| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| 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 standard table lookups and z-score calculations. Part (a) is direct standardization and table reading; part (b) involves working backwards from a probability but uses a symmetric interval from the mean, making it routine. Both parts are textbook exercises with no problem-solving insight required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(X > 25) = P\left(Z > \frac{25-20}{4}\right)\) | M1 | Standardise with 20 and 4; allow numerator \(20-25\) |
| \(= P(Z > 1.25)\) | M1 | \(1-\) probability for second M1 |
| \(= 1 - 0.8944 = 0.1056\) | A1 | Anything that rounds to \(0.106\); correct answer with no working award 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(X < 20) = 0.5\) so \(P(X < d) = 0.5 + 0.4641 = 0.9641\) | B1 | \(0.9641\) seen or implied by \(1.80\) |
| \(P(Z < z) = 0.9641,\ z = 1.80\) | B1 | \(1.80\) seen |
| \(\frac{d-20}{4} = 1.80\) | M1 | Standardise with 20 and 4 and equate to \(z\) value; \(Z=0.8315\) is M0 |
| \(d = 27.2\) | A1 | Anything that rounds to \(27.2\); correct answer with no working award 4/4 |
## Question 6:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(X > 25) = P\left(Z > \frac{25-20}{4}\right)$ | M1 | Standardise with 20 and 4; allow numerator $20-25$ |
| $= P(Z > 1.25)$ | M1 | $1-$ probability for second M1 |
| $= 1 - 0.8944 = 0.1056$ | A1 | Anything that rounds to $0.106$; correct answer with no working award 3/3 |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(X < 20) = 0.5$ so $P(X < d) = 0.5 + 0.4641 = 0.9641$ | B1 | $0.9641$ seen or implied by $1.80$ |
| $P(Z < z) = 0.9641,\ z = 1.80$ | B1 | $1.80$ seen |
| $\frac{d-20}{4} = 1.80$ | M1 | Standardise with 20 and 4 and equate to $z$ value; $Z=0.8315$ is M0 |
| $d = 27.2$ | A1 | Anything that rounds to $27.2$; correct answer with no working award 4/4 |
---6. The random variable $X$ has a normal distribution with mean 20 and standard deviation 4 .
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 25 )$.
\item Find the value of $d$ such that $\mathrm { P } ( 20 < X < d ) = 0.4641$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2007 Q6 [7]}}