Question

Goofy's fast food center wishes to estimate the proportion of people in its city that will purchase its products. Suppose the true proportion is 0.06. If 280 are sampled, what is the probability that the sample proportion will be less than 0.09? Round your answer to four decimal places.

Answer 1

`\mu_(\hat p)= 0.06`

`\sigma_(\hat p)= \sqrt{(0.06*(1-0.06))/(280)}= 0.0142`

And we want to find this probability:

`p(\hat p <0.09)`

`z = (p \mu_(\hat p))/(\sigma_(\hat p))`

`z = (0.09-0.06)/(0.0142)= 2.114`

And we can find this probability:

`P(z<2.114)`

And using the normal table or excel we got:

`P(z<2.114)= 0.9827`

Explanation:

For this case we have the following info given:

`n = 280` represent the sample size

`p =0.06` represent the true proportion

The sample proportion can be approximated with this distribution:

`\hat p \sim N (p ,\sqrt{(p(1-p))/(n)})`

The mean is given by:

`\mu_(\hat p)= 0.06`

And the deviation is given by:

`\sigma_(\hat p)= \sqrt{(0.06*(1-0.06))/(280)}= 0.0142`

And we want to find this probability:

`p(\hat p <0.09)`

And we can use the z score formula given by:

`z = (p \mu_(\hat p))/(\sigma_(\hat p))`

And replacing we got:

`z = (0.09-0.06)/(0.0142)= 2.114`

And we can find this probability:

`P(z<2.114)`

And using the normal table or excel we got:

`P(z<2.114)= 0.9827`