Answer:
[tex]P(x=3)=0.2269[/tex]
Mean=2.1
Standard deviation=1.21
Step-by-step explanation:
We are given that
n=7
Probability of success, p=0.3
q=1-p=1-0.3=0.7
We have to find the probability of 3 success for the binomial experiment and find the mean and standard deviation.
Binomial distribution formula
[tex]P(X=x)=nC_xp^{x}q^{n-x}[/tex]
Using the formula
[tex]P(x=3)=7C_3(0.3)^3(0.7)^{7-3}[/tex]
[tex]P(x=3)=7C_3(0.3)^3(0.7)^{4}[/tex]
[tex]P(x=3)=\frac{7!}{3!4!}(0.3)^3(0.7)^{4}[/tex]
[tex]P(x=3)=\frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}(0.3)^{3}(0.7)^{4}[/tex]
Using the formula
[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]
[tex]P(x=3)=0.2269[/tex]
Now,
Mean, [tex]\mu=np=7\times 0.3=2.1[/tex]
Standard deviation, [tex]\sigma=\sqrt{np(1-p)}[/tex]
Standard deviation, [tex]\sigma=\sqrt{7\times 0.3\times 0.7}[/tex]
Standard deviation, [tex]\sigma=1.21[/tex]
