OCR MEI Further Statistics Major Specimen — Question 5 7 marks

Exam BoardOCR MEI
ModuleFurther Statistics Major (Further Statistics Major)
SessionSpecimen
Marks7
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Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeSingle sum threshold probability
DifficultyStandard +0.3 This is a straightforward application of linear combinations of normal random variables with standard formulas. Part (i) requires finding the distribution of a sum (mean = 5×508, variance = 5×3.3²) and computing a single probability. Part (ii) requires forming L - 3S and standardizing. Both are routine textbook exercises requiring only direct application of taught methods with no novel insight or complex multi-step reasoning.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

5 A particular brand of pasta is sold in bags of two different sizes. The mass of pasta in the large bags is advertised as being 1500 g ; in fact it is Normally distributed with mean 1515 g and standard deviation 4.7 g . The mass of pasta in the small bags is advertised as being 500 g ; in fact it is Normally distributed with mean 508 g and standard deviation 3.3 g .
  1. Find the probability that the total mass of pasta in 5 randomly selected small bags is less than 2550 g .
  2. Find the probability that the mass of pasta in a randomly selected large bag is greater than three times the mass of pasta in a randomly selected small bag.

Question 5:
AnswerMarks Guidance
5(i) Mass of 5 small bags ~N(5(cid:117)508,5(cid:117)3.32)
~N(2540,54.45)
P
AnswerMarks
P(X < 2550) = 0.9123E
B2
B1
AnswerMarks
[3]3.3
1.1
AnswerMarks
3.4B1 For Normal and mean,
B1 For variance
BC
AnswerMarks Guidance
FT their mean and varianceDistribution must be stated
5(ii) S
Mean of L(cid:16)3S (cid:32)1515(cid:16)3(cid:117)508(cid:32)(cid:16)9
Variance of L(cid:16)3S(cid:32)4.72(cid:14)9(cid:117)3.32
L(cid:16)3S ~N((cid:16)9,120.1)
AnswerMarks
P(L(cid:16)3S (cid:33)0)(cid:32)0.2058M1
M1
A1
A1
AnswerMarks
[4]3.3
1.1
1.1
AnswerMarks
3.4Mean
Method for variance
Correct variance
BC
Question 5:
5 | (i) | Mass of 5 small bags ~N(5(cid:117)508,5(cid:117)3.32)
~N(2540,54.45)
P
P(X < 2550) = 0.9123 | E
B2
B1
[3] | 3.3
1.1
3.4 | B1 For Normal and mean,
B1 For variance
BC
FT their mean and variance | Distribution must be stated
5 | (ii) | S
Mean of L(cid:16)3S (cid:32)1515(cid:16)3(cid:117)508(cid:32)(cid:16)9
Variance of L(cid:16)3S(cid:32)4.72(cid:14)9(cid:117)3.32
L(cid:16)3S ~N((cid:16)9,120.1)
P(L(cid:16)3S (cid:33)0)(cid:32)0.2058 | M1
M1
A1
A1
[4] | 3.3
1.1
1.1
3.4 | Mean
Method for variance
Correct variance
BC
5 A particular brand of pasta is sold in bags of two different sizes. The mass of pasta in the large bags is advertised as being 1500 g ; in fact it is Normally distributed with mean 1515 g and standard deviation 4.7 g . The mass of pasta in the small bags is advertised as being 500 g ; in fact it is Normally distributed with mean 508 g and standard deviation 3.3 g .\\
(i) Find the probability that the total mass of pasta in 5 randomly selected small bags is less than 2550 g .\\
(ii) Find the probability that the mass of pasta in a randomly selected large bag is greater than three times the mass of pasta in a randomly selected small bag.

\hfill \mbox{\textit{OCR MEI Further Statistics Major  Q5 [7]}}