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 probabil- ity that among 6 companies surveyed at random, the number that give employees 4 weeks of vacation after 15 years of employment is

Answer 1

Answer:

a

$P(2\leq X\leq 5)=0.875$

b

$P(X$

Step-by-step explanation:

Let the random X variable representing the 6 companies that give 4 weeks of vacation after 15 years of employment:

-let p=0.5 be the probability of vacation. Since the companies are independent, X assumes a binomial random variable:$P(X=x)=b(x;6,0.5)\\\\={6 \choose x}(0.5)^x(0.5)^{6-x}\\\\={6 \choose x}(0.5)^6, \ x=0,1,2,3...\ \ \ eqtn1$

#Probability that the number of companies that give vacation is anywhere from 2 to 5:

We use equation 1;

$P(2\leq X\leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\={6 \choose 2}(0.5)^6+{6 \choose 3}(0.5)^6{6 \choose 4}(0.5)^6+{6 \choose 5}(0.5)^6\\\\=0.0156(15+20+15+6)\\\\=0.875$

Hence the probability that between 2 and 5 companies give vacation is 0.875

b. The probability that fewer than 3 companies give vacation is calculated as:

From equation one we get:

$P(X$

Hence the probability that less than three companies give vacation is 0.3432