Answer: 0.0384
Step-by-step explanation:
Given : The proportion of the households in Kansas City prefer the new package : p= 0.40
sample size : n= 300
Now, the probability that Catherine's random sample of 300 households will have a sample proportion greater than 0.45 will be :-
[tex]P(p>0.45)=P(z>\dfrac{0.45-0.40}{\sqrt{\dfrac{0.40(1-0.40)}{300}}})\\\\=P(z>1.77)[/tex]
[∵ [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]]
[tex]=1-P(z\leq1.77)[/tex] [∵ P(Z>z)=1-P(Z≤z)]
[tex]=1-0.9616=0.0384[/tex] [using p-value table for z]
Hence, the required probability = 0.0384