| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Standard +0.3 Part (i) is a straightforward normal distribution probability calculation requiring standardization and table lookup. Part (ii) adds a binomial probability layer (finding P(loss) first, then using binomial with n=4, r=1), which is a standard combination of topics but requires recognizing the two-stage approach. This is slightly above average difficulty due to the multi-step nature and combining two distributions, but remains a textbook-style question. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z_1 = \frac{12-6.4}{5.2} = 1.077\) | M1 | Standardising, can be all in thousands, no mix, no cc, no sq rt, no sq |
| \(z_2 = \frac{10-6.4}{5.2} = 0.692\) | M1 | \(\Phi_2 - \Phi_1\), \(\Phi_2\) must be \(> \Phi_1\) |
| \(\Phi(z_1) - \Phi(z_2) = 0.8593 - 0.7556 = 0.104\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{loss}) = P\!\left(z < \frac{0-6.4}{5.2}\right) = P(z < -1.231)\) | M1 | Standardising using \(x=0\), accept \(\frac{0.5-6.4}{5.2}\) |
| \(= 1 - 0.8909\) | ||
| \(= 0.109\) | A1 | Correct probability |
| \(P(1) = (0.1091)^1(0.8909)^3 \times {}_4C_1\) | M1 | Binomial term \({}_4C_x p^x(1-p)^{4-x}\) any \(p\), \(x \neq 0\) |
| \(= 0.309\) or \(0.308\) | A1 [4] | Correct answer |
## Question 2:
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $z_1 = \frac{12-6.4}{5.2} = 1.077$ | M1 | Standardising, can be all in thousands, no mix, no cc, no sq rt, no sq |
| $z_2 = \frac{10-6.4}{5.2} = 0.692$ | M1 | $\Phi_2 - \Phi_1$, $\Phi_2$ must be $> \Phi_1$ |
| $\Phi(z_1) - \Phi(z_2) = 0.8593 - 0.7556 = 0.104$ | A1 [3] | Correct answer |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{loss}) = P\!\left(z < \frac{0-6.4}{5.2}\right) = P(z < -1.231)$ | M1 | Standardising using $x=0$, accept $\frac{0.5-6.4}{5.2}$ |
| $= 1 - 0.8909$ | | |
| $= 0.109$ | A1 | Correct probability |
| $P(1) = (0.1091)^1(0.8909)^3 \times {}_4C_1$ | M1 | Binomial term ${}_4C_x p^x(1-p)^{4-x}$ any $p$, $x \neq 0$ |
| $= 0.309$ or $0.308$ | A1 [4] | Correct answer |
---2 The random variable $X$ is the daily profit, in thousands of dollars, made by a company. $X$ is normally distributed with mean 6.4 and standard deviation 5.2.\\
(i) Find the probability that, on a randomly chosen day, the company makes a profit between $\$ 10000$ and $\$ 12000$.\\
(ii) Find the probability that the company makes a loss on exactly 1 of the next 4 consecutive days.
\hfill \mbox{\textit{CAIE S1 2012 Q2 [7]}}