| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find p then binomial probability |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution with standard z-score calculations and a simple binomial probability. Part (a)(i) requires finding P(X > 142) using z-tables, part (a)(ii) is a routine binomial calculation with the probability from (a)(i), and part (b) is an inverse normal problem using the 90th percentile. All techniques are standard S1 procedures with no problem-solving insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 142) = P\left(Z > \frac{142-125}{24}\right)\) | M1 | Substitution of correct values into the \(\pm\)Standardisation formula, allow continuity correction, not \(\sigma^2\), \(\sqrt{\sigma}\) |
| \([= P(Z > 0.7083) =] 1 - 0.7604\) | M1 | Appropriate numerical area \(\Phi\), from final process, must be probability, expect \(p < 0.5\) |
| \(0.2396\) | A1 | \(0.239 \leqslant p \leqslant 0.240\) to at least 3sf |
| *Their* \(0.2396 \times 365\ [= 87.454]\) | M1 | FT *their* 4sf (or better) probability |
| \(87\) or \(88\) | A1 FT | Final answer must be positive integer, no indication of approximation/rounding, only dependent on previous M mark. SC B1 FT for *their* 3sf probability \(\times 365 =\) integer value, condone 0.24 used |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(0,1) = 0.7604^{10} + {}^{10}C_1 \times 0.2396^1 \times 0.7604^9\) | M1 | One term: \({}^{10}C_x p^x(1-p)^{10-x}\) for \(0 < x < 10\), any \(p\) |
| \([= 0.064628 + 0.20364]\) | A1 FT | Correct unsimplified expression using *their* probability to at least 3sf from (a)(i) or correct |
| \(0.268\) | A1 | AWRT, WWW |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z = \pm 1.282\) | B1 | Correct value only, critical value |
| \(\dfrac{t - 125}{24} = -1.282\) | M1 | Use of \(\pm\) Standardisation formula with correct values substituted, allow continuity correction, \(\sigma^2\), \(\sqrt{\sigma}\), to form an equation with a \(z\)-value and not probability |
| \(t = 94.2\) | A1 | AWRT, condone AWRT 94.3. Not dependent on B mark |
| 3 |
## Question 7(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 142) = P\left(Z > \frac{142-125}{24}\right)$ | **M1** | Substitution of correct values into the $\pm$Standardisation formula, allow continuity correction, not $\sigma^2$, $\sqrt{\sigma}$ |
| $[= P(Z > 0.7083) =] 1 - 0.7604$ | **M1** | Appropriate numerical area $\Phi$, from final process, must be probability, expect $p < 0.5$ |
| $0.2396$ | **A1** | $0.239 \leqslant p \leqslant 0.240$ to at least 3sf |
| *Their* $0.2396 \times 365\ [= 87.454]$ | **M1** | **FT** *their* 4sf (or better) probability |
| $87$ or $88$ | **A1 FT** | Final answer must be positive integer, no indication of approximation/rounding, only dependent on previous **M** mark. **SC B1 FT** for *their* 3sf probability $\times 365 =$ integer value, condone 0.24 used |
| | **5** | |
## Question 7(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(0,1) = 0.7604^{10} + {}^{10}C_1 \times 0.2396^1 \times 0.7604^9$ | **M1** | One term: ${}^{10}C_x p^x(1-p)^{10-x}$ for $0 < x < 10$, any $p$ |
| $[= 0.064628 + 0.20364]$ | **A1 FT** | Correct unsimplified expression using *their* probability to at least 3sf from **(a)(i)** or correct |
| $0.268$ | **A1** | AWRT, WWW |
| | **3** | |
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = \pm 1.282$ | **B1** | Correct value only, critical value |
| $\dfrac{t - 125}{24} = -1.282$ | **M1** | Use of $\pm$ Standardisation formula with correct values substituted, allow continuity correction, $\sigma^2$, $\sqrt{\sigma}$, to form an equation with a $z$-value and not probability |
| $t = 94.2$ | **A1** | AWRT, condone AWRT 94.3. Not dependent on **B** mark |
| | **3** | |7 The times, in minutes, that Karli spends each day on social media are normally distributed with mean 125 and standard deviation 24.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item On how many days of the year ( 365 days) would you expect Karli to spend more than 142 minutes on social media?
\item Find the probability that Karli spends more than 142 minutes on social media on fewer than 2 of 10 randomly chosen days.
\end{enumerate}\item On $90 \%$ of days, Karli spends more than $t$ minutes on social media.
Find the value of $t$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q7 [11]}}