Pre-U Pre-U 9794/1 Specimen — Question 10 9 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
SessionSpecimen
Marks9
TopicNormal Distribution
TypeMixed calculations with boundaries
DifficultyStandard +0.3 This is a straightforward normal distribution question requiring standard z-score calculations and inverse normal lookups. Part (i) is routine standardization, part (ii) involves finding a parameter using inverse normal, and part (iii) uses symmetry to find mean and variance from two percentiles. All techniques are standard A-level material with no novel problem-solving required, making it slightly easier than average.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

10 Cheeky Cola is sold in bottles of two sizes, small and large. For each size, the content of a randomly chosen bottle is normally distributed with mean and standard deviation, in litres, as given in the table.
MeanStandard deviation
Small bottle0.50.01
Large bottle1.5\(x\)
  1. Find the probability that a randomly chosen small bottle contains more than 0.51 litres.
  2. Find \(x\) if the probability that a randomly chosen large bottle contains less than 1.45 litres is 0.1 . The manufacturer introduces a new size of bottle of Cheeky Cola, called the mega bottle. It is found that the probabilities that a randomly chosen mega bottle contains less than 2.97 litres or more than 3.05 litres are both 0.05 .
  3. Assuming that the contents of the mega bottle are normally distributed, find the mean and variance of the distribution.

(i) Calculate \(z = \frac{0.51 - 0.5}{0.1} = 1\) [B1]
\(1 - \Phi(1) = 1 - 0.8413 = 0.159\) [B1] 2 marks
(ii) Use of \(\Phi(z) = 1 - 0.1 = 0.9\) [M1]
Attempt to solve \(z = 1.282 = \frac{0.05}{x}\) [A1]
Obtain \(x = 0.039\) [A1] 3 marks
(iii) Using the symmetry about the mean [M1]
\(\mu = 3.01\) [A1]
Solve \(1.645 = \frac{0.04}{\sigma}\) [A1]
Variance \(= \sigma^2 = 5.91 \times 10^{-4}\) [A1] 4 marks
(i) Calculate $z = \frac{0.51 - 0.5}{0.1} = 1$ [B1]
$1 - \Phi(1) = 1 - 0.8413 = 0.159$ [B1] **2 marks**

(ii) Use of $\Phi(z) = 1 - 0.1 = 0.9$ [M1]
Attempt to solve $z = 1.282 = \frac{0.05}{x}$ [A1]
Obtain $x = 0.039$ [A1] **3 marks**

(iii) Using the symmetry about the mean [M1]
$\mu = 3.01$ [A1]
Solve $1.645 = \frac{0.04}{\sigma}$ [A1]
Variance $= \sigma^2 = 5.91 \times 10^{-4}$ [A1] **4 marks**
10 Cheeky Cola is sold in bottles of two sizes, small and large. For each size, the content of a randomly chosen bottle is normally distributed with mean and standard deviation, in litres, as given in the table.

\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
 & Mean & Standard deviation \\
\hline
Small bottle & 0.5 & 0.01 \\
\hline
Large bottle & 1.5 & $x$ \\
\hline
\end{tabular}
\end{center}

(i) Find the probability that a randomly chosen small bottle contains more than 0.51 litres.\\
(ii) Find $x$ if the probability that a randomly chosen large bottle contains less than 1.45 litres is 0.1 .

The manufacturer introduces a new size of bottle of Cheeky Cola, called the mega bottle. It is found that the probabilities that a randomly chosen mega bottle contains less than 2.97 litres or more than 3.05 litres are both 0.05 .\\
(iii) Assuming that the contents of the mega bottle are normally distributed, find the mean and variance of the distribution.

\hfill \mbox{\textit{Pre-U Pre-U 9794/1  Q10 [9]}}