Respuesta :
Answer:
E) .0863
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
For this case we can find the sample proportion for each observation with the following formula:
[tex]\hat p = \frac{X}{n}[/tex]
Where X represent the number of returns and n =100 for each case since the standard value used. Using the last formula we got:
[tex]p_1 = \frac{10}{100}=0.1[/tex]
[tex]p_2 = \frac{9}{100}=0.09[/tex]
[tex]p_3 = \frac{11}{100}=0.11[/tex]
[tex]p_4 = \frac{7}{100}=0.07[/tex]
[tex]p_5 = \frac{3}{100}=0.03[/tex]
[tex]p_6 = \frac{12}{100}=0.12[/tex]
[tex]p_7 = \frac{8}{100}=0.08[/tex]
[tex]p_8 = \frac{4}{100}=0.04[/tex]
[tex]p_9 = \frac{6}{100}=0.06[/tex]
[tex]p_{10} = \frac{11}{100}=0.11[/tex]
And now witht those values we can find the sample mean of proportions with the following formula:
[tex]hat p = \frac{\sum_{i=1}^n p_i}{n}= \frac{0.1+0.09+0.11+0.07+0.03+0.12+0.08+0.04+0.06+0.11}{10}=0.081[/tex]
And we can find the standard error with the following formula:
[tex]SE= \sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.081(1-0.081)}{10}}=0.0863[/tex]
So then the best option on this case is given by:
E) .0863
