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

To find the probability that among 6 randomly surveyed companies, the number that give employees 4 weeks of vacation after 15 years of employment is x, we can use the binomial probability formula. This formula is given by P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials (in this case, 6), k is the number of successes (in this case, x), p is the probability of success (in this case, 0.5), and C(n,k) is the number of combinations of n and k. Since we want to find the probability for all possible values of x (0 to 6), we can calculate the probability for each value and add them up.

Step-by-step explanation:

To find the probability that among 6 randomly surveyed companies, the number that give employees 4 weeks of vacation after 15 years of employment is x, we can use the binomial probability formula. This formula is given by P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials (in this case, 6), k is the number of successes (in this case, x), p is the probability of success (in this case, 0.5), and C(n,k) is the number of combinations of n and k. Since we want to find the probability for all possible values of x (0 to 6), we can calculate the probability for each value and add them up. Let's calculate the probabilities for each value of x:

1. P(X=0) = C(6,0) * 0.5^0 * (1-0.5)^(6-0)
2. P(X=1) = C(6,1) * 0.5^1 * (1-0.5)^(6-1)
3. P(X=2) = C(6,2) * 0.5^2 * (1-0.5)^(6-2)
4. P(X=3) = C(6,3) * 0.5^3 * (1-0.5)^(6-3)
5. P(X=4) = C(6,4) * 0.5^4 * (1-0.5)^(6-4)
6. P(X=5) = C(6,5) * 0.5^5 * (1-0.5)^(6-5)
7. P(X=6) = C(6,6) * 0.5^6 * (1-0.5)^(6-6)

To calculate the probabilities, we can use the formula for combinations, which is given by C(n,k) = n! / (k!(n-k)!), where n! represents the factorial of n. Once we calculate the probabilities for each value of x, we can simply add them up to find the total probability.

Answer 2

a

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

b

`P(X<3)=0.3432`

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<3)=P(X=0)+P(X=1)+P(X=2)\\\\={6 \choose 0}(0.5)^6+{6 \choose 1}(0.5)^6{6 \choose 2}(0.5)^6\\\\=0.0156(1+6+15)\\\\=0.3432`

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