A doctor is measuring the mean systolic blood pressure of female students at a large college. Systolic blood pressure is known to have a skewed distribution. The doctor collects systolic blood pressure measurements from random sample of 28 female students. The resulting 90% confidence interval is (100.4, 159.6). Units of systolic blood pressure are mmHg.

Which one of the following conclusions is valid? Group of answer choices

a. 90% of the female students at the college have a systolic blood pressure between 100.4 mmHg and 159.6 mmHg.
b. The sampling distribution of means will probably not follow a normal distribution, so we cannot draw a conclusion.
c. We are 90% confident that the mean systolic blood pressure for female students at the college is between 100.4 mmHg and 159.6 mmHg.

c

step-by-step explanation:

i think because what else can be a sample of every member

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497