| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution with standard bookwork techniques: standardizing to find probabilities, using symmetry for two-tailed intervals, and reverse standardization. All three parts follow routine procedures with clearly stated parameters and require only table lookups and basic algebraic manipulation, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(< 700) = P\left(z < \frac{700 - 830}{120}\right) = P(z < -1.083)\) | M1 | Using \(\pm\) standardisation formula, no continuity correction, not \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(= 1 - 0.8606\) | M1 | Appropriate area \(\Phi\) from standardisation formula \(P(z<\ldots)\) in final probability solution, (\(<0.5\) if \(z\) is \(-\)ve, \(>0.5\) if \(z\) is \(+\)ve) |
| \(= 0.1394\) | A1 | Correct final probability rounding to 0.139 |
| Expected number of female adults \(= 430 \times \textit{their } 0.1394 = 59.9\) So 59 or 60 | B1 | FT their 3 or 4 SF probability, rounded or truncated to integer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{giraffe} < 830+w) = 95\%\) so \(z = 1.645\) | B1 | \(\pm 1.645\) seen (critical value) |
| \(\frac{(830 + w) - 830}{120} = \frac{w}{120} = 1.645\) | M1 | An equation using the standardisation formula with a \(z\)-value (not \(1-z\)), condone \(\sigma^2\) or \(\sqrt{\sigma}\), not 0.8519, 0.8289 |
| \(w = 197\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{male} > 950) = 0.834\), so \(z = -0.97\) | B1 | \(\pm 0.97\) seen |
| \(\frac{950 - 1190}{\sigma} = -0.97\) | M1 | Using \(\pm\) standardisation formula, condone continuity correction, \(\sigma^2\) or \(\sqrt{\sigma}\), condone equating with non \(z\)-value not 0.834, 0.166 |
| \(\sigma = 247\) | A1 | Condone \(-\sigma = -247\). www |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(< 700) = P\left(z < \frac{700 - 830}{120}\right) = P(z < -1.083)$ | M1 | Using $\pm$ standardisation formula, no continuity correction, not $\sigma^2$ or $\sqrt{\sigma}$ |
| $= 1 - 0.8606$ | M1 | Appropriate area $\Phi$ from standardisation formula $P(z<\ldots)$ in final probability solution, ($<0.5$ if $z$ is $-$ve, $>0.5$ if $z$ is $+$ve) |
| $= 0.1394$ | A1 | Correct final probability rounding to 0.139 |
| Expected number of female adults $= 430 \times \textit{their } 0.1394 = 59.9$ So 59 or 60 | B1 | **FT** their 3 or 4 SF probability, rounded or truncated to integer |
---
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{giraffe} < 830+w) = 95\%$ so $z = 1.645$ | B1 | $\pm 1.645$ seen (critical value) |
| $\frac{(830 + w) - 830}{120} = \frac{w}{120} = 1.645$ | M1 | An equation using the standardisation formula with a $z$-value (not $1-z$), condone $\sigma^2$ or $\sqrt{\sigma}$, not 0.8519, 0.8289 |
| $w = 197$ | A1 | Correct answer |
---
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{male} > 950) = 0.834$, so $z = -0.97$ | B1 | $\pm 0.97$ seen |
| $\frac{950 - 1190}{\sigma} = -0.97$ | M1 | Using $\pm$ standardisation formula, condone continuity correction, $\sigma^2$ or $\sqrt{\sigma}$, condone equating with non $z$-value not 0.834, 0.166 |
| $\sigma = 247$ | A1 | Condone $-\sigma = -247$. www |
---7 The weight of adult female giraffes has a normal distribution with mean 830 kg and standard deviation 120 kg .\\
(i) There are 430 adult female giraffes in a particular game reserve. Find the number of these adult female giraffes which can be expected to weigh less than 700 kg .\\
(ii) Given that $90 \%$ of adult female giraffes weigh between $( 830 - w ) \mathrm { kg }$ and $( 830 + w ) \mathrm { kg }$, find the value of $w$.\\
The weight of adult male giraffes has a normal distribution with mean 1190 kg and standard deviation $\sigma \mathrm { kg }$.\\
(iii) Given that $83.4 \%$ of adult male giraffes weigh more than 950 kg , find the value of $\sigma$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q7 [10]}}