| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Standard +0.3 Part (i) requires standardizing a normal distribution and using inverse normal tables (routine but slightly above basic recall). Part (ii) is a straightforward binomial calculation using the probability from part (i). This is a standard two-part question testing normal distribution and binomial application with no novel insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = 0.674\) | B1 | \(z\) value \(\pm 0.674\) |
| \(0.674 = \frac{0--3}{\sigma}\) | M1 | \(\pm\)Standardising with 0 and equating to a \(z\)-value |
| \(\sigma = 4.45\) | A1 | Correct answer; not ignoring a minus sign |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(0,1)\) | M1 | Any bin of form \(^8C_x(0.75)^x(0.25)^{8-x}\) any \(x\) |
| \(= (0.75)^8 + ^8C_1(0.25)(0.75)^7\) | M1 | Correct unsimplified answer, may be implied by numerical values |
| \(0.1001 + 0.2670 = 0.367\) | A1 | Correct answer |
| Method 2: \(1 - P(8,7,6,5,4,3,2) = 1-(0.25)^8 - ^8C_1(0.75)(0.25)^7 - \ldots\) | M1 | Any bin of form \(^8C_x(0.75)^x(0.25)^{8-x}\) any \(x\) |
| \(- ^8C_2(0.75)^6(0.25)^2\) | M1 | Correct unsimplified answer |
| \(= 0.367\) | A1 | Correct answer |
| Total: 3 |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = 0.674$ | B1 | $z$ value $\pm 0.674$ |
| $0.674 = \frac{0--3}{\sigma}$ | M1 | $\pm$Standardising with 0 and equating to a $z$-value |
| $\sigma = 4.45$ | A1 | Correct answer; not ignoring a minus sign |
| **Total: 3** | | |
## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0,1)$ | M1 | Any bin of form $^8C_x(0.75)^x(0.25)^{8-x}$ any $x$ |
| $= (0.75)^8 + ^8C_1(0.25)(0.75)^7$ | M1 | Correct unsimplified answer, may be implied by numerical values |
| $0.1001 + 0.2670 = 0.367$ | A1 | Correct answer |
| **Method 2:** $1 - P(8,7,6,5,4,3,2) = 1-(0.25)^8 - ^8C_1(0.75)(0.25)^7 - \ldots$ | M1 | Any bin of form $^8C_x(0.75)^x(0.25)^{8-x}$ any $x$ |
| $- ^8C_2(0.75)^6(0.25)^2$ | M1 | Correct unsimplified answer |
| $= 0.367$ | A1 | Correct answer |
| **Total: 3** | | |2 The random variable $X$ has the distribution $\mathrm { N } \left( - 3 , \sigma ^ { 2 } \right)$. The probability that a randomly chosen value of $X$ is positive is 0.25 .\\
(i) Find the value of $\sigma$.\\
(ii) Find the probability that, of 8 random values of $X$, fewer than 2 will be positive.\\
\hfill \mbox{\textit{CAIE S1 2018 Q2 [6]}}