| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.8 This is a straightforward normal distribution question requiring only standard table lookups and inverse normal calculations. Part (a) involves standardizing to find P(Z < 1.24) and part (b) requires finding the inverse normal for the 75th percentile. Both are routine S1 procedures with no problem-solving or conceptual challenges beyond basic application of the normal distribution formula. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 21) = P\left(z < \frac{21-15.8}{4.2}\right) = \Phi(1.238)\) | M1 | |
| \(0.892\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z = \pm 0.674\) | B1 | |
| \(\frac{k-15.8}{4.2} = 0.674\) | M1 | |
| \(18.6\) | A1 |
## Question 3:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 21) = P\left(z < \frac{21-15.8}{4.2}\right) = \Phi(1.238)$ | M1 | |
| $0.892$ | A1 | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = \pm 0.674$ | B1 | |
| $\frac{k-15.8}{4.2} = 0.674$ | M1 | |
| $18.6$ | A1 | |
---3 In a certain town, the time, $X$ hours, for which people watch television in a week has a normal distribution with mean 15.8 hours and standard deviation 4.2 hours.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen person from this town watches television for less than 21 hours in a week.
\item Find the value of $k$ such that $\mathrm { P } ( X < k ) = 0.75$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q3 [5]}}