Question

Twenty employees have applied for promotions. This year 7 promotions were awarded. If a random sample of 5 applications was selected.

a. What is the probability that at most 2 employees were promoted?
b. What is the probability that no employees were promoted?
c. Find the value of the expected value and the variance of this distribution.

Answer 1

a)0.764

b)0.116

c)Expected value =1.75

Variance=1.1375

Explanation:

Total Number of employees = 20

Number of promotions awarded = 7

Probability of success p =
`(7)/(20)`

Probability of failure q =
`1-(7)/(20)=(13)/(20)`

We randomly select 5 applications

a) What is the probability that at most 2 employees were promoted?

We will use binomial

`P(X=x)=^nC_r p^r q ^(n-r)\\P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)\\P(X\leq 2)=^(5)C_0 ((7)/(20))^0 ((13)/(20))^(5-0)+^(5)C_1 ((7)/(20))^1 ((13)/(20))^(5-1)+^(5)C_2 ((7)/(20))^2 ((13)/(20))^(5-2)\\P(X\leq 2)=(5!)/(0!(5-0)!) ((7)/(20))^0 ((13)/(20))^(5-0)+(5!)/(1!(5-1)!)((7)/(20))^1 ((13)/(20))^(5-1)+(5!)/(2!(5-2)!) ((7)/(20))^2 ((13)/(20))^(5-2)\\P(X\leq 2)=0.764`

b)What is the probability that no employees were promoted?

`P(X=0)=^(5)C_0 ((7)/(20))^0 ((13)/(20))^(5-0)\\P(X=0)=(5!)/(0!(5-0)!) ((7)/(20))^0 ((13)/(20))^(5-0)\\P(X=0)=0.116`

c)

`E(X)=np=5 * ((7)/(20))=1.75\\Var(X)=npq=5 * ((7)/(20)) * (13)/(20)=1.1375`