Question

.According to a survey by the Administrative Management Society, one-half of U.S. companies give employees 4 weeks of vacation after they have been with the company for 15 years. Find the probability that among G companies surveyed at random, the number that give employees 4 weeks of vacation after 15 years of employment is (a) anywhere from 2 to 5; (b) fewer than 3.

Answer 1

Let $X=$ the number of the companies that give employees 4 weeks of vacation after 15 years of employment: $X\sim B(n;p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x},x=0, 1, 2,...,n$$

Given that $n=G, p=0.5$

(a) Find the probability that among G companies surveyed at random, the number that give employees 4 weeks of vacation after 15 years of employment is anywhere from 2 to 5

$$P(2\leq X\leq5)=P(X=2)+P(X=3)+$$ $$+P(X=4)+P(X=5)=$$ $$=\binom{G}{2}0.5^2(1-0.5)^{G-2}+\binom{G}{3}0.5^3(1-0.5)^{G-3}+$$ $$+\binom{G}{4}0.5^4(1-0.5)^{G-4}+\binom{G}{5}0.5^4(1-0.5)^{G-4}=$$ $$=0.5^G\bigg({G! \over 2!(G-2)!}+{G! \over 3!(G-3)!}+{G! \over 4!(G-4)!}+{G! \over 5!(G-5)!}\bigg),$$ $$G\geq5$$

If $G=6$

$$P(2\leq X\leq5)=$$ $$=0.5^6\bigg({6! \over 2!(6-2)!}+{6! \over 3!(6-3)!}+{6! \over 4!(6-4)!}+{6! \over 5!(6-5)!}\bigg)=$$ $$=0.875$$

(b) Find the probability that among G companies surveyed at random, the number that give employees 4 weeks of vacation after 15 years of employment is fewer than 3.

$$P(X<3)=P(X=0)+P(X=1)+P(X=2)=$$ $$=\binom{G}{0}0.5^0(1-0.5)^{G-0}+\binom{G}{1}0.5^1(1-0.5)^{G-1}+$$ $$+\binom{G}{2}0.5^2(1-0.5)^{G-2}=$$ $$=0.5^G\bigg({G! \over 0!(G-0)!}+{G! \over 1!(G-1)!}+{G! \over 2!(G-2)!}\bigg),x\geq2$$

If $G=6$

$$P(X<3)=0.5^6\bigg({6! \over 0!(6-0)!}+{6! \over 1!(6-1)!}+{6! \over 2!(6-2)!}\bigg)=$$ $$=0.34375$$