Respuesta :
Answer:
(a) [tex]\mu_X=1.2[/tex], [tex]\sigma_X=0.8485[/tex], [tex]\sigma^{2} _X=0.72[/tex]
(b) [tex]\mu_X=4[/tex], [tex]\sigma_X=0.8944[/tex], [tex]\sigma^{2} _X=0.8[/tex]
(c) [tex]\mu_X=1.5[/tex], [tex]\sigma_X=1.0247[/tex], [tex]\sigma^{2} _X=1.05[/tex]
(d) [tex]\mu_X=3.2[/tex], [tex]\sigma_X=0.8[/tex], [tex]\sigma^{2} _X=0.64[/tex]
Step-by-step explanation:
A binomial random variable counts how often a particular event occurs in a fixed number of tries or trials.
For a binomial random variable [tex]X[/tex], we can use these formulas for mean, standard deviation, and variance:
[tex]\mu_X=np[/tex]
[tex]\sigma_X=\sqrt{np(1-p)}[/tex]
[tex]\sigma^{2} _X=np(1-p)[/tex]
Applying the above formulas we get,
(a) n = 3, p = 0.4
[tex]\mu_X=3\cdot 0.4=1.2[/tex]
[tex]\sigma_X=\sqrt{3\cdot 0.4(1-0.4)}=0.8485[/tex]
[tex]\sigma^{2} _X=3\cdot \:0.4\left(1-0.4\right)=0.72[/tex]
(b) n = 5, p = 0.8
[tex]\mu_X=5\cdot 0.8=4[/tex]
[tex]\sigma_X=\sqrt{5\cdot 0.8(1-0.8)}=0.8944[/tex]
[tex]\sigma^{2} _X=5\cdot \:0.8\left(1-0.8\right)=0.8[/tex]
(c) n = 5, p = 0.3
[tex]\mu_X=5\cdot 0.3=1.5[/tex]
[tex]\sigma_X=\sqrt{5\cdot 0.3(1-0.3)}=1.0247[/tex]
[tex]\sigma^{2} _X=5\cdot \:0.3\left(1-0.3\right)=1.05[/tex]
(d) n = 4, p = 0.8
[tex]\mu_X=4\cdot 0.8=3.2[/tex]
[tex]\sigma_X=\sqrt{4\cdot 0.8(1-0.8)}=0.8[/tex]
[tex]\sigma^{2} _X=4\cdot \:0.8\left(1-0.8\right)=0.64[/tex]
