Question

a) It has been estimated that 35% of all new employees of the WindThroft company have a first degree. In a sample of 10 new employees.

(i) What is the probability that exactly 4 of them have first degrees? [2 marks]

(ii) What is the probability that at least three (3) of them have first degrees? [8 marks]

(iii) Calculate the Mean, Variance and Standard deviation of the distribution [5 marks]

Answer 1

Solution:

(i)

p=35%=0.35

q=1-p=1-0.35=0.65

$P(X=x)=\ \ ^nC_xp^xq^{n-x}$

$P(X=4)=\ \ ^{10}C_4(0.35)^4(0.65)^6=0.2376$

(ii)

$P(X\ge3)=1-P(X<3) \\=1-[P(X=0)+P(X=1)+P(X=2)] \\=1-[\ \ ^{10}C_0(0.35)^0(0.65)^{10}+\ \ ^{10}C_1(0.35)^1(0.65)^9+\ \ ^{10}C_2(0.35)^2(0.65)^8] \\=1-[0.0134+0.0724+0.1756] \\=0.7386$

(iii)

Mean = np $=10\times0.35=3.5$

Variance = npq $=10\times0.35\times0.65=2.275$

SD $=\sqrt{variance}=\sqrt{2.275}=1.508$