In a Gallup survey of 1,012 randomly selected U. S. adults (age 18 and over), 53% said that they were dissatisfied with the quality of education students receive in kindergarten through grade 12. Use the normal distribution to construct and interpret a 90% confidence interval for the proportion of U. S. adults who are dissatisfied with the education students receive in kindergarten through grade 12. The standard error is SE

The standard error is 0.0157.

The 90% confidence interval for the proportion of U.S. adults who are dissatisfied with the education students receive in kindergarten through grade 12 is (0.5042, 0.5558). This means that we are 90% sure that the true proportion of U.S. adults who are dissatisfied with the education students receive in kindergarten through grade 12 is between 0.5042 and 0.5558.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

is the standard error.

In a Gallup survey of 1,012 randomly selected U.S. adults (age 18 and over), 53% said that they were dissatisfied with the quality of education students receive in kindergarten through grade 12.

This means that

The standard error is:

The standard error is 0.0157.

90% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval for the proportion of U.S. adults who are dissatisfied with the education students receive in kindergarten through grade 12 is (0.5042, 0.5558). This means that we are 90% sure that the true proportion of U.S. adults who are dissatisfied with the education students receive in kindergarten through grade 12 is between 0.5042 and 0.5558.

use formula for the circle a=3.14

step-by-step explanation:

its c or d   take a lucky guess my dude