| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Multinomial probability calculation |
| Difficulty | Standard +0.3 Part (i) is a straightforward binomial probability calculation with small n=8, requiring summing P(X=4)+P(X=5)+P(X=6). Part (ii) requires recognizing that p=0.13 (calculating 1-0.36-0.22-0.29), then applying normal approximation to B(300,0.13) with continuity correction. Both parts are standard textbook exercises testing routine application of binomial calculations and normal approximation—slightly above average difficulty only due to the two-part structure and the need to identify the correct probability in part (ii). |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(4, 5, 6) = (0.22)^y(0.78)^8C4 + (0.22)^y(0.78)^8C5 + (0.22)^y(0.78)^8C6\) | M1, M1 | Bin term with \(_nC_r p^r(1-p)^{n-r}\) seen \(r \neq 0\) any \(p < 1\); Summing 2 or 3 bin probs \(p = 0.22, n = 8\) |
| \(= 0.0763\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| var \(= 300 \times 0.13 \times 0.87 = 33.93\) | B1, B1 ft | Correct prob can be implied; Correct unsimplified np and npq ft wrong 0.13 |
| \(P(30 < x < 50) = P\left(\frac{30.5 - 39}{\sqrt{33.93}} < z < \frac{49.5 - 39}{\sqrt{33.93}}\right)\) | M1 | Standardising a value need sq rt |
| \(M1\) | Cont correction 30.5 / 31.5 or 48.5/49.5 only | |
| \(= P(-1.4592 < z < 1.8026)\) | M1 | Correct area \(\Phi_1 + \Phi_2 - 1\) oe; Rounding to correct answer |
| \(= \Phi(1.8026) + \Phi(1.4592) - 1 = 0.9643 + 0.9278 - 1 = 0.892\) | A1 [6] | SC \(P(31,...49)=300C31(0.13)^{31}(0.87)^{69} + \ldots +300C49\) etc.) B1B1 |
**(i)** $P(4, 5, 6) = (0.22)^y(0.78)^8C4 + (0.22)^y(0.78)^8C5 + (0.22)^y(0.78)^8C6$ | M1, M1 | Bin term with $_nC_r p^r(1-p)^{n-r}$ seen $r \neq 0$ any $p < 1$; Summing 2 or 3 bin probs $p = 0.22, n = 8$
$= 0.0763$ | A1 [3] | Correct answer
**(ii)** prob $= 0.13$
mean $= 300 \times 0.13 = 39$
var $= 300 \times 0.13 \times 0.87 = 33.93$ | B1, B1 ft | Correct prob can be implied; Correct unsimplified np and npq ft wrong 0.13
$P(30 < x < 50) = P\left(\frac{30.5 - 39}{\sqrt{33.93}} < z < \frac{49.5 - 39}{\sqrt{33.93}}\right)$ | M1 | Standardising a value need sq rt
$M1$ | Cont correction 30.5 / 31.5 or 48.5/49.5 only
$= P(-1.4592 < z < 1.8026)$ | M1 | Correct area $\Phi_1 + \Phi_2 - 1$ oe; Rounding to correct answer
$= \Phi(1.8026) + \Phi(1.4592) - 1 = 0.9643 + 0.9278 - 1 = 0.892$ | A1 [6] | SC $P(31,...49)=300C31(0.13)^{31}(0.87)^{69} + \ldots +300C49$ etc.) B1B15 On trains in the morning rush hour, each person is either a student with probability 0.36 , or an office worker with probability 0.22 , or a shop assistant with probability 0.29 or none of these.\\
(i) 8 people on a morning rush hour train are chosen at random. Find the probability that between 4 and 6 inclusive are office workers.\\
(ii) 300 people on a morning rush hour train are chosen at random. Find the probability that between 31 and 49 inclusive are neither students nor office workers nor shop assistants.
\hfill \mbox{\textit{CAIE S1 2013 Q5 [9]}}