When 319 college students are randomly selected and surveyed , it is found that 120 own a car . find a 99% confidence interval for the true proportion of all college students who own a car.

a.0. b.0. c.0. d.0.332< 0.421

Step-by-step explanation:

The confidence interval for a population proportion is calculated using the formula;

In the above expression;

p represents the sample proportion calculated as, number of successes/sample size;

p = 120/319 = 0.3762

z represents the confidence coefficient associated with the given level of significance. In our case the confidence level required is 99%. The z-score associated with this level of significance is +/-2.576

n represents the sample size under evaluation,  319 college students.

Using the values obtained above, the 99% confidence interval for the true proportion of all college students who own a car is;

60.9774

step-by-step explanation:

because when you add 47.6 plus 7.32 equals 54.92

and when you add .0574 + 6= 6.0574

and when you add 54.92 and 6.0574 you get the answer of 60.9774.

plzz mark brainlest last one got deleted. cause i didn't explain it but i did but i understand what the mod meant