \(60\)) |
| \(= 1 - \Phi(1.369)\) | ||
| \(= 0.0854\) or \(0.0855\) | A1 [5] | Correct value |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(O\ \text{given}\ +) = \frac{0.37}{0.83}\ (0.4458)$ | B1 | $0.83$ seen or implied |
| | A1 | Attempt to find $P(O\ \text{given}\ +)$ using conditional probability fraction |
| | M1 | Binomial term ${}^9C_r p^r(1-p)^{9-r},\ r \neq 0$ or $9$ |
| $P(0,1,2) = (0.4458)^0(0.5542)^9 +\ {}^9C_1(0.4458)^1(0.5542)^8 +\ {}^9C_2(0.4458)^2(0.5542)^7$ | M1 | Binomial expression $P(0,1,2)$ or $P(0,1,2,3)$ powers summing to 9 any $0 < p < 1$ |
| | A1 | Correct unsimplified expression |
| $= 0.156$ | A1 [6] | Correct final answer |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu = 150 \times 0.35 = 52.5$ | B1 | $150 \times 0.35\ (52.5)$ and $150 \times 0.35 \times 0.65\ (34.125)$ seen |
| $\sigma^2 = 150 \times 0.35 \times 0.65 = 34.125$ | M1 | Standardising, using sd not variance |
| | M1 | Using continuity correction, $59.5$ or $60.5$ |
| $P(>60.5) = P\!\left(z > \pm\frac{60.5-52.5}{\sqrt{34.125}}\right)$ | M1 | Correct area ($< 0.5$, for mean $<$ their $60$) |
| $= 1 - \Phi(1.369)$ | | |
| $= 0.0854$ or $0.0855$ | A1 [5] | Correct value |6 Human blood groups are identified by two parts. The first part is $\mathrm { A } , \mathrm { B } , \mathrm { AB }$ or O and the second part (the Rhesus part) is + or - . In the UK, $35 \%$ of the population are group $\mathrm { A } + , 8 \%$ are $\mathrm { B } + , 3 \%$ are $\mathrm { AB } +$, $37 \%$ are $\mathrm { O } + , 7 \%$ are $\mathrm { A } - , 2 \%$ are $\mathrm { B } - , 1 \%$ are $\mathrm { AB } -$ and $7 \%$ are $\mathrm { O } -$.\\
(i) A random sample of 9 people in the UK who are Rhesus + is taken. Find the probability that fewer than 3 are group $\mathrm { O } +$.\\
(ii) A random sample of 150 people in the UK is taken. Find the probability that more than 60 people are group A+.
\hfill \mbox{\textit{CAIE S1 2011 Q6 [11]}}