Respuesta :
Answer:
For this case we need to check the conditions in order to use the normal approximation:
1) [tex]np = 60*0.3= 18>10[/tex]
2) [tex]n(1-p) = 60*(1-0.3)= 42>10[/tex]
Since both conditions are satisfied and the independence condition is assumed we can use the normal approximation given by:
[tex]\hat p \sim N (\hat p, \sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
The mean would be given by:
[tex]\mu_{\hat p}=0.3[/tex]
And the deviation is given by:
[tex]\sigma_{\hat p}= \sqrt{\frac{0.3*(1-0.3)}{60}}= 0.0592[/tex]
Step-by-step explanation:
For this case we know the following info:
n =60 represent the sample size
[tex]\hat p = 0.3[/tex] represent the estimated proportion of people that will buy a packet of crackers after tasting
For this case we need to check the conditions in order to use the normal approximation:
1) [tex]np = 60*0.3= 18>10[/tex]
2) [tex]n(1-p) = 60*(1-0.3)= 42>10[/tex]
Since both conditions are satisfied and the independence condition is assumed we can use the normal approximation given by:
[tex]\hat p \sim N (\hat p, \sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
The mean would be given by:
[tex]\mu_{\hat p}=0.3[/tex]
And the deviation is given by:
[tex]\sigma_{\hat p}= \sqrt{\frac{0.3*(1-0.3)}{60}}= 0.0592[/tex]
