beginning of the lay. 1) Given the data 11.4, 11.3, 11.2 and 1 1.1 cm, determine the average and its standard deviation using Equation [2]. 1 average = 11.25 11.25 = 0.13 cm 11.25. 11.4 =-0.15 = 0.0225 11.25-11.3 = -0.05= 0.0DLS 0.05 11.25 - 11.20.05 = 0.0015 3 = 0.01467 =0.129 = 0.13 11. 25-11.1 :15 = 0.0225 tau are from two different sources and