| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Moderate -0.3 Part (a) is a straightforward binomial calculation with n=8, requiring cumulative probability P(X<6). Part (b) applies the standard normal approximation to binomial with continuity correction for n=120, p=0.7. Both parts are routine applications of textbook methods with no conceptual challenges—slightly easier than average due to clear setup and standard technique, though the two-part structure and need for continuity correction prevents it from being trivial. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - P(6,7,8) = 1 - ({}^8C_6\ 0.7^6 0.3^2 + {}^8C_7\ 0.7^7 0.3^1 + 0.7^8)\) | M1 | One term \({}^8C_x\ p^x(1-p)^{8-x}\), \(0 < p < 1\), \(x \neq 0\) |
| \(= 1 - 0.55177\) | A1 | Correct unsimplified expression, or better |
| \(= 0.448\) | A1 | |
| Alternative: \(P(0,1,2,3,4,5) = 0.3^8 + {}^8C_1\ 0.7^1 0.3^7 + {}^8C_2\ 0.7^2 0.3^6 + {}^8C_3\ 0.7^3 0.3^5 + {}^8C_4\ 0.7^4 0.3^4 + {}^8C_5\ 0.7^5 0.3^3\) | M1 | One term \({}^8C_x\ p^x(1-p)^{8-x}\), \(0 < p < 1\), \(x \neq 0\) |
| A1 | Correct unsimplified expression, or better | |
| \(= 0.448\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 120 \times 0.7 = 84\); Var \(= 120 \times 0.7 \times 0.3 = 25.2\) | B1 | Correct mean and variance, allow unsimplified |
| \(P(\text{more than } 75) = P\left(z > \frac{75.5 - 84}{\sqrt{25.2}}\right)\) | M1 | Substituting *their* \(\mu\) and \(\sigma\) into \(\pm\) standardising formula (any number), not \(\sigma^2\), not \(\sqrt{\sigma}\) |
| M1 | Using continuity correction 75.5 or 74.5 | |
| \(P(z > -1.693)\) | M1 | Appropriate area \(\Phi\), from final process, must be a probability |
| \(= 0.955\) | A1 | Allow \(0.9545 < p \leqslant 0.955\) |
| Total: 5 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - P(6,7,8) = 1 - ({}^8C_6\ 0.7^6 0.3^2 + {}^8C_7\ 0.7^7 0.3^1 + 0.7^8)$ | M1 | One term ${}^8C_x\ p^x(1-p)^{8-x}$, $0 < p < 1$, $x \neq 0$ |
| $= 1 - 0.55177$ | A1 | Correct unsimplified expression, or better |
| $= 0.448$ | A1 | |
| **Alternative:** $P(0,1,2,3,4,5) = 0.3^8 + {}^8C_1\ 0.7^1 0.3^7 + {}^8C_2\ 0.7^2 0.3^6 + {}^8C_3\ 0.7^3 0.3^5 + {}^8C_4\ 0.7^4 0.3^4 + {}^8C_5\ 0.7^5 0.3^3$ | M1 | One term ${}^8C_x\ p^x(1-p)^{8-x}$, $0 < p < 1$, $x \neq 0$ |
| | A1 | Correct unsimplified expression, or better |
| $= 0.448$ | A1 | |
| **Total: 3** | | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 120 \times 0.7 = 84$; Var $= 120 \times 0.7 \times 0.3 = 25.2$ | B1 | Correct mean and variance, allow unsimplified |
| $P(\text{more than } 75) = P\left(z > \frac{75.5 - 84}{\sqrt{25.2}}\right)$ | M1 | Substituting *their* $\mu$ and $\sigma$ into $\pm$ standardising formula (any number), not $\sigma^2$, not $\sqrt{\sigma}$ |
| | M1 | Using continuity correction 75.5 or 74.5 |
| $P(z > -1.693)$ | M1 | Appropriate area $\Phi$, from final process, must be a probability |
| $= 0.955$ | A1 | Allow $0.9545 < p \leqslant 0.955$ |
| **Total: 5** | | |5 In Greenton, 70\% of the adults own a car. A random sample of 8 adults from Greenton is chosen.\\[0pt]
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the number of adults in this sample who own a car is less than 6 . [3]\\
A random sample of 120 adults from Greenton is now chosen.
\item Use an approximation to find the probability that more than 75 of them own a car.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q5 [8]}}