| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Moderate -0.3 This is a straightforward two-part question combining normal distribution with binomial probability. Part (i) requires standard z-score calculations (routine for S1), and part (ii) is a direct application of binomial probability with n=4. The question involves no conceptual challenges—just methodical application of standard techniques taught in Statistics 1, 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 |
|---|---|---|
| \(z_1 = \pm\frac{90-120}{24} = -\frac{5}{4},\quad z_2 = \pm\frac{140-120}{24} = \frac{5}{6}\) | M1 | At least one standardisation, no cc, no sq rt, no sq using 120 and 24 and either 90 or 140 |
| \(= \Phi\!\left(\frac{20}{24}\right) - \Phi\!\left(-\frac{30}{24}\right)\) | A1 | \(-5/4\) and \(5/6\) unsimplified |
| \(= \Phi(0.8333) - (1 - \Phi(1.25)) = 0.7975 - (1 - 0.8944)\) or \(0.8944 - 0.2025 = 0.6919\) | M1 | Correct area \(\Phi - \Phi\) legitimately obtained and evaluated from phi(their \(z_2\)) – phi(their \(z_1\)) |
| \(= 0.692\) AG | A1 | Correct answer obtained from 0.7975 and 0.1056 or to 4sf or 0.6919 seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Any binomial term of form \(4C_x p^x (1-p)^{4-x}\), \(x \neq 0\) or \(4\) | M1 | Any binomial term of form \(4C_x p^x (1-p)^{4-x}\), \(x \neq 0\) or \(4\) |
| \(P(2,3,4) = 0.692^2(1-0.692)^2 \times {}^4C_2 + 0.692^3(1-0.692) \times {}^4C_3 + 0.692^4\) | B1 | One correct bin term with \(n=4\) and \(p=0.692\) |
| \(= 0.27256 + 0.40825 + 0.22931\) | M1 | Correct unsimplified expression using \(0.692\) or better |
| \(= 0.910\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - P(0,1) =\) | M1 | Any binomial term of form \(4C_x p^x (1-p)^{4-x}\), \(x \neq 0\) or \(4\) |
| \(1 - 0.692^0(1-0.692)^4 \times {}^4C_0 - 0.692^1(1-0.692)^3 \times {}^4C_1\) | B1 | One correct bin term with \(n=4\) and \(p=0.692\) |
| \(= 1 - 0.00899 - 0.0808757\) | M1 | Correct unsimplified expression using \(0.692\) or better |
| \(= 0.910\) | A1 | Correct answer |
## Question 5(i):
$z_1 = \pm\frac{90-120}{24} = -\frac{5}{4},\quad z_2 = \pm\frac{140-120}{24} = \frac{5}{6}$ | M1 | At least one standardisation, no cc, no sq rt, no sq using 120 and 24 and either 90 or 140
$= \Phi\!\left(\frac{20}{24}\right) - \Phi\!\left(-\frac{30}{24}\right)$ | A1 | $-5/4$ and $5/6$ unsimplified
$= \Phi(0.8333) - (1 - \Phi(1.25)) = 0.7975 - (1 - 0.8944)$ or $0.8944 - 0.2025 = 0.6919$ | M1 | Correct area $\Phi - \Phi$ legitimately obtained and evaluated from phi(their $z_2$) – phi(their $z_1$)
$= 0.692$ AG | A1 | Correct answer obtained from 0.7975 and 0.1056 or to 4sf or 0.6919 seen
## Question 5(ii):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Any binomial term of form $4C_x p^x (1-p)^{4-x}$, $x \neq 0$ or $4$ | M1 | Any binomial term of form $4C_x p^x (1-p)^{4-x}$, $x \neq 0$ or $4$ |
| $P(2,3,4) = 0.692^2(1-0.692)^2 \times {}^4C_2 + 0.692^3(1-0.692) \times {}^4C_3 + 0.692^4$ | B1 | One correct bin term with $n=4$ and $p=0.692$ |
| $= 0.27256 + 0.40825 + 0.22931$ | M1 | Correct unsimplified expression using $0.692$ or better |
| $= 0.910$ | A1 | Correct answer |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - P(0,1) =$ | M1 | Any binomial term of form $4C_x p^x (1-p)^{4-x}$, $x \neq 0$ or $4$ |
| $1 - 0.692^0(1-0.692)^4 \times {}^4C_0 - 0.692^1(1-0.692)^3 \times {}^4C_1$ | B1 | One correct bin term with $n=4$ and $p=0.692$ |
| $= 1 - 0.00899 - 0.0808757$ | M1 | Correct unsimplified expression using $0.692$ or better |
| $= 0.910$ | A1 | Correct answer |
---5 The weights of apples sold by a store can be modelled by a normal distribution with mean 120 grams and standard deviation 24 grams. Apples weighing less than 90 grams are graded as 'small'; apples weighing more than 140 grams are graded as 'large'; the remainder are graded as 'medium'.\\
(i) Show that the probability that an apple chosen at random is graded as medium is 0.692 , correct to 3 significant figures.\\
(ii) Four apples are chosen at random. Find the probability that at least two are graded as medium. [4]\\
\hfill \mbox{\textit{CAIE S1 2018 Q5 [8]}}