| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from summary statistics |
| Difficulty | Moderate -0.8 Part (a) is a routine application of coded mean and variance formulas with straightforward arithmetic. Part (b) is a standard normal distribution calculation requiring only a z-score computation and table lookup. Both parts are textbook exercises with no problem-solving or conceptual challenge beyond basic recall. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = 80 - \dfrac{147}{30} = 80 - 4.9 = 75.1\) | M1 | For \(-147/30\) oe seen |
| A1 | Correct answer | |
| \(\text{sd} = \sqrt{\dfrac{952}{30} - \left(\dfrac{147}{30}\right)^2} = \sqrt{7.72\ldots}\) | M1 | \(952/30 - (\pm \text{ their coded mean})^2\) |
| \(\text{sd} = 2.78\) | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(x > 160) = P\!\left(z > \dfrac{160 - 148.6}{18.5}\right)\) | M1 | Standardising, no cc, no sq rt |
| \(= P(z > 0.616)\) | ||
| \(= 1 - 0.7310\) | M1 | \(1 - \Phi\) |
| \(= 0.269\) | A1 [3] | Correct answer |
## Question 4 (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 80 - \dfrac{147}{30} = 80 - 4.9 = 75.1$ | M1 | For $-147/30$ oe seen |
| | A1 | Correct answer |
| $\text{sd} = \sqrt{\dfrac{952}{30} - \left(\dfrac{147}{30}\right)^2} = \sqrt{7.72\ldots}$ | M1 | $952/30 - (\pm \text{ their coded mean})^2$ |
| $\text{sd} = 2.78$ | A1 [4] | Correct answer |
## Question 4 (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(x > 160) = P\!\left(z > \dfrac{160 - 148.6}{18.5}\right)$ | M1 | Standardising, no cc, no sq rt |
| $= P(z > 0.616)$ | | |
| $= 1 - 0.7310$ | M1 | $1 - \Phi$ |
| $= 0.269$ | A1 [3] | Correct answer |
---4
\begin{enumerate}[label=(\alph*)]
\item Amy measured her pulse rate while resting, $x$ beats per minute, at the same time each day on 30 days. The results are summarised below.
$$\Sigma ( x - 80 ) = - 147 \quad \Sigma ( x - 80 ) ^ { 2 } = 952$$
Find the mean and standard deviation of Amy's pulse rate.
\item Amy's friend Marok measured her pulse rate every day after running for half an hour. Marok's pulse rate, in beats per minute, was found to have a mean of 148.6 and a standard deviation of 18.5. Assuming that pulse rates have a normal distribution, find what proportion of Marok's pulse rates, after running for half an hour, were above 160 beats per minute.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2015 Q4 [7]}}