| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single normal population sample mean |
| Difficulty | Standard +0.3 This is a straightforward application of standard normal distribution techniques: standardizing individual values, using the sampling distribution of means (dividing SD by √n), and combining independent normal variables by adding means and variances. All steps are routine calculations 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 transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | P(70 < weight < 80) = 0.789 (0.78870…) |
| [1] | 1.1 | BC |
| 5 | (b) | Mean weight ⁓ N(130, 64/10) |
| P(Mean ≥ 125) = 0.976 (0.97594…) | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | For mean |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | (i) |
| [1] | 2.4 | |
| 5 | (c) | (ii) |
Total weight ⁓ N(985, 356)
| Answer | Marks |
|---|---|
| P(Total weight < 1000) = 0.787 (0.78669…) | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | For method for mean |
Question 5:
5 | (a) | P(70 < weight < 80) = 0.789 (0.78870…) | B1
[1] | 1.1 | BC
5 | (b) | Mean weight ⁓ N(130, 64/10)
P(Mean ≥ 125) = 0.976 (0.97594…) | M1
A1
B1
[3] | 3.1b
1.1
1.1 | For mean
For variance
BC
OR total weight ⁓ N(10×130,
10×64)
5 | (c) | (i) | Weights of individual pies must be independent | E1
[1] | 2.4
5 | (c) | (ii) | Mean = 4 × (75 + 130) + 165
Variance = 4 × (16 + 64) + 36
Total weight ⁓ N(985, 356)
P(Total weight < 1000) = 0.787 (0.78669…) | M1
M1
A1
B1
[4] | 3.1b
1.1
1.1
1.1 | For method for mean
For method for variance
For both correct
BC5 A food company makes mini apple pies. The weight of pastry in a pie is Normally distributed with mean 75 g and standard deviation 4 g . The weight of filling in a pie is Normally distributed with mean 130 g and standard deviation 8 g . You should assume that the weights of pastry and filling in a pie are independent.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the weight of pastry in a randomly chosen pie is between 70 g and 80 g .
\item Find the probability that the mean weight of filling in 10 randomly chosen pies is at least 125 g.
The pies are sold in packs of 4 . The weight of the packaging is Normally distributed with mean 165 g and standard deviation 6 g .
\item In order to find the probability that the total weight of a pack of 4 pies is less than 1 kg , you must assume that the weight of the packaging is independent of the weight of the pies.
\begin{enumerate}[label=(\roman*)]
\item State another necessary assumption.
\item Given that the assumptions are valid, calculate this probability.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2021 Q5 [9]}}