Respuesta :
Solution: We are given:
[tex]\hat{p}=0.55,p=0.51,n=900[/tex]
We have to find [tex]P(\hat{p}\geq 0.55)[/tex]
Now we have to find the z score corresponding to [tex]\hat{p}=0.55[/tex]
[tex]z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]
[tex]=\frac{0.55-0.51}{\sqrt{\frac{0.51(1-0.51)}{900} } }[/tex]
[tex]=\frac{0.04}{0.0167}[/tex]
[tex]=2.40[/tex]
[tex]\therefore P(\hat{p}\geq 0.55}=P(z\geq 2.4)[/tex]
Using the standard normal table, we have:
[tex]P(z\geq 2.4)=0.0082[/tex]
Therefore, the probability of observing a sample proportion of voters 0.55 or higher supporting the incumbent governor is 0.0082
Answer:
0.0081
Explanation:
Let p be the population proportion of voters in the state support the incumbent governor .
According to the given information , we have
Population proportion : [tex]p=0.51[/tex]
Sample size (Number of respondents): n= 900
Sample proportion of voters in the state support the incumbent governor : [tex]\hat{p}=0.55[/tex]
Further Explanation:
Formula to find z-score (for proportion) :
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z=\dfrac{0.55-0.51}{\sqrt{\dfrac{0.51(1-0.51)}{900}}}=2.40048014405\approx2.40[/tex] [Rounded to the nearest two decimal places]
To find the probability of observing a sample proportion of voters 0.55 or higher supporting the incumbent governor , we find the p-value below :
[tex]P(p>0.55)=P(z>2.40)\ \ \text{[Area on under the right side of the normal curve greater than z=2.40]}\\\\=1-P(z<2.40)\ \ [\because P(Z>z)=1-P(Z<z)]\\\\=1- 0.9918024\ \ [\text{By using online z-value calculator}]\\\\=0.0081976\approx0.0081[/tex] [Rounded to the nearest 4 decimal places.]
Therefore , the probability of observing a sample proportion of voters 0.55 or higher supporting the incumbent governor = 0.0081
Refer the given attachment for further understanding.
Learn more :
- https://brainly.com/question/13716713 [Answered by Jeanashupp]
Keywords :
sample proportion , z-value , z-table , Normal curve.

