| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Linear relationship μ = kσ |
| Difficulty | Standard +0.3 This is a straightforward normal distribution problem requiring standardization and inverse normal lookup, followed by a routine binomial probability calculation. The constraint μ = 3σ simplifies to a single-variable problem, and both parts use standard techniques with no novel insight required. |
| 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/Working | Mark | Guidance |
| \(z = 0.38\) | B1 | \(\pm 0.38\) (0) seen or implied |
| \(\pm\frac{25-\mu}{\mu/3} = 0.38\) | M1 | Standardising attempt resulting in \(z =\) some \(\mu/\sigma\)/both, no continuity correction |
| (substituting to eliminate \(\mu\) or \(\sigma\)) | M1 | Substituting to eliminate \(\mu\) or \(\sigma\) and attempt to solve linear equation |
| \(\mu = 22.2,\ \sigma = 7.40\) | A1 [4] | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(4) = {}^6C_4(0.352)^4(0.648)^2\) | M1 | \({}^6C_r \times (p)^r \times (1-p)^{6-r},\ r = 2\) or \(4\) |
| \(= 0.0967\) | A1 [2] | Correct answer |
## Question 1:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 0.38$ | B1 | $\pm 0.38$ (0) seen or implied |
| $\pm\frac{25-\mu}{\mu/3} = 0.38$ | M1 | Standardising attempt resulting in $z =$ some $\mu/\sigma$/both, no continuity correction |
| (substituting to eliminate $\mu$ or $\sigma$) | M1 | Substituting to eliminate $\mu$ or $\sigma$ and attempt to solve linear equation |
| $\mu = 22.2,\ \sigma = 7.40$ | A1 [4] | Both correct |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(4) = {}^6C_4(0.352)^4(0.648)^2$ | M1 | ${}^6C_r \times (p)^r \times (1-p)^{6-r},\ r = 2$ or $4$ |
| $= 0.0967$ | A1 [2] | Correct answer |
---1 The random variable $X$ is normally distributed and is such that the mean $\mu$ is three times the standard deviation $\sigma$. It is given that $\mathrm { P } ( X < 25 ) = 0.648$.\\
(i) Find the values of $\mu$ and $\sigma$.\\
(ii) Find the probability that, from 6 random values of $X$, exactly 4 are greater than 25 .
\hfill \mbox{\textit{CAIE S1 2011 Q1 [6]}}