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Question
The set \(P\) contains all prime numbers less than 2500.
The set \(Q\) is the set of all subsets of \(P\).
The set \(S\) contains all positive integers less than 2500.
The function \(f:{\text{ }}S \to Q\) is defined by \(f(s)\) as the set of primes exactly dividing \(s\), for \(s \in S\).
For example \(f(4) = \{ 2\} ,{\text{ }}f(45) = \{ 3,{\text{ }}5\} \).
Explain why only one of the following statements is true
(i) \(17 \subset P\);
(ii) \(\{ 7,{\text{ }}17,{\text{ }}37,{\text{ }}47,{\text{ }}57\} \in Q\);
(iii) \(\phi \subset Q\) and \(\phi \in Q\), where \(\phi \) is the empty set.
(i) State the value of \(f(1)\), giving a reason for your answer.
(ii) Find \(n\left( {f(2310)} \right)\).
Determine whether or not \(f\) is
(i) injective;
(ii) surjective.
Answer/Explanation
Markscheme
(i) 17 is an element not a subset of \(P\) R1
(ii) 57 is not a prime number R1
(iii) any demonstration that this is the true statement A1
because every set contains the empty set as a subset R1
[4 marks]
(i) \(f(1) = \phi \) A1
because 1 has no prime factors R1
(ii) \(f(2310) = f(2 \times 3 \times 5 \times 7 \times 11){\text{ }}\left( { = \{ 2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11\} } \right)\) A1
\(n\left( {f(2310)} \right) = 5\) A1
[4 marks]
(i) not injective A1
because, for example, \(f(2) = f(4) = \{ 2\} \) R1
(ii) not surjective A1
\({f^{ – 1}}(2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11,{\text{ }}13)\) does not belong to S because
\(2 \times 3 \times 5 \times 7 \times 11 \times 13 > 2500\) R1
Note: Accept any appropriate example.
[4 marks]
Question
A sample of size 100 is taken from a normal population with unknown mean μ and known variance 36.
Another investigator decides to use the same data to test the hypotheses H0 : μ = 65 , H1 : μ = 67.9.
An investigator wishes to test the hypotheses H0 : μ = 65, H1 : μ > 65.
He decides on the following acceptance criteria:
Accept H0 if the sample mean \(\bar x\) ≤ 66.5
Accept H1 if \(\bar x\) > 66.5
Find the probability of a Type I error.
She decides to use the same acceptance criteria as the previous investigator. Find the probability of a Type II error.
Find the critical value for \({\bar x}\) if she wants the probabilities of a Type I error and a Type II error to be equal.
Answer/Explanation
Markscheme
\(\bar X \sim {\text{N}}\left( {\mu ,\,\frac{{{\sigma ^2}}}{n}} \right)\)
\(\bar X \sim {\text{N}}\left( {65,\,\frac{{36}}{{100}}} \right)\) (A1)
P(Type I Error) \( = {\text{P}}\left( {\bar X > 66.5} \right)\) (M1)
= 0.00621 A1
[3 marks]
P(Type II Error) = P(accept H0 | H1 is true)
\( = {\text{P}}\left( {\bar X \leqslant 66.5\left| {\mu = 67.9} \right.} \right)\) (M1)
\( = {\text{P}}\left( {\bar X \leqslant 66.5} \right)\) when \(\bar X \sim {\text{N}}\left( {67.9,\,\frac{{36}}{{100}}} \right)\) (M1)
= 0.00982 A1
[3 marks]
the variances of the distributions given by H0 and H1 are equal, (R1)
by symmetry the value of \({\bar x}\) lies midway between 65 and 67.9 (M1)
\( \Rightarrow \bar x = \frac{1}{2}\left( {65 + 67.9} \right) = 66.45\) A1
[3 marks]