Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50 Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight year period. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight–year period.
Define the random variable in X and P in words.
Which distribution should you use in this problem?

Step-by-step explanation:

a) Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 451

x = 1.5/100 × 451 = 7

p = 7/451 = 0.02

q = 1 - 0.02 = 0.98

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.97 = 0.1

α/2 = 0.01/2 = 0.03

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.03 = 0.97

The z score corresponding to the area on the z table is 2.17. Thus, Thus, the z score for a confidence level of 97% is 2.17

Therefore, the 97% confidence interval is

0.02 ± 2.17√(0.02)(0.98)/451

= 0.02 ± 0.014

b) x represents the number of members of the 50 Plus Fitness Association who ran and died in the same eight–year period.

P represents the proportion of members of the 50 Plus Fitness Association who ran and died in the same eight–year period.

The distribution that should be used is the normal distribution

the answer is d. (1.2, -4.7) on edgn

hope this !