| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution probability then binomial/normal approximation on sample |
| Difficulty | Standard +0.3 This is a straightforward normal distribution question requiring standard z-score calculations and a normal approximation to binomial. Part (a) is direct standardization, part (b) requires finding a percentile given cumulative probability, and part (c) applies a routine approximation with continuity correction. All techniques are standard S1 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X < 76) =] P\left(Z < \frac{76 - 80.5}{6.6}\right)\) | M1 | Use of \(\pm\) standardisation formula with \(76\), \(80.5\) and \(6.6\), condone \(6.6^2\) or \(\sqrt{6.6}\), no continuity correction |
| \([= \Phi(-0.6818) = 1 - \Phi(0.6818) =]\) \(1 - 0.7524 = 0.2476\) | M1 | Calculating the appropriate probability area leading to final answer |
| \(24.8\%\) | A1 | \(24.75\% < \text{ans} \leq 24.8\%\) (percentage value required). If A0 scored, SC B1 for \(24.75\% < \text{ans} \leq 24.8\%\) www |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \([\% \text{ of large eggs} = 100 - 40 - 24.76 = 35.24]\) | B1 | \(0.378 \leq z < 0.3791\) or \(-0.3791 < z \leq -0.378\) seen |
| \(\left[P\left(Z > \frac{x - 80.5}{6.6}\right) = 0.40 + 0.2476 = 0.6476\right]\) \(\frac{x - 80.5}{6.6} = 0.378\) | M1 | Use of \(\pm\) standardisation formula with \(x\), \(80.5\), \(6.6\) and a \(z\)-value (not \(0.6476\), \(0.3524\), \(0.4\), \(0.2476\)); treat \(\pm 0.38\) as a \(z\)-value, not \(6.6^2\), not \(\sqrt{6.6}\), no continuity correction |
| \(x = 83[.0]\) | A1 | awrt \(83.0\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Mean} = 150 \times 0.4 = 60\) \(\text{Var} = 150 \times 0.4 \times 0.6 = 36\) | B1 | \(60\) and \(36\) seen, allow unsimplified |
| \(P(X > 68) = P\left(Z > \frac{68.5 - 60}{\sqrt{36}}\right)\) | M1 | Substituting their \(60\) and their \(6\) into \(\pm\) standardisation formula (any number for \(68.5\)), condone their \(\sigma^2\) and their \(\sqrt{\sigma}\) |
| M1 | Using continuity correction \(67.5\) or \(68.5\) in their standardisation formula | |
| \(P(Z > 1.417) = 1 - \Phi(1.417)\) \([= 1 - 0.9217]\) | M1 | Appropriate area \(\Phi\), from final process, must be a probability |
| \(0.0783\) | A1 | \(0.07825 < p \leq 0.0783\). If A0 scored, SC B1 for \(0.07825 < p \leq 0.0783\) |
| 5 |
## Question 3(a):
$[P(X < 76) =] P\left(Z < \frac{76 - 80.5}{6.6}\right)$ | M1 | Use of $\pm$ standardisation formula with $76$, $80.5$ and $6.6$, condone $6.6^2$ or $\sqrt{6.6}$, no continuity correction
$[= \Phi(-0.6818) = 1 - \Phi(0.6818) =]$ $1 - 0.7524 = 0.2476$ | M1 | Calculating the appropriate probability area leading to final answer
$24.8\%$ | A1 | $24.75\% < \text{ans} \leq 24.8\%$ (percentage value required). If A0 scored, SC B1 for $24.75\% < \text{ans} \leq 24.8\%$ www
| **3** |
---
## Question 3(b):
$[\% \text{ of large eggs} = 100 - 40 - 24.76 = 35.24]$ | B1 | $0.378 \leq z < 0.3791$ or $-0.3791 < z \leq -0.378$ seen
$\left[P\left(Z > \frac{x - 80.5}{6.6}\right) = 0.40 + 0.2476 = 0.6476\right]$ $\frac{x - 80.5}{6.6} = 0.378$ | M1 | Use of $\pm$ standardisation formula with $x$, $80.5$, $6.6$ and a $z$-value (not $0.6476$, $0.3524$, $0.4$, $0.2476$); treat $\pm 0.38$ as a $z$-value, not $6.6^2$, not $\sqrt{6.6}$, no continuity correction
$x = 83[.0]$ | A1 | awrt $83.0$
| **3** |
---
## Question 3(c):
$\text{Mean} = 150 \times 0.4 = 60$ $\text{Var} = 150 \times 0.4 \times 0.6 = 36$ | B1 | $60$ and $36$ seen, allow unsimplified
$P(X > 68) = P\left(Z > \frac{68.5 - 60}{\sqrt{36}}\right)$ | M1 | Substituting their $60$ and their $6$ into $\pm$ standardisation formula (any number for $68.5$), condone their $\sigma^2$ and their $\sqrt{\sigma}$
| M1 | Using continuity correction $67.5$ or $68.5$ in their standardisation formula
$P(Z > 1.417) = 1 - \Phi(1.417)$ $[= 1 - 0.9217]$ | M1 | Appropriate area $\Phi$, from final process, must be a probability
$0.0783$ | A1 | $0.07825 < p \leq 0.0783$. If A0 scored, SC B1 for $0.07825 < p \leq 0.0783$
| **5** |
---3 A farmer sells eggs. The weights, in grams, of the eggs can be modelled by a normal distribution with mean 80.5 and standard deviation 6.6. Eggs are classified as small, medium or large according to their weight. A small egg weighs less than 76 grams and $40 \%$ of the eggs are classified as medium.
\begin{enumerate}[label=(\alph*)]
\item Find the percentage of eggs that are classified as small.
\item Find the least possible weight of an egg classified as large.\\
150 of the eggs for sale last week were weighed.
\item Use an approximation to find the probability that more than 68 of these eggs were classified as medium.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q3 [11]}}