| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find p then binomial probability |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard normal distribution techniques (finding σ from a percentile) followed by routine binomial probability calculation. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 16) = P\left(Z < \frac{16-28}{\sigma}\right) = 0.1\) | B1 | \(\pm 1.282\) seen, cao – critical value |
| \(\frac{16-28}{\sigma} = -1.282\) | M1 | Use of \(\pm\)standardisation formula with 16, 28, \(\sigma\) and a z-value (not 0.1, 0.9, 0.282, 0.5398, 0.8159) equated to a z-value. Condone continuity correct \(\pm 0.5\), not \(\sigma^2, \sqrt{\sigma}\). Condone \(\pm\frac{12}{\sigma} = -1.282\) |
| \(\sigma = 9.36\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([1 - P(0,1,2) =] 1 - ({}^{12}C_0(0.1)^0(0.9)^{12} + {}^{12}C_1(0.1)^1(0.9)^{11} + {}^{12}C_2(0.1)^2(0.9)^{10})\) | M1 | One term \({}^{12}C_x\,(p)^x(1-p)^{12-x}\), \(0 < p < 1\), \(x \neq 0,1,2\) |
| \([1-(0.2824 + 0.3766 + 0.2301)]\) | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| \(0.111\) | B1 | \(0.1108699\ldots\) rounded to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(3,4,5,6,7,8,9,10,11,12) = {}^{12}C_3(0.1)^3(0.9)^9 + {}^{12}C_4(0.1)^4(0.9)^8 + \ldots + {}^{12}C_{11}(0.1)^{11}(0.9)^1 + {}^{12}C_{12}(0.1)^{12}(0.9)^0\) | M1 | One term \({}^{12}C_x\,(p)^x(1-p)^{12-x}\), \(0 < p < 1\), \(x \neq 0,1,2\) |
| \([0.08523 + 0.02131 + \ldots + 1.08\times10^{-10} + 1\times10^{-12}]\) | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| \(0.111\) | B1 | \(0.1108699\ldots\) rounded to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(-1.3 < Z < 1.3)\) | B1 | Identifying at least one of \(-1.3\) or \(1.3\) as the appropriate z-values |
| \(= 2\,\Phi(1.3) - 1 = 2 \times 0.9032 - 1\) | M1 | Calculating the appropriate probability area from 2 symmetrical z-values (leading to final answer, expect \(> 0.5\)) |
| \(= 0.806,\ \frac{504}{625}\) | A1 | \(0.8064,\ 0.806 \leqslant p < 0.8065\) |
| In 365 days \(0.8064 \times 365 = 294\) or \(295\) | B1 FT | Strict FT on their at least 4-figure probability (not z-value). Final answer must be positive integer, no approximation or rounding stated |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 16) = P\left(Z < \frac{16-28}{\sigma}\right) = 0.1$ | B1 | $\pm 1.282$ seen, cao – critical value |
| $\frac{16-28}{\sigma} = -1.282$ | M1 | Use of $\pm$standardisation formula with 16, 28, $\sigma$ and a z-value (not 0.1, 0.9, 0.282, 0.5398, 0.8159) equated to a z-value. Condone continuity correct $\pm 0.5$, not $\sigma^2, \sqrt{\sigma}$. Condone $\pm\frac{12}{\sigma} = -1.282$ |
| $\sigma = 9.36$ | A1 | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[1 - P(0,1,2) =] 1 - ({}^{12}C_0(0.1)^0(0.9)^{12} + {}^{12}C_1(0.1)^1(0.9)^{11} + {}^{12}C_2(0.1)^2(0.9)^{10})$ | M1 | One term ${}^{12}C_x\,(p)^x(1-p)^{12-x}$, $0 < p < 1$, $x \neq 0,1,2$ |
| $[1-(0.2824 + 0.3766 + 0.2301)]$ | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| $0.111$ | B1 | $0.1108699\ldots$ rounded to at least 3SF |
**Alternative Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(3,4,5,6,7,8,9,10,11,12) = {}^{12}C_3(0.1)^3(0.9)^9 + {}^{12}C_4(0.1)^4(0.9)^8 + \ldots + {}^{12}C_{11}(0.1)^{11}(0.9)^1 + {}^{12}C_{12}(0.1)^{12}(0.9)^0$ | M1 | One term ${}^{12}C_x\,(p)^x(1-p)^{12-x}$, $0 < p < 1$, $x \neq 0,1,2$ |
| $[0.08523 + 0.02131 + \ldots + 1.08\times10^{-10} + 1\times10^{-12}]$ | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| $0.111$ | B1 | $0.1108699\ldots$ rounded to at least 3SF |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(-1.3 < Z < 1.3)$ | B1 | Identifying at least one of $-1.3$ or $1.3$ as the appropriate z-values |
| $= 2\,\Phi(1.3) - 1 = 2 \times 0.9032 - 1$ | M1 | Calculating the appropriate probability area from 2 symmetrical z-values (leading to final answer, expect $> 0.5$) |
| $= 0.806,\ \frac{504}{625}$ | A1 | $0.8064,\ 0.806 \leqslant p < 0.8065$ |
| In 365 days $0.8064 \times 365 = 294$ or $295$ | B1 FT | Strict FT on their at least 4-figure probability (not z-value). Final answer must be positive integer, no approximation or rounding stated |6 The mass of grapes sold per day by a large shop can be modelled by a normal distribution with mean 28 kg . On $10 \%$ of days less than 16 kg of grapes are sold.
\begin{enumerate}[label=(\alph*)]
\item Find the standard deviation of the mass of grapes sold per day.\\
The mass of grapes sold on any day is independent of the mass sold on any other day.
\item 12 days are chosen at random.
Find the probability that less than 16 kg of grapes are sold on more than 2 of these 12 days.
\item In a random sample of 365 days, on how many days would you expect the mass of grapes sold to be within 1.3 standard deviations of the mean?
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q6 [10]}}