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6. [-/1 Points] DETAILS ILLOWSKYINTROSTAT1 8.HW. 118. MY NOTES Suppose that insurance companies did a survey. They randomly surveyed 400 drivers and found that 300 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) Part (a) (i) Enter an exact number as an integer, fraction, or decimal. X = (ii) Enter an exact number as an integer, fraction, or decimal. n= (iii) Round your answer to four decimal places. p'= (rounded to four decimal places) Part (b) In words, define the random variables X and P. O X is the number of people who do not buckle up, and P' is the proportion of people in the sample who do not buckle up. O X is the number of people who claim they buckle up, and P' is the proportion of people in the sample who buckle up. O X is the proportion of people in the sample who claim they buckle up, and P' is the number of people who buckle up. O X is the proportion of people in the sample who do not buckle up, and P' is the number of people who do not buckle up. Part (c) Which distribution should you use for this problem? (Round your answer to four decimal places.) P'~