| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard two probabilities given |
| Difficulty | Standard +0.3 This is a standard S1 inverse normal problem requiring students to set up two equations from given probabilities (P(X ≥ 1.78) = 0.2 and P(X ≥ 1.65) = 0.7), find z-scores from tables, solve simultaneous equations for μ and σ, then calculate a final probability. While it involves multiple steps, it follows a well-practiced routine with no conceptual surprises, making it slightly easier than the average A-level question. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.04g Normal distribution properties: empirical rule (68-95-99.7), points of inflection |
| Answer | Marks |
|---|---|
| 2 separate sketches OK. | B1 |
| Bell Shape | B1 |
| 1.78 & 0.2 or 1.65 & 0.3 | B1 |
| Accept clear alternatives to 0.3: 0.7/0.5/0.2 |
| Answer | Marks |
|---|---|
| \(\frac{1.78 - \mu}{\sigma} = 0.8416 \Rightarrow 1.78 - \mu = 0.8416\sigma\) | M1 |
| either for method | |
| 0.8416 | B1 |
| \(\frac{1.65 - \mu}{\sigma} = -0.5244 \Rightarrow 1.65 - \mu = -0.5244\sigma\) | B1 |
| (-))0.5244 | |
| Solving gives \(\mu = 1.70\), \(\sigma = 0.095\) | M1A1A1 |
| N.B. awrt 0.84, 0.52 B1B0 | |
| awrt 1.7, 0.095 cao |
| Answer | Marks |
|---|---|
| \(P(\text{height} \geq 1.74) = 1 - P(\text{height} < 1.74)\) | M1 |
| 'one minus' | |
| \(= 1 - P\left(Z < \frac{1.74-1.70}{0.095}\right)\) | M1 |
| standardise with their mu and sigma | |
| \(= 1 - P(Z < 0.42) = 0.3372\) | A1 |
| awrt 0.337 |
**Part (a):**
| 2 separate sketches OK. | B1 |
| Bell Shape | B1 |
| 1.78 & 0.2 or 1.65 & 0.3 | B1 |
| Accept clear alternatives to 0.3: 0.7/0.5/0.2 | |
**Part (b):**
| $\frac{1.78 - \mu}{\sigma} = 0.8416 \Rightarrow 1.78 - \mu = 0.8416\sigma$ | M1 |
| either for method | |
| 0.8416 | B1 |
| $\frac{1.65 - \mu}{\sigma} = -0.5244 \Rightarrow 1.65 - \mu = -0.5244\sigma$ | B1 |
| (-))0.5244 | |
| Solving gives $\mu = 1.70$, $\sigma = 0.095$ | M1A1A1 |
| N.B. awrt 0.84, 0.52 B1B0 | |
| awrt 1.7, 0.095 cao | |
**Part (c):**
| $P(\text{height} \geq 1.74) = 1 - P(\text{height} < 1.74)$ | M1 |
| 'one minus' | |
| $= 1 - P\left(Z < \frac{1.74-1.70}{0.095}\right)$ | M1 |
| standardise with their mu and sigma | |
| $= 1 - P(Z < 0.42) = 0.3372$ | A1 |
| awrt 0.337 | |
**Total 12 marks**
---5. From experience a high-jumper knows that he can clear a height of at least 1.78 m once in 5 attempts. He also knows that he can clear a height of at least 1.65 m on 7 out of 10 attempts.
Assuming that the heights the high-jumper can reach follow a Normal distribution,
\begin{enumerate}[label=(\alph*)]
\item draw a sketch to illustrate the above information,
\item find, to 3 decimal places, the mean and the standard deviation of the heights the high-jumper can reach,
\item calculate the probability that he can jump at least 1.74 m .
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2006 Q5 [12]}}