| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from percentile/frequency data |
| Difficulty | Standard +0.8 Parts (i) and (ii) are routine normal distribution calculations using tables. Part (iii) requires solving simultaneous equations from two percentiles (9.6th and 84.8th percentiles), which is more demanding than standard inverse normal problems and requires algebraic manipulation of z-scores—this elevates it above average difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X<45) = P\left(Z < \frac{45-40}{8}\right) = P(Z < 0.625)\) | M1 | \(\pm\) Standardise, no continuity correction, \(\sigma^2\) or \(\sqrt{\sigma}\), formula must be seen |
| \(0.734(0)\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - 2(1-(i)) = 2(i)-1 = 2((i)-0.5)\) | M1 | Use result of part (i) or recalculated to find area. OE |
| \(0.468\) | A1ft | \(0 <\) FT from (i) \(< 1\) or correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X<10) = 48/500 = 0.096\); \(z = -1.305\) | B1 | \(z = \pm 1.305\) |
| \(P(X>24) = 76/500 = 0.152\); \(z = 1.028\) | B1 | \(z = \pm 1.028\) |
| \(10 - \mu = -1.305\sigma\); \(24 - \mu = 1.028\sigma\) | M1 | Form 1 equation using 10 or 24 with \(\mu, \sigma, z\)-value. Allow continuity correction, not \(\sigma^2\), \(\sqrt{\sigma}\) |
| \(14 = 2.333\sigma\) | M1 | OE. Solve two equations in \(\sigma\) and \(\mu\) to form equation in one variable |
| \(\sigma = 6.[00]\), \(\mu = 17.8[3]\) | A1 | CAO, WWW |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X<45) = P\left(Z < \frac{45-40}{8}\right) = P(Z < 0.625)$ | M1 | $\pm$ Standardise, no continuity correction, $\sigma^2$ or $\sqrt{\sigma}$, formula must be seen |
| $0.734(0)$ | A1 | CAO |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 2(1-(i)) = 2(i)-1 = 2((i)-0.5)$ | M1 | Use result of **part (i)** or recalculated to find area. OE |
| $0.468$ | A1ft | $0 <$ FT from **(i)** $< 1$ or correct |
---
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X<10) = 48/500 = 0.096$; $z = -1.305$ | B1 | $z = \pm 1.305$ |
| $P(X>24) = 76/500 = 0.152$; $z = 1.028$ | B1 | $z = \pm 1.028$ |
| $10 - \mu = -1.305\sigma$; $24 - \mu = 1.028\sigma$ | M1 | Form 1 equation using 10 or 24 with $\mu, \sigma, z$-value. Allow continuity correction, not $\sigma^2$, $\sqrt{\sigma}$ |
| $14 = 2.333\sigma$ | M1 | OE. Solve two equations in $\sigma$ and $\mu$ to form equation in one variable |
| $\sigma = 6.[00]$, $\mu = 17.8[3]$ | A1 | CAO, WWW |
---6 The heights, in metres, of fir trees in a large forest have a normal distribution with mean 40 and standard deviation 8 .\\
(i) Find the probability that a fir tree chosen at random in this forest has a height less than 45 metres.\\
(ii) Find the probability that a fir tree chosen at random in this forest has a height within 5 metres of the mean.\\
In another forest, the heights of another type of fir tree are modelled by a normal distribution. A scientist measures the heights of 500 randomly chosen trees of this type. He finds that 48 trees are less than 10 m high and 76 trees are more than 24 m high.\\
(iii) Find the mean and standard deviation of the heights of trees of this type.\\
\hfill \mbox{\textit{CAIE S1 2019 Q6 [9]}}