Answer:
Check the explanation
Step-by-step explanation:
p=12/20=0.6
H_0:p=0.4,H_a:p>0.4
[tex]t^*[/tex]=[tex]\frac{} \frac{0.6-0.4}/{\sqrt{\frac{0.6*0.4}{20}}[/tex]=1.83
pvalue=P(t>1.83,19dof)=0.042
Hence, we reject H0 and say the proportion of peanuts is more than 40%
b)E[weight]=0.2*1+0.3*1.5+0.1*2.5+0.4*0.5=1.1
V[weight]=0.2(1-1.1)^2+0.3(1.5-1.1)^2+0.1(2.5-1.1)^2+0.4(0.5-1.1)^2=0.39
SD[weight]=sqrt(0.39)=0.6245
[tex]P(\bar{X}<1)=P(Z<\frac{1-1.1}{0.6245/\sqrt{100}}[/tex]=-1.6013)=0.0547=5.47\%