| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| 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 S1 normal distribution question requiring only standard table lookups and inverse normal calculations. Part (a) is direct standardization and table reading; part (b) is a routine inverse normal problem. Both are textbook exercises with no problem-solving or conceptual challenges beyond basic procedure. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks |
|---|---|
| \((z = \pm) \frac{15 - 16.12}{1.6} (= -0.70)\) | M1 |
| P(\(Z < -0.70\)) = 1 − 0.7580 = 0.2420 (awrt 0.242) | M1, A1 |
| Answer | Marks |
|---|---|
| [P(\(T < t\)) = 0.30 implies] \(z = \frac{t - 16.12}{1.6} = -0.5244\) | M1 A1 |
| \(\frac{t - 16.12}{1.6} = -0.5244 \Rightarrow t = 16.12 - 1.6 \times "0.5244"\) | M1 |
| \(t = \text{awrt } 15.28\) (allow awrt 15.28/9) | A1 |
## (a)
$(z = \pm) \frac{15 - 16.12}{1.6} (= -0.70)$ | M1 |
P($Z < -0.70$) = 1 − 0.7580 = 0.2420 (awrt 0.242) | M1, A1 |
## (b)
[P($T < t$) = 0.30 implies] $z = \frac{t - 16.12}{1.6} = -0.5244$ | M1 A1 |
$\frac{t - 16.12}{1.6} = -0.5244 \Rightarrow t = 16.12 - 1.6 \times "0.5244"$ | M1 |
$t = \text{awrt } 15.28$ (allow awrt 15.28/9) | A1 |
**Notes:**
- Allow slips e.g. 16.2 for 16.12 for 1st M1 in (a) and (b)
- **1st M1** for standardising expression with 15, 16.12 and 1.6 − allow ±
- **2nd M1** for 1 - a probability (> 0.5) from tables or calculator based on their standardised value. **Correct answer only scores 3/3**
- **In part (b)** they can use any letter or symbol instead of t
- **1st M1** for standardising with t (o.e.), 16.12 and 1.6, allow ±, and setting equal to a z value
- **1st A1** for an equation with $z = \pm 0.5244$ or better. e.g. $\frac{t - 16.12}{1.6} = \pm 0.52$ (or 0.525) scores M1 (but A0)
- **2nd M1** for solving their linear equation as far as $t = a \pm b \times 1.6$. Not dependent on 1st
- e.g. solving $\frac{t - 16.12}{1.6} = 0.3$ to give $t = 16.12 + 1.6 \times 0.3$ scores this M1
- Allow $\frac{t - 16.12}{1.6^2} = 0.3$ to give $t = 16.12 + 1.6^2 \times 0.3$ to score M1 too
- **2nd A1** dependent on both M marks. Allow awrt 15.28 or awrt 15.29. Condone awrt 15.3 if a correct expression for $t = ...$ is seen. **Answers with no working: 15.28 is M1A1M1A1, 15.29 is M1A0M1A1, 15.3 is M1A0M1A0**
---Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60
A child from the school is selected at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that this child runs 100 m in less than 15 s. [3]
\end{enumerate}
On sports day the school awards certificates to the fastest 30\% of the children in the 100 m race.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Estimate, to 2 decimal places, the slowest time taken to run 100 m for which a child will be awarded a certificate. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2011 Q4 [7]}}