| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard two probabilities given |
| Difficulty | Standard +0.3 This is a standard normal distribution problem requiring z-score lookups and simultaneous equations. While it has multiple parts (11 marks total), each step follows a routine procedure: converting percentages to z-scores, writing equations, and solving. The 68.3% in part (d) is a well-known property (μ±σ). No novel insight required, just careful application of standard techniques, 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 |
|---|---|---|
| \(P(3 < 2.55) = 0.025\) | \(\frac{235-\mu}{\sigma} = -1.96\) | |
| \(\therefore \frac{235-\mu}{\sigma} = -1.96\) | M1 | \(\frac{235-\mu}{\sigma}\) |
| \(\therefore \mu - 235 = 1.96\sigma\) | A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(3 > 2\mu) = 0.15\) | \(\frac{2\mu-\mu}{\sigma} = -e\); M1 | |
| \(\therefore 2\mu - \mu = 1.0364\sigma\) | M1 | 1.0364 |
| \(\therefore 2\mu - \mu = 1.0364\sigma\) | B1 | |
| A1/(3) |
| Answer | Marks | Guidance |
|---|---|---|
| Solving for \(\mu\) and \(\sigma\) | M1 | |
| Substituting for other unknowns | M1 | |
| \(\mu = 268.30\)...; \(\sigma = 17.0204\)... | AWRT 268; AWRT 17 | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu \pm \sigma = 268.3 \pm 17.02 = (251.2, 285)\) | M1 | \(\mu + \text{three} \sigma\) |
| 3sf | A1/(2) |
## Part (a)
$P(3 < 2.55) = 0.025$ | | $\frac{235-\mu}{\sigma} = -1.96$
$\therefore \frac{235-\mu}{\sigma} = -1.96$ | M1 | $\frac{235-\mu}{\sigma}$
$\therefore \mu - 235 = 1.96\sigma$ | A1 (2) |
## Part (b)
$P(3 > 2\mu) = 0.15$ | | $\frac{2\mu-\mu}{\sigma} = -e$; M1
$\therefore 2\mu - \mu = 1.0364\sigma$ | M1 | 1.0364
$\therefore 2\mu - \mu = 1.0364\sigma$ | B1 |
| A1/(3) |
## Part (c)
Solving for $\mu$ and $\sigma$ | M1 |
Substituting for other unknowns | M1 |
$\mu = 268.30$...; $\sigma = 17.0204$... | AWRT 268; AWRT 17 | A1 (4)
## Part (d)
$\mu \pm \sigma = 268.3 \pm 17.02 = (251.2, 285)$ | M1 | $\mu + \text{three} \sigma$
| 3sf | A1/(2)
---The duration of the pregnancy of a certain breed of cow is normally distributed with mean $\mu$ days and standard deviation $\sigma$ days. Only 2.5\% of all pregnancies are shorter than 235 days and 15\% are longer than 286 days.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mu - 235 = 1.96\sigma$. [2]
\item Obtain a second equation in $\mu$ and $\sigma$. [3]
\item Find the value of $\mu$ and the value of $\sigma$. [4]
\item Find the values between which the middle 68.3\% of pregnancies lie. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2002 Q5 [11]}}