Answer: 0.4402
Step-by-step explanation:
Given : The proportion of the registered voters in a country are Republican = P=0.50
Sample space = 36
The test statistic for proportion :-
[tex]z=\dfrac{p-P}{\sqrt{\dfrac{P(1-P)}{n}}}[/tex]
For p= 0.477
[tex]z=\dfrac{0.477-0.50}{\sqrt{\dfrac{0.50(1-0.50)}{36}}}\approx-0.276[/tex]
For p= 0.58
[tex]z=\dfrac{0.58-0.50}{\sqrt{\dfrac{0.50(1-0.50)}{36}}}\approx0.96[/tex]
Now, the probability that the proportion of freshmen in the sample is between 0.477 and 0.58 (by using the standard normal distribution table):-
[tex]P(0.477<x<0.58)=P(-0.276<z<0.96)\\\\=P(z<0.96)-P(z<-0.276)\\\\=0.8314724- 0.391274=0.4401984\approx0.4402[/tex]
Hence, the probability that the proportion of freshmen in the sample is between 0.477 and 0.580 = 0.4402