Answer:
"0.0808" is the appropriate response.
Step-by-step explanation:
Given:
n = 524
[tex]\hat{P}[/tex] = 41%
or,
= 0.41
[tex]1-\hat{P}=1-0.41[/tex]
[tex]=0.59[/tex]
[tex]\mu \hat{P}=\hat{P}[/tex]
[tex]=0.41[/tex]
Now,
⇒ [tex]6 \hat{P}=\sqrt{\frac{\hat {P}(1-\hat{P})}{n} }[/tex]
[tex]=\sqrt{\frac{0.41\times 0.59}{524} }[/tex]
[tex]=0.0215[/tex]
[tex]P(\hat {P}>44 \ percent)[/tex]
or,
[tex]P(\hat{P}>0.44)[/tex]
[tex]=1-P(\hat{P}<0.44)[/tex]
[tex]=1-P(\frac{\hat{P}-\mu \hat{P}}{6 \hat{P}} <\frac{0.44-0.41}{0.0215} )[/tex]
[tex]=1-P(z<1.40)[/tex]
By using the standard normal table, we get
[tex]=1-0.9192[/tex]
[tex]=0.0808[/tex]
