Suppose that insurance companies did a survey. They randomly surveyed 420 drivers and found that 340 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 tdistribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) 3 Part (a) 3] Part (b) El Part (c) Which distribution should you use for this problem? (Round your answer to four decimal places.) \"CH-X ,x ) Explain your choice. 0 The binomial distribution should be used because there are two outcomes, buckle up or do not buckle up. 0 The normal distribution should be used because we are interested in proportions and the sample size is large. 0 The Student's t-distribution should be used because \\/ npq s 10, which implies a small sample. 0 The Student's t-distribution should be used because we do not know the standard deviation. Construct a 95% confidence interval for the population proportion who claim they always buckle up. (i) State the confidence interval. (Round your answers to four decimal places.) (ii) Sketch the graph. C.L. = = P' (iii) Calculate the error bound. (Round your answer to four decimal places.)