(a) Five female and seven male teachers applied... | WAEC SSCE - Further Mathematics - 2008 (Essay)

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(a) Five female and seven male teachers applied for 4 vacancies in a Junior High School. The teachers are equally qualified. Find the number of ways of employing the 4 teachers, if : (i) there is no restriction ; (ii) at least 2 of them are females.

(b) The table shows the positions awarded to 7 contestants by Judges X and Y in a competition.

Contestant	P	Q	R	S	T	U	V
Judge X	2	7	1	3	6	5	4
Judge Y	4	6	2	3	1	1	5

(i) Calculate, correct to one decimal place, the Spearman's rank correlation coefficient.

(ii) Interpret your answer in b(i) above.

Answer

(a) Selecting 4 teachers from 5 female + 7 male = 12 teachers.

(i) No restriction. Choose any 4 of the 12:

\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \mathbf{495 \text{ ways}} \]

(ii) At least 2 females. The cases are exactly 2, 3 or 4 females (the rest male):

\[ \binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} + \binom{5}{4}\binom{7}{0} \] \[ = (10)(21) + (10)(7) + (5)(1) = 210 + 70 + 5 = \mathbf{285 \text{ ways}} \]

(b) Spearman's rank correlation for 7 contestants (\(n = 7\)). Judge Y has a tie (T and U both awarded position 1), so they share positions 1 and 2, each taking the average rank \(1.5\); the remaining Y values are then re-ranked \(2\to3, 3\to4, 4\to5, 5\to6, 6\to7\). Judge X already has distinct positions 1 to 7.

Contestant	\(R_X\)	\(R_Y\)	\(d\)	\(d^2\)
P	2	5	-3	9
Q	7	7	0	0
R	1	3	-2	4
S	3	4	-1	1
T	6	1.5	4.5	20.25
U	5	1.5	3.5	12.25
V	4	6	-2	4
Total \(\sum d^2\)				50.5

\[ r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 50.5}{7(49 - 1)} = 1 - \frac{303}{336} \approx 1 - 0.902 = \mathbf{0.1} \]

(ii) Interpretation. \(r_s \approx 0.1\) is very close to zero, so there is only a very weak positive correlation. The two judges show almost no agreement in the way they ranked the contestants.

Read lesson note on Permutation And Combinations (WAEC)

Answer Details

(a) Selecting 4 teachers from 5 female + 7 male = 12 teachers.

(i) No restriction. Choose any 4 of the 12:

\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \mathbf{495 \text{ ways}} \]

(ii) At least 2 females. The cases are exactly 2, 3 or 4 females (the rest male):

\[ \binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} + \binom{5}{4}\binom{7}{0} \] \[ = (10)(21) + (10)(7) + (5)(1) = 210 + 70 + 5 = \mathbf{285 \text{ ways}} \]

Contestant	\(R_X\)	\(R_Y\)	\(d\)	\(d^2\)
P	2	5	-3	9
Q	7	7	0	0
R	1	3	-2	4
S	3	4	-1	1
T	6	1.5	4.5	20.25
U	5	1.5	3.5	12.25
V	4	6	-2	4
Total \(\sum d^2\)				50.5

\[ r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 50.5}{7(49 - 1)} = 1 - \frac{303}{336} \approx 1 - 0.902 = \mathbf{0.1} \]

(ii) Interpretation. \(r_s \approx 0.1\) is very close to zero, so there is only a very weak positive correlation. The two judges show almost no agreement in the way they ranked the contestants.

Read lesson note on Permutation And Combinations (WAEC)

Permutation And Combinations

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(a) Five female and seven male teachers applied for 4 vacancies in a Junior High School. The teachers are equally qualified. Find the number of ways of empl...

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(a) Five female and seven male teachers applied for 4 vacancies in a Junior High School. The teachers are equally qualified. Find the number of ways of empl...

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