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Further Mathematics Paper 2, Nov/Dec. 2013  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
General Comments
Weakness/Remedies
Strength
Question 13

 

Question 13

(a)        A committee of 6 is to be selected at random from a group of 7 boys and 4 girls.
            (i)         In how many ways can this be done if there are no restrictions?
            (ii)        Find the probability that the committee contains at least two girls.
(b)        Eight students scored the following marks in theory and practical tests in Chemistry.

Students

A

B

C

D

E

F

G

H

Score in Theory

6

7

5.5

4

8

9

5

3

Score in Practical

9

7

4

3

6

8

6.5

5

            
            Calculate spearman’s rank correlation coefficient between scores in the two tests.

_____________________________________________________________________________________________________
Observation

This question was reported to be well attempted by majority of the candidates.  Their performance in this question was also reported to be good.  Candidates were reported to have shown that if there were no restrictions, then number of ways of selecting 6 person from a group of 7 boys and 4 girls was     11C6 = 462 ways.
The probability of selecting at least two girls implied that either 2 or 3 or 4 girls could be selected.  It does not include 5 and 6 since there were only 4 girls in the group i.e. Pr(at least 2 girls)                      =  +  +  =  + +   or 0.80
Many candidates were reported not to have interpreted “at least 2 girls” correctly.

In part(b), candidates were expected to rank the students’ score in each subject, find the squared difference in their ranks and apply the formula r = 1 -  where d = difference in ranks and n = number of observations, r = correlation coefficient, as shown:

Student

Rank in Theory

Rank in Practical

d

d2

A

4

1

3

9

B

3

3

0

0

C

5

7

-2

4

D

7

8

-1

1

E

2

5

-3

9

F

1

2

-1

1

G

6

4

2

4

       H

             8

             6

   2

4

 

 

 

∑d²=32

 

 

Therefore r = 1   = 0.
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