OCR MEI Further Statistics Major 2019 June — Question 3 9 marks

Exam BoardOCR MEI
ModuleFurther Statistics Major (Further Statistics Major)
Year2019
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeMixed sum threshold probability
DifficultyStandard +0.3 This is a straightforward application of linear combinations of normal random variables with clear parameters. Part (a) requires summing 5 independent normals, part (b) applies a simple linear transformation (scaling), and part (c) combines multiple normals with given means and variances. All parts use standard techniques with no conceptual challenges beyond routine variance calculations and normal table lookups, making it slightly easier than average for Further Maths Statistics.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

3 The weights of bananas sold by a supermarket are modelled by a Normal distribution with mean 205 g and standard deviation 11 g .
  1. Find the probability that the total weight of 5 randomly selected bananas is at least 1 kg . When a banana is peeled the change in its weight is modelled as being a reduction of \(35 \%\).
  2. Find the probability that the weight of a randomly selected peeled banana is at most 150 g Andy makes smoothies. Each smoothie is made using 2 peeled bananas and 20 strawberries from the supermarket, all the items being randomly chosen. The weight of a strawberry is modelled by a Normal distribution with mean 22.5 g and standard deviation 2.7 g .
  3. Find the probability that the total weight of a smoothie is less than 700 g .

Question 3:
AnswerMarks Guidance
3(a) Total weight : N(5 × 205, 5 × 112)
P(Total ≥ 1000 g) = 0.8453M1
A1
AnswerMarks
[2]For distribution
BC
AnswerMarks Guidance
3(b) Peeled weight : N(0.65 × 205, 0.652 × 112)
P(weight ≤ 150 g) = 0.9904B1
M1
A1
AnswerMarks
[3]For N and mean
For variance
AnswerMarks
BC(mean = 133.25)
(variance =
51.1225)
AnswerMarks Guidance
3(c) Weight of smoothie
: N(2×133.25 + 20×22.5, 2×51.1225 + 20×2.72)
N(716.5, 248.045)
AnswerMarks
P(weight < 700 g) = 0.1474M1
M1
A1
A1
AnswerMarks
[4]Method for mean FT their part (b)
Method for variance FT their part (b)
For both correct
BC
Question 3:
3 | (a) | Total weight : N(5 × 205, 5 × 112)
P(Total ≥ 1000 g) = 0.8453 | M1
A1
[2] | For distribution
BC
3 | (b) | Peeled weight : N(0.65 × 205, 0.652 × 112)
P(weight ≤ 150 g) = 0.9904 | B1
M1
A1
[3] | For N and mean
For variance
BC | (mean = 133.25)
(variance =
51.1225)
3 | (c) | Weight of smoothie
: N(2×133.25 + 20×22.5, 2×51.1225 + 20×2.72)
N(716.5, 248.045)
P(weight < 700 g) = 0.1474 | M1
M1
A1
A1
[4] | Method for mean FT their part (b)
Method for variance FT their part (b)
For both correct
BC
3 The weights of bananas sold by a supermarket are modelled by a Normal distribution with mean 205 g and standard deviation 11 g .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the total weight of 5 randomly selected bananas is at least 1 kg .

When a banana is peeled the change in its weight is modelled as being a reduction of $35 \%$.
\item Find the probability that the weight of a randomly selected peeled banana is at most 150 g

Andy makes smoothies. Each smoothie is made using 2 peeled bananas and 20 strawberries from the supermarket, all the items being randomly chosen. The weight of a strawberry is modelled by a Normal distribution with mean 22.5 g and standard deviation 2.7 g .
\item Find the probability that the total weight of a smoothie is less than 700 g .
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2019 Q3 [9]}}