1. Can you please show mw in detail that how do you find the Standard deviation 'SD' for this question to solve .

Suppose the following data are selected randomly from a population of normally distributed values. 44 51 43 48 41 57 54 39 4O 48 45 39 41 Construct a 95% condence interval to estimate the population mean. Appendix; ( Round the intermediate values to 2 decimal places, e.g. 25.31. Round your answers to 2 decimal places, e3. 25.31.) 41.85 5 ,U 5 48.92 According to Runzheimer International, the average cost of a domestic trip for business travelers in the financial industry is $1,250. Suppose another travel industry research company takes a random sample of 51 business travelers in the financial industry and determines that the sample average cost of a domestic trip is $1,192, with a sample standard deviation of $279. Construct a 98% confidence interval for the population mean from these sample data. Assume that the data are normally distributed in the population. Now go back and examine the $1,250 figure published by Runzheimer International. Does it fall into the confidence interval computed from the sample data? What does this tell you? Appendix A Statistical Tables (Round your answers to 2 decimal places.) The 98% confidence interval: 1098.12 SUS 1285.88Solution n = 51 x = 1,192 S = 279 df = n - 1 = 50 98% CI and a /2 = .01 1 01,50 = 2.403 S 279 x +t In = 1, 192 + 2.403 = 1, 192 + 93.88 = 1, 098.12