Respuesta :
Answer:
a)0.764
b)0.116
c)Expected value =1.75
Variance=1.1375
Step-by-step explanation:
Total Number of employees = 20
Number of promotions awarded = 7
Probability of success p =[tex]\frac{7}{20}[/tex]
Probability of failure q = [tex]1-\frac{7}{20}=\frac{13}{20}[/tex]
We randomly select 5 applications
a) What is the probability that at most 2 employees were promoted?
We will use binomial
[tex]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 (\frac{7}{20})^0 (\frac{13}{20})^{5-0}+^{5}C_1 (\frac{7}{20})^1 (\frac{13}{20})^{5-1}+^{5}C_2 (\frac{7}{20})^2 (\frac{13}{20})^{5-2}\\P(X\leq 2)=\frac{5!}{0!(5-0)!} (\frac{7}{20})^0 (\frac{13}{20})^{5-0}+\frac{5!}{1!(5-1)!}(\frac{7}{20})^1 (\frac{13}{20})^{5-1}+\frac{5!}{2!(5-2)!} (\frac{7}{20})^2 (\frac{13}{20})^{5-2}\\P(X\leq 2)=0.764[/tex]
b)What is the probability that no employees were promoted?
[tex]P(X=0)=^{5}C_0 (\frac{7}{20})^0 (\frac{13}{20})^{5-0}\\P(X=0)=\frac{5!}{0!(5-0)!} (\frac{7}{20})^0 (\frac{13}{20})^{5-0}\\P(X=0)=0.116[/tex]
c)
[tex]E(X)=np=5 \times (\frac{7}{20})=1.75\\Var(X)=npq=5 \times (\frac{7}{20}) \times \frac{13}{20}=1.1375[/tex]
