| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Independent binomial samples with compound probability |
| Difficulty | Standard +0.3 This is a straightforward multi-part binomial question requiring standard calculations: (a) direct binomial probability with n=7, (b) compound probability using result from (a), and (c) normal approximation to binomial. All techniques are routine S1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \([P(5,6,7) =]\ {^7C_5}\ 0.7^5\ 0.3^2 + {^7C_6}\ 0.7^6\ 0.3^1 + 0.7^7\) | M1 | One term \(^7C_x\ (p)^x(1-p)^{7-x}\), with \(0 < p < 1\), \(x \neq 0\) or 7. |
| \([= 0.31765 + 0.24706 + 0.08235]\) | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer. |
| \(= 0.647\) | B1 | \(0.647 \leqslant p < 0.6475\) |
| Method 2: \(1 - \{0.3^7 + {^7C_1}\ 0.7^1\ 0.3^6 + {^7C_2}\ 0.7^2\ 0.3^5 + {^7C_3}\ 0.7^3\ 0.3^4 + {^7C_4}\ 0.7^4\ 0.3^3\}\) | (M1) | One term \(^7C_x\ (p)^x(1-p)^{7-x}\), with \(0 < p < 1\), \(x \neq 0\) or 7. |
| Correct expression | (A1) | Condone omission of final bracket '\(\}\)'. If other brackets omitted, allow recovery if \(1 - 0.35294\) seen. |
| \(= 0.647\) | (B1) | \(0.647 \leqslant p < 0.6475\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \([1 - P(0\ \text{white weeks}) =]\ 1 - (1-0.647)^3\) | M1 | \(1 - p^3\), \(0 < p < 1\), \(p = 1 - their\ \textbf{(a)}\), or correct. |
| \(0.956\) | A1 | |
| Method 2: \([P(1,2,3\ \text{white weeks}) =]\ 3 \times 0.647 \times 0.353^2 + 3 \times 0.647^2 \times 0.353 + 0.647^3\) | (M1) | \(3 \times q \times (1-q)^2 + 3 \times q^2 \times (1-q) + q^3\), \(q = their\ \textbf{(a)}\), or correct. |
| \(0.956\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 60 \times 0.8 = 48\), Variance \(= 60 \times 0.8 \times 0.2 = 9.6\) | B1 | 48 and 9.6, \(9\frac{3}{5}\), \(\frac{48}{5}\) seen, allow unsimplified. May be seen in standardisation formula. \([\sigma =] 3.098 \leqslant \sigma \leqslant 3.1[0]\) implies correct variance. Incorrect notation penalised but values can be used as anticipated in remainder of question. |
| \(P(X < 47) = P\left(Z < \frac{46.5 - 48}{\sqrt{9.6}}\right)\) | M1 | Substituting *their* \(\mu\) and \(\sigma\) into \(\pm\) standardising formula (any number for 46.5), not *their* \(\sigma^2\) or \(\sqrt{\text{their}\ \sigma}\) |
| M1 | Use continuity correction 46.5 or 47.5 in *their* standardised formula. Note: \(\frac{\pm 1.5}{\sqrt{9.6}}\) or \(\frac{\pm 1.5}{3.098}\) seen gains M2 BOD. | |
| \([P(Z < -0.4841) = 1 - \Phi(0.4841)]\) \(1 - 0.6858\) | M1 | Appropriate area \(\Phi\), from final process, must be a probability. Expect final answer \(< 0.5\). Note: appropriate final answer implies this M1. |
| \(= 0.314\) | A1 | \(0.314 \leqslant p < 0.3145\) |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $[P(5,6,7) =]\ {^7C_5}\ 0.7^5\ 0.3^2 + {^7C_6}\ 0.7^6\ 0.3^1 + 0.7^7$ | M1 | One term $^7C_x\ (p)^x(1-p)^{7-x}$, with $0 < p < 1$, $x \neq 0$ or 7. |
| $[= 0.31765 + 0.24706 + 0.08235]$ | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer. |
| $= 0.647$ | B1 | $0.647 \leqslant p < 0.6475$ |
| **Method 2:** $1 - \{0.3^7 + {^7C_1}\ 0.7^1\ 0.3^6 + {^7C_2}\ 0.7^2\ 0.3^5 + {^7C_3}\ 0.7^3\ 0.3^4 + {^7C_4}\ 0.7^4\ 0.3^3\}$ | (M1) | One term $^7C_x\ (p)^x(1-p)^{7-x}$, with $0 < p < 1$, $x \neq 0$ or 7. |
| Correct expression | (A1) | Condone omission of final bracket '$\}$'. If other brackets omitted, allow recovery if $1 - 0.35294$ seen. |
| $= 0.647$ | (B1) | $0.647 \leqslant p < 0.6475$ |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $[1 - P(0\ \text{white weeks}) =]\ 1 - (1-0.647)^3$ | M1 | $1 - p^3$, $0 < p < 1$, $p = 1 - their\ \textbf{(a)}$, or correct. |
| $0.956$ | A1 | |
| **Method 2:** $[P(1,2,3\ \text{white weeks}) =]\ 3 \times 0.647 \times 0.353^2 + 3 \times 0.647^2 \times 0.353 + 0.647^3$ | (M1) | $3 \times q \times (1-q)^2 + 3 \times q^2 \times (1-q) + q^3$, $q = their\ \textbf{(a)}$, or correct. |
| $0.956$ | (A1) | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 60 \times 0.8 = 48$, Variance $= 60 \times 0.8 \times 0.2 = 9.6$ | B1 | 48 and 9.6, $9\frac{3}{5}$, $\frac{48}{5}$ seen, allow unsimplified. May be seen in standardisation formula. $[\sigma =] 3.098 \leqslant \sigma \leqslant 3.1[0]$ implies correct variance. Incorrect notation penalised but values can be used as anticipated in remainder of question. |
| $P(X < 47) = P\left(Z < \frac{46.5 - 48}{\sqrt{9.6}}\right)$ | M1 | Substituting *their* $\mu$ and $\sigma$ into $\pm$ standardising formula (any number for 46.5), not *their* $\sigma^2$ or $\sqrt{\text{their}\ \sigma}$ |
| | M1 | Use continuity correction 46.5 or 47.5 in *their* standardised formula. Note: $\frac{\pm 1.5}{\sqrt{9.6}}$ or $\frac{\pm 1.5}{3.098}$ seen gains M2 BOD. |
| $[P(Z < -0.4841) = 1 - \Phi(0.4841)]$ $1 - 0.6858$ | M1 | Appropriate area $\Phi$, from final process, must be a probability. Expect final answer $< 0.5$. Note: appropriate final answer implies this M1. |
| $= 0.314$ | A1 | $0.314 \leqslant p < 0.3145$ |
---5 In a certain area in the Arctic the probability that it snows on any given day is 0.7 , independent of all other days.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that in a week (7 days) it snows on at least five days.\\
A week in which it snows on at least five days out of seven is called a 'white' week.
\item Find the probability that in three randomly chosen weeks at least one is a white week.\\
In a different area in the Arctic, the probability that a week is a white week is 0.8 .
\item Use a suitable approximation to find the probability that in 60 randomly chosen weeks fewer than 47 are white weeks.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q5 [10]}}