Answer :
Answer:
We need to check the conditions in order to use the normal approximation.
1) [tex]np=100*0.97=97 > 10[/tex]
2) [tex]n(1-p)=100*(1-0.97)=3 < 10[/tex]
For this case the first condition is satisfied but the second NO, so then is not appropiate for this case use the normal approximation given by:
[tex]p \sim N(p , \sqrt{\frac{p(1-p)}{n}})[/tex]
Step-by-step explanation:
For this case we have the following info given:
n = 100 represent the sample size
x= 97 represent the people selected who indicates that they were in favor of the proposal
The estimated proportion for this case is given by:
[tex]\hat p =\frac{x}{n}= \frac{97}{100}=0.97[/tex]
We need to check the conditions in order to use the normal approximation.
1) [tex]np=100*0.97=97 > 10[/tex]
2) [tex]n(1-p)=100*(1-0.97)=3 < 10[/tex]
For this case the first condition is satisfied but the second NO, so then is not appropiate for this case use the normal approximation given by:
[tex]p \sim N(p , \sqrt{\frac{p(1-p)}{n}})[/tex]
Answer:
Since n(1-p) < 5, it is not appropriate to assume that the sampling distribution of the sample proportion is approximately normal.
Step-by-step explanation:
Binomial probability distribution:
Probability of x sucesses on n repeated trials, with p probability.
Can be approximated to the normal distribution if:
np > 5 and n(1-p) > 5
In this problem:
A random sample of 100 residents was selected, and 97 of those selected indicated that they were in favor of the proposal.
This means that [tex]n = 100, p = \frac{97}{100} = 0.97[/tex]
Is it appropriate to assume that the sampling distribution of the sample proportion is approximately normal?
np = 100*0.97 = 97 > 5
n(1-p) = 100*0.03 = 3 < 5
Since n(1-p) < 5, it is not appropriate to assume that the sampling distribution of the sample proportion is approximately normal.