Answer:
a) The percentage of households in the town with three or more largescreen TVs is estimated as :
The best estimation for the population proportion is :

And that represent the 1.4%.
b) And the 95% confidence interval would be given (0.00370;0.0243).
And the % would be between 0.37% and 2.43%.
Step-by-step explanation:
Data given and notation  
n=500 represent the random sample taken    
X=7 represent the households with three or more large-screen TVs
 estimated proportion of households with three or more large-screen TVs
 estimated proportion of households with three or more large-screen TVs
 represent the significance level (no given, but is assumed)
 represent the significance level (no given, but is assumed)    
z would represent the statistic (variable of interest)    
p= population proportion of households with three or more large-screen TVs
Part a
The percentage of households in the town with three or more largescreen TVs is estimated as :
The best estimation for the population proportion is :

And that represent the 1.4%.
Part b
Yes is possible. We hav that  and
 and  so we have the assumption of normality to find the interval.
 so we have the assumption of normality to find the interval.
The confidence interval would be given by this formula
 
For the 95% confidence interval the value of  and
 and  , with that value we can find the quantile required for the interval in the normal standard distribution.
, with that value we can find the quantile required for the interval in the normal standard distribution.
 
And replacing into the confidence interval formula we got:
 
 
And the 95% confidence interval would be given (0.00370;0.0243).
And the % would be between 0.37% and 2.43%.