| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find p then binomial probability |
| Difficulty | Standard +0.3 This is a straightforward S1 normal distribution question with standard techniques: inverse normal calculation to find σ, probability calculations using tables, and a simple binomial probability. Part (d) tests conceptual understanding of mean/variance but requires only basic reasoning. Slightly above average due to multiple parts and the conceptual component, but all techniques are routine for S1. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(X > 4.5) = 0.625\), so \(P\left(Z > \frac{-0.7}{\sigma}\right) = 0.625\) | M1 A1 | |
| \(\frac{-0.7}{\sigma} = -0.32\), \(\sigma = 2.1875 \approx 2.2\) | M1 A1 A1 | |
| (b) \(P(4 < X < 7) = P(-0.55 < Z < 0.82)\) | M1 A1 | |
| \(= 0.794 - (1 - 0.709) = 0.503\), i.e. \(50.3\%\) | M1 A1 | |
| (c) \(P(X < 5) = P(Z < -0.091) = 1 - 0.536 = 0.464\) | M1 A1 | |
| so assuming independence, probability = \(0.464^2 = 0.215\) | B1 A1 | |
| (d) (i) unchanged at \(5.2\) | B1 | |
| (ii) decreases, as average deviation from mean is less | B2 | |
| Not normal as shape of curve changes, and \(P(X = 5.2) > 0\) | B1 | Total: 17 marks |
(a) $P(X > 4.5) = 0.625$, so $P\left(Z > \frac{-0.7}{\sigma}\right) = 0.625$ | M1 A1 |
$\frac{-0.7}{\sigma} = -0.32$, $\sigma = 2.1875 \approx 2.2$ | M1 A1 A1 |
(b) $P(4 < X < 7) = P(-0.55 < Z < 0.82)$ | M1 A1 |
$= 0.794 - (1 - 0.709) = 0.503$, i.e. $50.3\%$ | M1 A1 |
(c) $P(X < 5) = P(Z < -0.091) = 1 - 0.536 = 0.464$ | M1 A1 |
so assuming independence, probability = $0.464^2 = 0.215$ | B1 A1 |
(d) (i) unchanged at $5.2$ | B1 |
(ii) decreases, as average deviation from mean is less | B2 |
Not normal as shape of curve changes, and $P(X = 5.2) > 0$ | B1 | **Total: 17 marks**7. The times taken by a large number of people to read a certain book can be modelled by a normal distribution with mean $5 \cdot 2$ hours. It is found that $62 \cdot 5 \%$ of the people took more than $4 \cdot 5$ hours to read the book.
\begin{enumerate}[label=(\alph*)]
\item Show that the standard deviation of the times is approximately $2 \cdot 2$ hours.
\item Calculate the percentage of the people who took between 4 and 7 hours to read the book.
\item Calculate the probability that two of the people chosen at random both took less than 5 hours to read the book, stating any assumption that you make.
\item If a number of extra people were taken into account, all of whom took exactly $5 \cdot 2$ hours to read the book, state with reasons what would happen to (i) the mean, (ii) the variance and explain briefly why the distribution would no longer be normal.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q7 [17]}}