(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. (i) Calculate, correct to one decimal place, the Spearman's rank correlation coefficient. (ii) Interpret your answer in b(i) above.

Wednesday, 09 April 2025

Register • Login

(a) Five female and seven male teachers applied for 4 vacancies in a Junior High ...

Trending Questions

** Free JAMB 2025 CBT Practice **

In a federal system,the essence of specifying the constitutional relationship between ...

Abraham circumcised his son,Isaac, on the eight days because

In mammals, the exchange of nutrients and metabolic products occurs in the

The problems of conducting census include?

Take Free Practice Test On 2025 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE

(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.

Explanation

(a) 5 female, 7 male = 12 in total.
(i) If no restriction, there are 12 teachers to fill up 4 vacancies.
Number of ways =

^{12} C_{4}

\frac{12!}{4! 8!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2}

495

ways.
(ii) No of ways if at least 2 of them are females
= No of ways [no restriction - 0 female - 1 female]
0 females implies 4 males
No of ways =

^{7} C_{4} = \frac{7!}{4! 3!}

\frac{7 \times 6 \times 5}{3 \times 2} = 35

ways.
1 female = 1 female + 3 males
No of ways =

^{5} C_{1} \times^{7} C_{3}

\frac{5!}{4! 1!} \times \frac{7!}{4! 3!}

5 \times 35 = 175

ways.

∴

No of ways if at least 2 are females = 495 - (35 + 175) = 285 ways.
(b)

X	Y	$R_{X}$	$R_{Y}$	$d = R_{X} - R_{Y}$	$d^{2}$
2	4	6	3	3	9
7	6	1	1	0	0
1	2	7	5	2	4
3	3	5	4	1	1
6	1	2	6.5	-4.5	20.25
5	1	3	6.5	-3.5	12.25
4	5	4	2	2	4
$\sum$					49.5

R_{S} = 1 - \frac{6 \sum d^{2}}{n (n^{2} - 1)}

1 - \frac{6 (49.5)}{7 (7^{2} - 1)}

1 - \frac{297}{336}

\frac{13}{112} = 0.116

(ii) This implies that X and Y are fairly positively correlated. As X increases, Y increases as well.

Prev Current Next

STAY UPDATED
FOLLOW US ON :

Facebook

Register • Login