| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct expected frequency calculation |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution with standard procedures: standardizing to z-scores, using tables, and reverse lookup. Part (d) requires finding a mean from a given probability, which is slightly less routine than forward calculations, but all parts follow textbook methods with no conceptual challenges or novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks |
|---|---|
| \(P(Z > \frac{706-704}{\sqrt{3.2}}) = P(Z > 1.12) = 0.1314\) | M2 A1 |
| Answer | Marks |
|---|---|
| \(P(\frac{703-704}{\sqrt{3.2}} < Z < \frac{708-704}{\sqrt{3.2}})\) | M1 |
| \(= P(-0.56 < Z < 2.24)\) | M1 |
| \(= P(Z < 2.24) - P(Z < -0.56)\) | M1 |
| \(= 0.9875 - 0.2877 = 0.6998\) | A1 |
| Answer | Marks |
|---|---|
| \(P(Z < \frac{700-704}{\sqrt{3.2}}) = P(Z < -2.24) = 0.0125\) | M1 A1 |
| expect \(0.0125 \times 1200 = 15\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Z < \frac{700-\mu}{\sqrt{3.2}}) = 0.01\) | M1 | |
| \(\frac{700-\mu}{\sqrt{3.2}} = -3.0902\) | M1 | |
| \(\mu = 700 + (3.0902 \times \sqrt{3.2}) = 705.5 \text{ ml (1dp)}\) | M1 A1 | (15) |
| Answer | Marks |
|---|---|
| Total | (75) |
**(a)**
$P(Z > \frac{706-704}{\sqrt{3.2}}) = P(Z > 1.12) = 0.1314$ | M2 A1 |
**(b)**
$P(\frac{703-704}{\sqrt{3.2}} < Z < \frac{708-704}{\sqrt{3.2}})$ | M1 |
$= P(-0.56 < Z < 2.24)$ | M1 |
$= P(Z < 2.24) - P(Z < -0.56)$ | M1 |
$= 0.9875 - 0.2877 = 0.6998$ | A1 |
**(c)**
$P(Z < \frac{700-704}{\sqrt{3.2}}) = P(Z < -2.24) = 0.0125$ | M1 A1 |
expect $0.0125 \times 1200 = 15$ | M1 A1 |
**(d)**
$P(Z < \frac{700-\mu}{\sqrt{3.2}}) = 0.01$ | M1 |
$\frac{700-\mu}{\sqrt{3.2}} = -3.0902$ | M1 |
$\mu = 700 + (3.0902 \times \sqrt{3.2}) = 705.5 \text{ ml (1dp)}$ | M1 A1 | (15)
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**Total** | (75)7. The volume of liquid in bottles of sparkling water from one producer is believed to be normally distributed with a mean of 704 ml and a variance of $3.2 \mathrm { ml } ^ { 2 }$.
Calculate the probability that a randomly chosen bottle from this producer contains
\begin{enumerate}[label=(\alph*)]
\item more than 706 ml ,
\item between 703 and 708 ml .
The bottles are labelled as containing 700 ml .
\item In a delivery of 1200 bottles, how many could be expected to contain less than the stated 700 ml ?
The bottling process can be adjusted so that the mean changes but the variance is unchanged.
\item What should the mean be changed to in order to have only a $0.1 \%$ chance of a bottle having less than 700 ml of sparkling water? Give your answer correct to 1 decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q7 [15]}}