Question

Consider a binomial experiment with n=11 and p=0.3. a. Compute (to 4 decimals). b. Compute (to 4 decimals). c. Compute (to 4 decimals). d. Compute (to 4 decimals). e. Compute (to 1 decimal). f. Compute and . (to 2 decimals) (to 2 decimals)

Answer 1

Answer:

a. Compute f(0) (to 4 decimals).

$P(x=0) =0.0198$

b. Compute f(7) (to 4 decimals).

$P(x=7) =0.0173$

c. Compute P(x≤1) (to 4 decimals).

$P(X\leq1)=0.1130$

d. Compute P(x≥3) (to 4 decimals).

$P(x\geq3)=0.4304$

e. Compute E(x) (to 1 decimal).

$E(x)=3.3$

f. Compute V(x) and σ. (to 2 decimals)

$V(x)=2.31\\\\\sigma=1.52$

Step-by-step explanation:

The question is incomplete:

a. Compute f(0) (to 4 decimals).

$P(x=0) = \binom{11}{0} p^0q^{11}=1*1*0.0198=0.0198$

b. Compute f(7) (to 4 decimals).

$P(x=7) = \binom{11}{7} p^{7}q^{4}=330*0.0002*0.2401=0.0173$

c. Compute P(x≤1) (to 4 decimals).

$P(X\leq1)=P(x=0)+P(x=1)=0.0198+0.0932=0.1130\\\\\\ P(x=0) = \binom{11}{0} p^{0}q^{11}=1*1*0.0198=0.0198\\\\P(x=1) = \binom{11}{1} p^{1}q^{10}=11*0.3*0.0282=0.0932$

d. Compute P(x≥3) (to 4 decimals).

$P(X\geq3)=1-(P(0)+P(1)+P(2)+P(3))\\\\P(x\geq3)=1-(0.0198+0.0932+0.1998+0.2568)=1-0.5696\\\\P(x\geq3)=0.4304\\\\\\P(x=0) = \binom{11}{0} p^{0}q^{11}=1*1*0.0198=0.0198\\\\ P(x=1) = \binom{11}{1} p^{1}q^{10}=11*0.3*0.0282=0.0932\\\\P(x=2) = \binom{11}{2} p^{2}q^{9}=55*0.09*0.0404=0.1998\\\\P(x=3) = \binom{11}{3} p^{3}q^{8}=165*0.027*0.0576=0.2568$

e. Compute E(x) (to 1 decimal).

$E(x)=np=11*0.3=3.3$

f. Compute V(x) and σ. (to 2 decimals)

$V(x)=npq=11*0.3*0.7=2.31\\\\\sigma=\sqrt{V(x)}=\sqrt{2.31}=1.52$