| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Single tail probability P(X < a) or P(X > a) |
| Difficulty | Standard +0.3 This is a straightforward S1 normal distribution question requiring standardization and z-table lookups. Parts (a) and (b) are routine single-tail probability calculations. Part (c) involves working backwards from a probability to find a mean, which is slightly less routine but still a standard S1 technique with clear structure. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Z < \frac{127 - 122.3}{2.6}) = P(Z < 1.81) = 0.9649\) | M2 A1 | |
| \(P(Z < \frac{121.5 - 122.3}{2.6}) = P(Z < -0.31) = 0.3783\) | M2 A1 | |
| \(P(Z < \frac{454 - \mu}{1.6}) = 0.05\) | M1 | |
| \(\frac{454 - \mu}{1.6} = -1.6449; \mu = 456.6\) (4sf) | M1 A2 | (10) |
$P(Z < \frac{127 - 122.3}{2.6}) = P(Z < 1.81) = 0.9649$ | M2 A1 |
$P(Z < \frac{121.5 - 122.3}{2.6}) = P(Z < -0.31) = 0.3783$ | M2 A1 |
$P(Z < \frac{454 - \mu}{1.6}) = 0.05$ | M1 |
$\frac{454 - \mu}{1.6} = -1.6449; \mu = 456.6$ (4sf) | M1 A2 | (10)4. A company produces jars of English Honey. The weight of the glass jars used is normally distributed with a mean of 122.3 g and a standard deviation of 2.6 g .
Calculate the probability that a randomly chosen jar will weigh
\begin{enumerate}[label=(\alph*)]
\item less than 127 g ,
\item less than 121.5 g .
The weight of honey put into each jar by a machine is normally distributed with a standard deviation of 1.6 g . The machine operator can adjust the mean weight of the honey put into each jar without changing the standard deviation.
\item Find, correct to 4 significant figures, the minimum that the mean weight can be set to such that at most 1 in 20 of the jars will contain less than 454 g .\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q4 [10]}}