Suppose that textbook weights are normally distributed. You measure 28 textbooks' weights, and find they have a mean weight of 76 ounces. Assume the population standard deviation is 12.3 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight. Round answers to 2 decimal places.

Answer:Step-by-step explanation:We want to find 95% confidence interval for the mean of the weight of of textbooks. Number of samples. n = 28 textbooks weightMean, u =76 ouncesStandard deviation, s = 12.3 ouncesFor a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.We will apply the formulaConfidence interval= mean +/- z ×standard deviation/√nIt becomes 76 +/- 1.96 × 12.3/√28= 76 +/- 1.96 × 2.3113= 76 +/- 4.53The lower end of the confidence interval is 76 - 4.53 =71.47The upper end of the confidence interval is 76 + 4.53 = 80.53 Therefore, with 95% confidence interval, the mean textbook weight is between 71.47 ounces and 80.53 ounces