Given values: 54, 65, 76, 43, 45, 76, 87 12, 23, 34, 67,65, 78, 45, 34. Given 95% confidence and sigma (population standard deviation) (12) Compute the confidence interval. Round to one decimal place. (A) (47.5, 59.7)(B) (12, 87) (C) (48.5, 58.7) (D) None of the above.

Answer:Option (D) None of the above.Step-by-step explanation:We are given the following data:54, 65, 76, 43, 45, 76, 87, 12, 23, 34, 67,65, 78, 45, 34Formula: [tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  [tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex] [tex]Mean =\displaystyle\frac{804}{15} = 53.6[/tex] Sum of squares of differences = 0.16 + 129.96 + 501.76 + 112.36 + 73.96 + 501.76 + 1115.56 + 1730.56 + 936.36 + 384.16 + 179.56 + 129.96 + 595.36 + 73.96 + 384.16 = 6849.6[tex]S.D = \sqrt{\frac{6849.6}{15}} = 21.4[/tex] Confidence interval: [tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex] Putting the values, we get, [tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex] [tex]53.6 \pm 1.96(\frac{21.4}{\sqrt{15}} ) = 53.6 \pm 10.83 = (42.8,64.4)[/tex]