| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Conditional probability with normal |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question requiring standard z-score calculations and understanding of conditional probability. Part (a) is routine, part (b) applies P(A|B) = P(A∩B)/P(B) with normal probabilities, and part (c) uses inverse normal tables. All techniques are standard S1 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(H > 170) = P\left(Z > \frac{170-160}{8}\right)\) \([= P(Z > 1.25)]\) | M1 | Attempt at standardising with 170, 160 and 8. Allow \(\pm\) i.e. for \(\pm\frac{170-160}{8}\) |
| \(= 1 - 0.8944\) | M1 | Attempting \(1-p\) where \(0.8 < p < 1\). Correct answer only 3/3 |
| \(= 0.1056\) (calc 0.1056498...) awrt 0.106 (accept 10.6%) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(H > 180) = P\left(Z > \frac{180-160}{8}\right)\) \([= 1 - 0.9938]\) | M1 | Standardising with 180, 160 and 8 |
| \(= 0.0062\) (calc 0.006209...) awrt 0.0062 or \(\frac{31}{5000}\) | A1 | For 0.0062 seen, maybe seen as part of another expression/calculation |
| \([P(H > 180 \mid H > 170)] = \frac{0.0062}{0.1056}\) | M1 | Using conditional probability with denom = their (a) and num < their denom. Values needed |
| \(= 0.0587\) (calc 0.0587760...) awrt 0.0587 or 0.0588 | A1 | For awrt 0.0587 or 0.0588. Condone 5.87% or 5.88% or \(\frac{31}{528}\). Correct answer only 4/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(H > h \mid H > 170) (= 0.5)\) or \(\frac{P(H>h)}{P(H>170)}(= 0.5)\) | M1 | For a correct conditional probability statement. Either line, don't insist on 0.5, ft (a) |
| \([P(H > h)] = 0.5 \times \text{"0.1056"} = 0.0528\) (calc 0.0528249...) or \([P(H < h)] = 0.9472\) | A1ft | For \([P(H>h)] = 0.5 \times \text{their}(a)\). Award M1A1ft for correct evaluation of \(0.5 \times \text{their}(a)\) or sight of 0.0528 or better |
| \(\frac{h-160}{8} = 1.62\) (calc 1.6180592...) | M1 B1 | 2nd M1 for attempt to standardise (\(\pm\)) with 160 and 8 and set equal to \(\pm z\) value \((1.56 < \ |
| \(h =\) awrt 173 cm awrt 173 | A1 | For awrt 173 but dependent on both M marks |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(H > 170) = P\left(Z > \frac{170-160}{8}\right)$ $[= P(Z > 1.25)]$ | M1 | Attempt at standardising with 170, 160 and 8. Allow $\pm$ i.e. for $\pm\frac{170-160}{8}$ |
| $= 1 - 0.8944$ | M1 | Attempting $1-p$ where $0.8 < p < 1$. Correct answer only 3/3 |
| $= 0.1056$ (calc 0.1056498...) **awrt 0.106** (accept 10.6%) | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(H > 180) = P\left(Z > \frac{180-160}{8}\right)$ $[= 1 - 0.9938]$ | M1 | Standardising with 180, 160 and 8 |
| $= 0.0062$ (calc 0.006209...) awrt 0.0062 or $\frac{31}{5000}$ | A1 | For 0.0062 seen, maybe seen as part of another expression/calculation |
| $[P(H > 180 \mid H > 170)] = \frac{0.0062}{0.1056}$ | M1 | Using conditional probability with denom = their (a) and num < their denom. Values needed |
| $= 0.0587$ (calc 0.0587760...) **awrt 0.0587 or 0.0588** | A1 | For awrt 0.0587 or 0.0588. Condone 5.87% or 5.88% or $\frac{31}{528}$. Correct answer only 4/4 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(H > h \mid H > 170) (= 0.5)$ or $\frac{P(H>h)}{P(H>170)}(= 0.5)$ | M1 | For a correct conditional probability statement. Either line, don't insist on 0.5, ft (a) |
| $[P(H > h)] = 0.5 \times \text{"0.1056"} = 0.0528$ (calc 0.0528249...) or $[P(H < h)] = 0.9472$ | A1ft | For $[P(H>h)] = 0.5 \times \text{their}(a)$. Award M1A1ft for correct evaluation of $0.5 \times \text{their}(a)$ or sight of 0.0528 or better |
| $\frac{h-160}{8} = 1.62$ (calc 1.6180592...) | M1 B1 | 2nd M1 for attempt to standardise ($\pm$) with 160 and 8 and set equal to $\pm z$ value $(1.56 < \|z\| < 1.68)$. B1 for $(z=)$ awrt $\pm 1.62$ (seen) |
| $h =$ awrt 173 cm **awrt 173** | A1 | For awrt 173 but dependent on **both** M marks |
---7. The heights of adult females are normally distributed with mean 160 cm and standard deviation 8 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly selected adult female has a height greater than 170 cm .
Any adult female whose height is greater than 170 cm is defined as tall.
An adult female is chosen at random. Given that she is tall,
\item find the probability that she has a height greater than 180 cm .
Half of tall adult females have a height greater than $h \mathrm {~cm}$.
\item Find the value of $h$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q7 [12]}}